NCERT Solutions · Class 9 Mathematics Exploring Algebraic Identities

Recognising a perfect square trinomial. An expression is \(\displaystyle (a+b)^{2}\) in disguise when the first and last terms are perfect squares, say \(\displaystyle a^{2}\) and \(\displaystyle b^{2}\), and the middle term is exactly \(\displaystyle 2ab\). So for each part: take the square roots of the outer two terms, then check that twice their product gives the middle term. Checking matters — without it you can factor something that is not actually a square.(i) \(\displaystyle 9x^{2}+24xy+16y^{2}\). Here \(\displaystyle 9x^{2}=(3x)^{2}\) and \(\displaystyle 16y^{2}=(4y)^{2}\). Check the middle: \(\displaystyle 2(3x)(4y)=24xy\). It matches. \[9x^{2}+24xy+16y^{2}=(3x+4y)^{2} \](ii) \(\displaystyle 4s^{2}+20st+25t^{2}\). Roots \(\displaystyle 2s\) and \(\displaystyle 5t\); check \(\displaystyle 2(2s)(5t)=20st\). It matches. \[4s^{2}+20st+25t^{2}=(2s+5t)^{2} \](iii) \(\displaystyle 49x^{2}+28xy+4y^{2}\). Roots \(\displaystyle 7x\) and \(\displaystyle 2y\); check \(\displaystyle 2(7x)(2y)=28xy\). It matches. \[49x^{2}+28xy+4y^{2}=(7x+2y)^{2} \](iv) \(\displaystyle 64p^{2}+\frac{32}{3}pq+\frac{4}{9}q^{2}\). Roots \(\displaystyle 8p\) and \(\displaystyle \frac{2}{3}q\); check \(\displaystyle 2(8p)\left(\frac{2}{3}q\right)=\frac{32}{3}pq\). It matches. \[64p^{2}+\tfrac{32}{3}pq+\tfrac{4}{9}q^{2}=\left(8p+\tfrac{2}{3}q\right)^{2}=\tfrac{1}{9}(24p+2q)^{2} \](v) \(\displaystyle 3a^{2}+4ab+\frac{4}{3}b^{2}\) (starred). Now \(\displaystyle 3a^{2}\) is not the square of a rational expression, so the identity cannot be applied as it stands. This is where the hint comes in: in Example $\displaystyle 7$ a common factor of \(\displaystyle 2\) was pulled out first. Do the same here and pull out \(\displaystyle 3\): \[3a^{2}+4ab+\tfrac{4}{3}b^{2}=3\left(a^{2}+\tfrac{4}{3}ab+\tfrac{4}{9}b^{2}\right). \] Inside the bracket the outer terms are \(\displaystyle a^{2}\) and \(\displaystyle \left(\frac{2}{3}b\right)^{2}\), and \(\displaystyle 2(a)\left(\frac{2}{3}b\right)=\frac{4}{3}ab\) — a genuine perfect square. \[3a^{2}+4ab+\tfrac{4}{3}b^{2}=3\left(a+\tfrac{2}{3}b\right)^{2}=\tfrac{1}{3}(3a+2b)^{2} \] (Check by expanding: \(\displaystyle \frac{1}{3}(9a^{2}+12ab+4b^{2})=3a^{2}+4ab+\frac{4}{3}b^{2}\).)(vi) \(\displaystyle \frac{9}{5}s^{2}+6sv+5v^{2}\) (starred). Again neither outer term is a perfect square of something rational. Pulling out \(\displaystyle \frac{1}{5}\) clears both denominators at once: \[\tfrac{9}{5}s^{2}+6sv+5v^{2}=\tfrac{1}{5}\left(9s^{2}+30sv+25v^{2}\right). \] Inside, \(\displaystyle 9s^{2}=(3s)^{2}\), \(\displaystyle 25v^{2}=(5v)^{2}\) and \(\displaystyle 2(3s)(5v)=30sv\). It matches. \[\tfrac{9}{5}s^{2}+6sv+5v^{2}=\tfrac{1}{5}(3s+5v)^{2} \] (Check: \(\displaystyle \frac{1}{5}(9s^{2}+30sv+25v^{2})=\frac{9}{5}s^{2}+6sv+5v^{2}\).)So the answer to the hint's question is yes: whenever the coefficients are awkward, taking out a suitable common factor — a whole number in (v), a fraction in (vi) — can turn the leftover expression into a perfect square.Answers: (i) \(\displaystyle (3x+4y)^{2}\); (ii) \(\displaystyle (2s+5t)^{2}\); (iii) \(\displaystyle (7x+2y)^{2}\); (iv) \(\displaystyle \left(8p+\frac{2}{3}q\right)^{2}\); (v) \(\displaystyle 3\left(a+\frac{2}{3}b\right)^{2}=\frac{1}{3}(3a+2b)^{2}\); (vi) \(\displaystyle \frac{1}{5}(3s+5v)^{2}\).

Squaring a difference. When a number sits just below a round number, write it as a subtraction and use \(\displaystyle (a-b)^{2}=a^{2}-2ab+b^{2}\). The middle term is subtracted; the last term is added, because \(\displaystyle (-b)^{2}=+b^{2}\).(i) \(\displaystyle (79)^{2}=(80-1)^{2}\) \[=80^{2}-2(80)(1)+1^{2}=6400-160+1=6241 \](ii) \(\displaystyle (193)^{2}=(200-7)^{2}\) \[=200^{2}-2(200)(7)+7^{2}=40000-2800+49=37249 \](iii) \(\displaystyle (299)^{2}=(300-1)^{2}\) \[=300^{2}-2(300)(1)+1^{2}=90000-600+1=89401 \]Answers: (i) \(\displaystyle 6241\); (ii) \(\displaystyle 37249\); (iii) \(\displaystyle 89401\).

Choose the identity by looking at the nearest round number. Both identities work for every number, so "easier" simply means: less arithmetic. Go to the nearest number whose square you know at a glance — a multiple of \(\displaystyle 10\), \(\displaystyle 100\) or \(\displaystyle 1000\). If the given number is above it, the gap is added, so use \(\displaystyle (a+b)^{2}=a^{2}+2ab+b^{2}\). If it is below it, use \(\displaystyle (a-b)^{2}=a^{2}-2ab+b^{2}\). Either way you want the gap \(\displaystyle b\) to be small, because you have to square it.

Number	Best split	Identity that is easier
\(\displaystyle 117^{2}\)	\(\displaystyle 120-3\)	\(\displaystyle (a-b)^{2}\)
\(\displaystyle 78^{2}\)	\(\displaystyle 80-2\)	\(\displaystyle (a-b)^{2}\)
\(\displaystyle 198^{2}\)	\(\displaystyle 200-2\)	\(\displaystyle (a-b)^{2}\)
\(\displaystyle 214^{2}\)	\(\displaystyle 200+14\)	\(\displaystyle (a+b)^{2}\)
\(\displaystyle 1104^{2}\)	\(\displaystyle 1100+4\)	\(\displaystyle (a+b)^{2}\)
\(\displaystyle 1120^{2}\)	\(\displaystyle 1100+20\)	\(\displaystyle (a+b)^{2}\)

(i) \(\displaystyle 117^{2}=(120-3)^{2}=14400-2(120)(3)+9=14400-720+9=13689\)(ii) \(\displaystyle 78^{2}=(80-2)^{2}=6400-2(80)(2)+4=6400-320+4=6084\)(iii) \(\displaystyle 198^{2}=(200-2)^{2}=40000-2(200)(2)+4=40000-800+4=39204\)(iv) \(\displaystyle 214^{2}=(200+14)^{2}=40000+2(200)(14)+196=40000+5600+196=45796\)(v) \(\displaystyle 1104^{2}=(1100+4)^{2}=1210000+2(1100)(4)+16=1210000+8800+16=1218816\)(vi) \(\displaystyle 1120^{2}=(1100+20)^{2}=1210000+2(1100)(20)+400=1210000+44000+400=1254400\)A remark on (iv) and (vi): the choice is not forced. For \(\displaystyle 214\) you could equally take \(\displaystyle 220-6\), giving \(\displaystyle 48400-2640+36=45796\) — the same answer, as it must be. And \(\displaystyle 1120^{2}\) is quickest of all if you notice \(\displaystyle 1120=112\times 10\), so \(\displaystyle 1120^{2}=112^{2}\times 100\). Different routes, one answer; that is the point of an identity.Answers: (i) \(\displaystyle 13689\); (ii) \(\displaystyle 6084\); (iii) \(\displaystyle 39204\); (iv) \(\displaystyle 45796\); (v) \(\displaystyle 1218816\); (vi) \(\displaystyle 1254400\).

Match the shape to the identity. Three-term expressions with two perfect squares point to \(\displaystyle (a\pm b)^{2}=a^{2}\pm 2ab+b^{2}\); six-term expressions with three perfect squares point to \(\displaystyle (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ca\). In the three-letter case the signs of the cross terms tell you which letters carry a minus sign.(i) \(\displaystyle 16y^{2}-24y+9\). Outer terms: \(\displaystyle (4y)^{2}\) and \(\displaystyle 3^{2}\). The middle term is negative, so try \(\displaystyle (4y-3)^{2}\); check \(\displaystyle 2(4y)(3)=24y\), which matches the \(\displaystyle -24y\). \[16y^{2}-24y+9=(4y-3)^{2} \](ii) \(\displaystyle \frac{9}{4}s^{2}+6st+4t^{2}\). Outer terms: \(\displaystyle \left(\frac{3}{2}s\right)^{2}\) and \(\displaystyle (2t)^{2}\). Check \(\displaystyle 2\left(\frac{3}{2}s\right)(2t)=6st\). It matches. \[\tfrac{9}{4}s^{2}+6st+4t^{2}=\left(\tfrac{3}{2}s+2t\right)^{2}=\tfrac{1}{4}(3s+4t)^{2} \](iii) \(\displaystyle \frac{m^{2}}{9}+\frac{mk}{3}+\frac{k^{2}}{4}+3nk+2mn+9n^{2}\). Six terms, and three of them are squares: \(\displaystyle \frac{m^{2}}{9}=\left(\frac{m}{3}\right)^{2}\), \(\displaystyle \frac{k^{2}}{4}=\left(\frac{k}{2}\right)^{2}\), \(\displaystyle 9n^{2}=(3n)^{2}\). All signs are positive, so try \(\displaystyle a=\frac{m}{3}\), \(\displaystyle b=\frac{k}{2}\), \(\displaystyle c=3n\) and check the three cross terms: \[2ab=2\cdot\tfrac{m}{3}\cdot\tfrac{k}{2}=\tfrac{mk}{3},\qquad 2bc=2\cdot\tfrac{k}{2}\cdot 3n=3nk,\qquad 2ca=2\cdot 3n\cdot\tfrac{m}{3}=2mn. \] All three match the printed terms. \[\tfrac{m^{2}}{9}+\tfrac{mk}{3}+\tfrac{k^{2}}{4}+3nk+2mn+9n^{2}=\left(\tfrac{m}{3}+\tfrac{k}{2}+3n\right)^{2} \](iv) \(\displaystyle \frac{p^{2}}{16}-2+\frac{16}{p^{2}}\). The outer terms are \(\displaystyle \left(\frac{p}{4}\right)^{2}\) and \(\displaystyle \left(\frac{4}{p}\right)^{2}\), and their product is neat: \(\displaystyle 2\cdot\frac{p}{4}\cdot\frac{4}{p}=2\). The middle term is \(\displaystyle -2\), so the sign inside is a minus. \[\tfrac{p^{2}}{16}-2+\tfrac{16}{p^{2}}=\left(\tfrac{p}{4}-\tfrac{4}{p}\right)^{2} \](v) \(\displaystyle 9a^{2}+4b^{2}+c^{2}-12ab+6ac-4bc\). The squares are \(\displaystyle (3a)^{2}\), \(\displaystyle (2b)^{2}\), \(\displaystyle c^{2}\). Two cross terms are negative and one is positive, and both negative ones involve \(\displaystyle b\) — so \(\displaystyle b\) is the term carrying the minus sign. Try \(\displaystyle a\!\to\!3a\), \(\displaystyle b\!\to\!-2b\), \(\displaystyle c\!\to\!c\): \[2(3a)(-2b)=-12ab,\qquad 2(-2b)(c)=-4bc,\qquad 2(c)(3a)=6ac. \] All three match. \[9a^{2}+4b^{2}+c^{2}-12ab+6ac-4bc=(3a-2b+c)^{2} \] (Since squaring kills an overall sign, \(\displaystyle (-3a+2b-c)^{2}\) is the same answer.)Answers: (i) \(\displaystyle (4y-3)^{2}\); (ii) \(\displaystyle \left(\frac{3}{2}s+2t\right)^{2}\); (iii) \(\displaystyle \left(\frac{m}{3}+\frac{k}{2}+3n\right)^{2}\); (iv) \(\displaystyle \left(\frac{p}{4}-\frac{4}{p}\right)^{2}\); (v) \(\displaystyle (3a-2b+c)^{2}\).

Use the identity once for each expression — the only work is deciding what \(\displaystyle a\), \(\displaystyle b\) and \(\displaystyle c\) are.\[(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ca \](i) \(\displaystyle (p+3q+7r)^{2}\)Match the expression to the identity: \(\displaystyle a=p\), \(\displaystyle b=3q\), \(\displaystyle c=7r\).Square each term:\[a^{2}=p^{2}, \qquad b^{2}=(3q)^{2}=9q^{2}, \qquad c^{2}=(7r)^{2}=49r^{2} \]Now the three cross terms. Each is twice a product, and the whole coefficient gets squared or multiplied — not just the letter:\[2ab=2(p)(3q)=6pq, \qquad 2bc=2(3q)(7r)=42qr, \qquad 2ca=2(7r)(p)=14rp \]Adding all six:\[(p+3q+7r)^{2}=p^{2}+9q^{2}+49r^{2}+6pq+42qr+14rp \](ii) \(\displaystyle (3x-2y+4z)^{2}\)Here \(\displaystyle a=3x\), \(\displaystyle b=-2y\), \(\displaystyle c=4z\). Take the minus sign into \(\displaystyle b\) rather than trying to remember a different identity — that is the whole trick with a subtraction.Squares (a negative squared is positive, so the \(\displaystyle y^{2}\) term is positive):\[a^{2}=9x^{2}, \qquad b^{2}=(-2y)^{2}=4y^{2}, \qquad c^{2}=16z^{2} \]Cross terms, keeping the sign of \(\displaystyle b\):\[2ab=2(3x)(-2y)=-12xy, \qquad 2bc=2(-2y)(4z)=-16yz, \qquad 2ca=2(4z)(3x)=24zx \]\[(3x-2y+4z)^{2}=9x^{2}+4y^{2}+16z^{2}-12xy-16yz+24zx \]Notice which terms went negative: the two that involve \(\displaystyle y\), because \(\displaystyle y\) is the only one carrying a minus. The \(\displaystyle zx\) term stays positive — two of the three products never touch \(\displaystyle y\).Answer: (i) \(\displaystyle p^{2}+9q^{2}+49r^{2}+6pq+42qr+14rp\) (ii) \(\displaystyle 9x^{2}+4y^{2}+16z^{2}-12xy-16yz+24zx\)

Test a claimed identity by expanding both sides. An identity must be true for every value of every letter. So there are two possible jobs: expand both sides and show they agree, or produce a single set of values where they disagree. Here, expanding shows they do not agree.Expand each square with \(\displaystyle (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ca\), reading the minus signs into the terms: \[(a+b-c)^{2}=a^{2}+b^{2}+c^{2}+2ab-2bc-2ca \] \[(a-b+c)^{2}=a^{2}+b^{2}+c^{2}-2ab-2bc+2ca \] \[(a-b-c)^{2}=a^{2}+b^{2}+c^{2}-2ab+2bc-2ca \]Now add. The three \(\displaystyle ab\) terms give \(\displaystyle 2ab-2ab-2ab=-2ab\); the three \(\displaystyle bc\) terms give \(\displaystyle -2bc-2bc+2bc=-2bc\); the three \(\displaystyle ca\) terms give \(\displaystyle -2ca+2ca-2ca=-2ca\). So \[(a+b-c)^{2}+(a-b+c)^{2}+(a-b-c)^{2}=3a^{2}+3b^{2}+3c^{2}-2ab-2bc-2ca. \]The right-hand side of the claim is \(\displaystyle 2a^{2}+2b^{2}+2c^{2}\), which is not the same expression.A single counter-example settles it. Take \(\displaystyle a=b=c=1\): \[\text{LHS}=(1+1-1)^{2}+(1-1+1)^{2}+(1-1-1)^{2}=1+1+1=3, \qquad \text{RHS}=2+2+2=6. \] Since \(\displaystyle 3\neq 6\), the statement fails for at least one set of values, so it is not an identity.A caution worth remembering. The equation is not never true — it just is not always true. For instance \(\displaystyle a=0\), \(\displaystyle b=c=1\) gives LHS \(\displaystyle =0+0+4=4\) and RHS \(\displaystyle =0+2+2=4\). Finding values that work is therefore not evidence of an identity; only an expansion valid for all values is. This is exactly why we expand instead of testing a few numbers.Answer: No, it is not an identity. The correct expansion is \(\displaystyle (a+b-c)^{2}+(a-b+c)^{2}+(a-b-c)^{2}=3a^{2}+3b^{2}+3c^{2}-2ab-2bc-2ca\), and \(\displaystyle a=b=c=1\) gives \(\displaystyle 3\) on the left but \(\displaystyle 6\) on the right.Note on the passage that follows the question. The remaining text belongs to the book's discussion, not to the question, and the figure it refers to is not reproduced here. The identity itself needs no figure: starting from \(\displaystyle a^{2}-b^{2}=(a+b)(a-b)\) and adding \(\displaystyle b^{2}\) to both sides gives \[(a+b)(a-b)+b^{2}=a^{2}-b^{2}+b^{2}=a^{2}. \] Śhrīdharāchārya's use of it is to choose \(\displaystyle b\) so that \(\displaystyle a+b\) and \(\displaystyle a-b\) are easy: \(\displaystyle 55^{2}=(55+5)(55-5)+5^{2}=60\times 50+25=3000+25=3025\).

Splitting the middle term. For a product \(\displaystyle (x+p)(x+q)=x^{2}+(p+q)x+pq\), so to factor a quadratic you hunt for two numbers whose product is the constant term and whose sum is the coefficient of the middle term. When the leading coefficient is not \(\displaystyle 1\), the same idea works on the product (leading coefficient) \(\displaystyle \times\) (constant term), and you then group in pairs.(i) \(\displaystyle s^{2}-11s+24\). Need two numbers with product \(\displaystyle +24\) and sum \(\displaystyle -11\). The product is positive and the sum negative, so both are negative: \(\displaystyle -3\) and \(\displaystyle -8\) work, since \(\displaystyle (-3)(-8)=24\) and \(\displaystyle -3-8=-11\). \[s^{2}-11s+24=(s-3)(s-8) \] Check: \(\displaystyle (s-3)(s-8)=s^{2}-8s-3s+24=s^{2}-11s+24\). ✔(ii) \(\displaystyle (\underline{\phantom{xxx}})(x+1)=3x^{2}-4x-7\). The missing factor must be linear, of the form \(\displaystyle 3x+k\), because \(\displaystyle 3x\cdot x=3x^{2}\). Its constant must satisfy \(\displaystyle k\times 1=-7\), so \(\displaystyle k=-7\). \[(3x-7)(x+1)=3x^{2}+3x-7x-7=3x^{2}-4x-7 \ \checkmark \] So the blank is \(\displaystyle 3x-7\).(iii) \(\displaystyle 10x^{2}-11x-6=(2x-\underline{\phantom{xx}})(\underline{\phantom{xx}}+2)\). Split the middle term: we need two numbers with product \(\displaystyle 10\times(-6)=-60\) and sum \(\displaystyle -11\); those are \(\displaystyle -15\) and \(\displaystyle +4\). \[10x^{2}-11x-6=10x^{2}-15x+4x-6=5x(2x-3)+2(2x-3)=(2x-3)(5x+2) \] Matching this against the printed shape, the first blank is \(\displaystyle 3\) and the second is \(\displaystyle 5x\).(iv) \(\displaystyle 6x^{2}+7x+2\). Two numbers with product \(\displaystyle 6\times 2=12\) and sum \(\displaystyle 7\): they are \(\displaystyle 3\) and \(\displaystyle 4\). \[6x^{2}+7x+2=6x^{2}+3x+4x+2=3x(2x+1)+2(2x+1)=(2x+1)(3x+2) \]Answers: (i) \(\displaystyle (s-3)(s-8)\); (ii) \(\displaystyle (3x-7)\); (iii) \(\displaystyle (2x-3)(5x+2)\), so the blanks are \(\displaystyle 3\) and \(\displaystyle 5x\); (iv) \(\displaystyle (2x+1)(3x+2)\).

Pick the identity that fits the shape of the numbers. Three identities cover everything here: \[(a+b)^{2}=a^{2}+2ab+b^{2},\qquad (a-b)^{2}=a^{2}-2ab+b^{2},\qquad (a+b)(a-b)=a^{2}-b^{2}, \] together with \(\displaystyle (x+p)(x+q)=x^{2}+(p+q)x+pq\) when the two numbers are not symmetric about a round number. The rule of thumb: find a round number nearby, and see whether the two numbers sit at equal distances from it (difference of squares) or at unequal distances (the \(\displaystyle (x+p)(x+q)\) form).(i) \(\displaystyle (41)^{2}\): \(\displaystyle 41=40+1\), so use \(\displaystyle (a+b)^{2}\). \[(40+1)^{2}=1600+2(40)(1)+1=1600+80+1=1681 \](ii) \(\displaystyle (27)^{2}\): \(\displaystyle 27=30-3\), so use \(\displaystyle (a-b)^{2}\). \[(30-3)^{2}=900-2(30)(3)+9=900-180+9=729 \](iii) \(\displaystyle 23\times 17\): both numbers are \(\displaystyle 3\) away from \(\displaystyle 20\), one above and one below — the signature of \(\displaystyle (a+b)(a-b)\). \[(20+3)(20-3)=400-9=391 \](iv) \(\displaystyle (135)^{2}\): use \(\displaystyle (a+b)^{2}\) with \(\displaystyle 135=130+5\). \[(130+5)^{2}=16900+2(130)(5)+25=16900+1300+25=18225 \] Śhrīdharāchārya's form is even quicker here: \(\displaystyle 135^{2}=(135+5)(135-5)+5^{2}=140\times 130+25=18200+25=18225\).(v) \(\displaystyle (97)^{2}\): \(\displaystyle 97=100-3\), so use \(\displaystyle (a-b)^{2}\). \[(100-3)^{2}=10000-2(100)(3)+9=10000-600+9=9409 \](vi) \(\displaystyle 18\times 29\): these are not equally spaced about a round number (\(\displaystyle 18=20-2\), \(\displaystyle 29=20+9\)), so use \(\displaystyle (x+p)(x+q)=x^{2}+(p+q)x+pq\) with \(\displaystyle x=20\), \(\displaystyle p=-2\), \(\displaystyle q=9\). \[18\times 29=400+(-2+9)(20)+(-2)(9)=400+140-18=522 \](vii) \(\displaystyle 34\times 43\): take \(\displaystyle x=40\), \(\displaystyle p=-6\), \(\displaystyle q=3\). \[34\times 43=1600+(-6+3)(40)+(-6)(3)=1600-120-18=1462 \](viii) \(\displaystyle (205)^{2}\): use \(\displaystyle (a+b)^{2}\) with \(\displaystyle 205=200+5\). \[(200+5)^{2}=40000+2(200)(5)+25=40000+2000+25=42025 \]Answers: (i) \(\displaystyle 1681\); (ii) \(\displaystyle 729\); (iii) \(\displaystyle 391\); (iv) \(\displaystyle 18225\); (v) \(\displaystyle 9409\); (vi) \(\displaystyle 522\); (vii) \(\displaystyle 1462\); (viii) \(\displaystyle 42025\).

Count the terms first. Six terms with three perfect squares means \(\displaystyle (a+b+c)^{2}\); three terms with two perfect squares means \(\displaystyle (a\pm b)^{2}\); three terms with no square pattern means splitting the middle term. In the three-letter cases, the signs of the cross terms tell you which letters are negative — and a useful check is that exactly the letters carrying a minus sign appear in the negative cross terms.(i) \(\displaystyle 9a^{2}+b^{2}+4c^{2}-6ab+12ac-4bc\). Squares: \(\displaystyle (3a)^{2}\), \(\displaystyle b^{2}\), \(\displaystyle (2c)^{2}\). The negative cross terms are the two containing \(\displaystyle b\), so \(\displaystyle b\) is the negative one. Take \(\displaystyle 3a\), \(\displaystyle -b\), \(\displaystyle 2c\): \[2(3a)(-b)=-6ab,\quad 2(-b)(2c)=-4bc,\quad 2(2c)(3a)=12ac \ \checkmark \] \[9a^{2}+b^{2}+4c^{2}-6ab+12ac-4bc=(3a-b+2c)^{2} \](ii) \(\displaystyle 16s^{2}+25t^{2}-40st\). Rearranged, \(\displaystyle 16s^{2}-40st+25t^{2}\): squares \(\displaystyle (4s)^{2}\) and \(\displaystyle (5t)^{2}\), and \(\displaystyle 2(4s)(5t)=40st\) matches the middle term, which is negative. \[16s^{2}+25t^{2}-40st=(4s-5t)^{2} \](iii) \(\displaystyle r^{2}-r-42\). No square pattern, so split the middle term: two numbers with product \(\displaystyle -42\) and sum \(\displaystyle -1\), namely \(\displaystyle -7\) and \(\displaystyle +6\). \[r^{2}-r-42=(r-7)(r+6) \] Check: \(\displaystyle r^{2}+6r-7r-42=r^{2}-r-42\). ✔(iv) \(\displaystyle 49g^{2}+14gh+h^{2}\). Squares \(\displaystyle (7g)^{2}\) and \(\displaystyle h^{2}\); check \(\displaystyle 2(7g)(h)=14gh\). \[49g^{2}+14gh+h^{2}=(7g+h)^{2} \](v) \(\displaystyle 64u^{2}+121v^{2}+4w^{2}-176uv-32uw+44vw\). Squares: \(\displaystyle (8u)^{2}\), \(\displaystyle (11v)^{2}\), \(\displaystyle (2w)^{2}\). The positive cross term is \(\displaystyle +44vw\), which pairs \(\displaystyle v\) with \(\displaystyle w\) — so \(\displaystyle v\) and \(\displaystyle w\) must have the same sign as each other and the opposite sign to \(\displaystyle u\). Take \(\displaystyle 8u\), \(\displaystyle -11v\), \(\displaystyle -2w\): \[2(8u)(-11v)=-176uv,\quad 2(-11v)(-2w)=+44vw,\quad 2(8u)(-2w)=-32uw \ \checkmark \] \[64u^{2}+121v^{2}+4w^{2}-176uv-32uw+44vw=(8u-11v-2w)^{2} \]Answers: (i) \(\displaystyle (3a-b+2c)^{2}\); (ii) \(\displaystyle (4s-5t)^{2}\); (iii) \(\displaystyle (r-7)(r+6)\); (iv) \(\displaystyle (7g+h)^{2}\); (v) \(\displaystyle (8u-11v-2w)^{2}\).

Factor the top, factor the bottom, cancel what is common. A rational expression is simplified exactly the way a fraction like \(\displaystyle \frac{18}{24}\) is: write both parts as products, then cancel the factors that appear in both. You may only cancel factors (things being multiplied), never individual terms.(i) \(\displaystyle \dfrac{3p^{2}-3pq-18q^{2}}{p^{2}+3pq-10q^{2}}\). Numerator: take out \(\displaystyle 3\), then split the middle term of \(\displaystyle p^{2}-pq-6q^{2}\) using \(\displaystyle -3\) and \(\displaystyle +2\): \[3p^{2}-3pq-18q^{2}=3(p-3q)(p+2q). \] Denominator: two numbers with product \(\displaystyle -10\) and sum \(\displaystyle +3\) are \(\displaystyle +5\) and \(\displaystyle -2\): \[p^{2}+3pq-10q^{2}=(p+5q)(p-2q). \] So \[\frac{3p^{2}-3pq-18q^{2}}{p^{2}+3pq-10q^{2}}=\frac{3(p-3q)(p+2q)}{(p+5q)(p-2q)}. \] Compare the four factors: \(\displaystyle p-3q\), \(\displaystyle p+2q\) on top and \(\displaystyle p+5q\), \(\displaystyle p-2q\) below. No factor is common, so nothing cancels and the expression is already in its simplest form. (Note \(\displaystyle p+2q\) and \(\displaystyle p-2q\) are different factors — a sign matters.)A word of caution about this part. Every other question in this set cancels neatly, so as printed this one looks like a misprint: changing the sign of a single middle term would make it cancel. If the numerator were \(\displaystyle 3p^{2}+3pq-18q^{2}=3(p+3q)(p-2q)\), the answer would be \(\displaystyle \frac{3(p+3q)}{p+5q}\); if instead the denominator were \(\displaystyle p^{2}-3pq-10q^{2}=(p-5q)(p+2q)\), the answer would be \(\displaystyle \frac{3(p-3q)}{p-5q}\). Since we cannot tell which was intended, the honest answer to the question as printed is the factored form above.(ii) \(\displaystyle \dfrac{n^{3}-3n^{2}m+3nm^{2}-m^{3}}{5m^{2}-10mn+5n^{2}}\). The numerator is exactly \(\displaystyle (a-b)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3}\) with \(\displaystyle a=n\), \(\displaystyle b=m\), so it is \(\displaystyle (n-m)^{3}\). The denominator is \(\displaystyle 5(m^{2}-2mn+n^{2})=5(m-n)^{2}\), and \(\displaystyle (m-n)^{2}=(n-m)^{2}\) because squaring removes the sign. \[\frac{(n-m)^{3}}{5(n-m)^{2}}=\frac{n-m}{5} \](iii) \(\displaystyle \dfrac{w^{3}-v^{3}+x^{3}+3wvx}{w^{2}+v^{2}+x^{2}-2wv-2vx+2wx}\). For the numerator use \(\displaystyle a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)\) with \(\displaystyle a=w\), \(\displaystyle b=-v\), \(\displaystyle c=x\). Then \(\displaystyle a^{3}+b^{3}+c^{3}=w^{3}-v^{3}+x^{3}\) and \(\displaystyle -3abc=-3(w)(-v)(x)=+3wvx\) — precisely the numerator. So \[w^{3}-v^{3}+x^{3}+3wvx=(w-v+x)\left(w^{2}+v^{2}+x^{2}+wv+vx-wx\right). \] The denominator is the square \(\displaystyle (w-v+x)^{2}\): check \(\displaystyle 2(w)(-v)=-2wv\), \(\displaystyle 2(-v)(x)=-2vx\), \(\displaystyle 2(w)(x)=+2wx\). Cancelling one factor of \(\displaystyle w-v+x\), \[\frac{w^{2}+v^{2}+x^{2}+wv+vx-wx}{w-v+x} \](iv) \(\displaystyle \dfrac{4y^{2}-20yz+25z^{2}}{25z^{2}-4y^{2}}\). Numerator \(\displaystyle =(2y-5z)^{2}=(5z-2y)^{2}\). Denominator \(\displaystyle =(5z)^{2}-(2y)^{2}=(5z-2y)(5z+2y)\). \[\frac{(5z-2y)^{2}}{(5z-2y)(5z+2y)}=\frac{5z-2y}{5z+2y} \](v) \(\displaystyle \dfrac{(x^{2}+x-6)(x^{2}-7x+12)}{(x^{2}-6x+8)(x^{2}-9)}\). Factor all four quadratics: \[x^{2}+x-6=(x+3)(x-2),\quad x^{2}-7x+12=(x-3)(x-4), \] \[x^{2}-6x+8=(x-2)(x-4),\quad x^{2}-9=(x-3)(x+3). \] Top and bottom contain exactly the same four factors, so everything cancels: \[\frac{(x+3)(x-2)(x-3)(x-4)}{(x-2)(x-4)(x-3)(x+3)}=1 \](vi) \(\displaystyle \dfrac{p^{4}-16}{p^{2}-4p+4}\). Apply the difference of squares twice on top: \(\displaystyle p^{4}-16=(p^{2})^{2}-4^{2}=(p^{2}-4)(p^{2}+4)=(p-2)(p+2)(p^{2}+4)\). The bottom is \(\displaystyle (p-2)^{2}\). \[\frac{(p-2)(p+2)(p^{2}+4)}{(p-2)^{2}}=\frac{(p+2)(p^{2}+4)}{p-2} \] (\(\displaystyle p^{2}+4\) is a sum of squares and does not factor further here.)Answers: (i) \(\displaystyle \frac{3(p-3q)(p+2q)}{(p+5q)(p-2q)}\) — nothing cancels, so this is already simplest; (ii) \(\displaystyle \frac{n-m}{5}\); (iii) \(\displaystyle \frac{w^{2}+v^{2}+x^{2}+wv+vx-wx}{w-v+x}\); (iv) \(\displaystyle \frac{5z-2y}{5z+2y}\); (v) \(\displaystyle 1\); (vi) \(\displaystyle \frac{(p+2)(p^{2}+4)}{p-2}\).

Read the shape, then choose the identity. A square of a binomial calls for \(\displaystyle (a\pm b)^{2}\); a product of a sum and a difference of the same two terms calls for \(\displaystyle (a+b)(a-b)=a^{2}-b^{2}\); a square of three terms calls for \(\displaystyle (a+b+c)^{2}\); a cube calls for \(\displaystyle (a-b)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3}\); and \(\displaystyle (a-b)(a^{2}+ab+b^{2})=a^{3}-b^{3}\).(i) \(\displaystyle (-3x+4)^{2}\). Take \(\displaystyle a=-3x\), \(\displaystyle b=4\): \(\displaystyle a^{2}=9x^{2}\), \(\displaystyle 2ab=2(-3x)(4)=-24x\), \(\displaystyle b^{2}=16\). \[(-3x+4)^{2}=9x^{2}-24x+16 \](ii) \(\displaystyle (2s+7)(2s-7)\) — a sum times the matching difference, so \(\displaystyle a^{2}-b^{2}\) with \(\displaystyle a=2s\), \(\displaystyle b=7\). \[(2s+7)(2s-7)=4s^{2}-49 \](iii) \(\displaystyle \left(p^{2}+\frac{1}{2}\right)\left(p^{2}-\frac{1}{2}\right)\), with \(\displaystyle a=p^{2}\), \(\displaystyle b=\frac{1}{2}\). \[=p^{4}-\tfrac{1}{4} \](iv) \(\displaystyle (2n+7)(2n-7)\) — the same as (ii) with \(\displaystyle n\) in place of \(\displaystyle s\). \[=4n^{2}-49 \](v) \(\displaystyle (s-2t)(s^{2}+2st+4t^{2})\). This is \(\displaystyle (a-b)(a^{2}+ab+b^{2})\) with \(\displaystyle a=s\), \(\displaystyle b=2t\): note \(\displaystyle ab=2st\) and \(\displaystyle b^{2}=4t^{2}\), exactly the bracket given. \[(s-2t)(s^{2}+2st+4t^{2})=s^{3}-8t^{3} \](vi) \(\displaystyle \left(\frac{1}{2r}-4r\right)^{2}\), with \(\displaystyle a=\frac{1}{2r}\), \(\displaystyle b=4r\): \(\displaystyle a^{2}=\frac{1}{4r^{2}}\), \(\displaystyle 2ab=2\cdot\frac{1}{2r}\cdot 4r=4\), \(\displaystyle b^{2}=16r^{2}\). \[\left(\tfrac{1}{2r}-4r\right)^{2}=\frac{1}{4r^{2}}-4+16r^{2} \](vii) \(\displaystyle (-3m+4k-l)^{2}\). Use \(\displaystyle (a+b+c)^{2}\) with \(\displaystyle a=-3m\), \(\displaystyle b=4k\), \(\displaystyle c=-l\): \[a^{2}+b^{2}+c^{2}=9m^{2}+16k^{2}+l^{2}, \] \[2ab=2(-3m)(4k)=-24km,\quad 2bc=2(4k)(-l)=-8kl,\quad 2ca=2(-l)(-3m)=6lm. \] \[(-3m+4k-l)^{2}=9m^{2}+16k^{2}+l^{2}-24km-8kl+6lm \](viii) \(\displaystyle \left(x-\frac{1}{3}y\right)^{3}\). Use \(\displaystyle (a-b)^{3}\) with \(\displaystyle a=x\), \(\displaystyle b=\frac{y}{3}\): \[a^{3}=x^{3},\quad 3a^{2}b=3x^{2}\cdot\tfrac{y}{3}=x^{2}y,\quad 3ab^{2}=3x\cdot\tfrac{y^{2}}{9}=\tfrac{xy^{2}}{3},\quad b^{3}=\tfrac{y^{3}}{27}. \] \[\left(x-\tfrac{1}{3}y\right)^{3}=x^{3}-x^{2}y+\frac{xy^{2}}{3}-\frac{y^{3}}{27} \](ix) \(\displaystyle \left(\frac{7}{2}k-\frac{2}{3}m\right)^{3}\), with \(\displaystyle a=\frac{7k}{2}\), \(\displaystyle b=\frac{2m}{3}\): \[a^{3}=\frac{343k^{3}}{8},\qquad 3a^{2}b=3\cdot\frac{49k^{2}}{4}\cdot\frac{2m}{3}=\frac{49k^{2}m}{2}, \] \[3ab^{2}=3\cdot\frac{7k}{2}\cdot\frac{4m^{2}}{9}=\frac{14km^{2}}{3},\qquad b^{3}=\frac{8m^{3}}{27}. \] \[\left(\tfrac{7}{2}k-\tfrac{2}{3}m\right)^{3}=\frac{343}{8}k^{3}-\frac{49}{2}k^{2}m+\frac{14}{3}km^{2}-\frac{8}{27}m^{3} \]Answers: (i) \(\displaystyle 9x^{2}-24x+16\); (ii) \(\displaystyle 4s^{2}-49\); (iii) \(\displaystyle p^{4}-\frac{1}{4}\); (iv) \(\displaystyle 4n^{2}-49\); (v) \(\displaystyle s^{3}-8t^{3}\); (vi) \(\displaystyle 16r^{2}-4+\frac{1}{4r^{2}}\); (vii) \(\displaystyle 9m^{2}+16k^{2}+l^{2}-24km-8kl+6lm\); (viii) \(\displaystyle x^{3}-x^{2}y+\frac{xy^{2}}{3}-\frac{y^{3}}{27}\); (ix) \(\displaystyle \frac{343}{8}k^{3}-\frac{49}{2}k^{2}m+\frac{14}{3}km^{2}-\frac{8}{27}m^{3}\).

Turn each number into a round number plus or minus a small one. For products, check whether the two numbers are equally spaced about a round number — then \(\displaystyle (a+b)(a-b)=a^{2}-b^{2}\) applies. For cubes, use \(\displaystyle (a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}\) or \(\displaystyle (a-b)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3}\).(i) \(\displaystyle 17\times 21\): both are \(\displaystyle 2\) away from \(\displaystyle 19\). \[(19-2)(19+2)=19^{2}-2^{2}=361-4=357 \](ii) \(\displaystyle 104\times 96\): both are \(\displaystyle 4\) away from \(\displaystyle 100\). \[(100+4)(100-4)=10000-16=9984 \](iii) \(\displaystyle 24\times 16\): both are \(\displaystyle 4\) away from \(\displaystyle 20\). \[(20+4)(20-4)=400-16=384 \](iv) \(\displaystyle 147^{3}=(150-3)^{3}\), with \(\displaystyle a=150\), \(\displaystyle b=3\): \[a^{3}=3375000,\quad 3a^{2}b=3(22500)(3)=202500,\quad 3ab^{2}=3(150)(9)=4050,\quad b^{3}=27. \] \[147^{3}=3375000-202500+4050-27=3176523 \](v) \(\displaystyle 199^{3}=(200-1)^{3}\): \[=8000000-3(40000)(1)+3(200)(1)-1=8000000-120000+600-1=7880599 \](vi) \(\displaystyle 127^{3}=(130-3)^{3}\), with \(\displaystyle a=130\), \(\displaystyle b=3\): \[a^{3}=2197000,\quad 3a^{2}b=3(16900)(3)=152100,\quad 3ab^{2}=3(130)(9)=3510,\quad b^{3}=27. \] \[127^{3}=2197000-152100+3510-27=2048383 \](vii) \(\displaystyle (-107)^{3}\). An odd power of a negative number is negative, so \(\displaystyle (-107)^{3}=-\left(107^{3}\right)\). Now \(\displaystyle 107^{3}=(100+7)^{3}\): \[=1000000+3(10000)(7)+3(100)(49)+343=1000000+210000+14700+343=1225043. \] \[(-107)^{3}=-1225043 \](viii) \(\displaystyle (-299)^{3}=-\left(299^{3}\right)\), and \(\displaystyle 299^{3}=(300-1)^{3}\): \[=27000000-3(90000)(1)+3(300)(1)-1=27000000-270000+900-1=26730899. \] \[(-299)^{3}=-26730899 \]Answers: (i) \(\displaystyle 357\); (ii) \(\displaystyle 9984\); (iii) \(\displaystyle 384\); (iv) \(\displaystyle 3176523\); (v) \(\displaystyle 7880599\); (vi) \(\displaystyle 2048383\); (vii) \(\displaystyle -1225043\); (viii) \(\displaystyle -26730899\).

Identify the pattern before you factor. Two terms that are both cubes point to \(\displaystyle a^{3}\pm b^{3}\); two terms that are both squares point to \(\displaystyle a^{2}-b^{2}\); three terms with two squares point to \(\displaystyle (a\pm b)^{2}\); four terms with two cubes point to \(\displaystyle (a\pm b)^{3}\); and four terms of the form \(\displaystyle a^{3}+b^{3}+c^{3}-3abc\) point to \(\displaystyle (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)\).(i) \(\displaystyle 4y^{2}+1+\frac{1}{16y^{2}}\). Outer squares \(\displaystyle (2y)^{2}\) and \(\displaystyle \left(\frac{1}{4y}\right)^{2}\); check \(\displaystyle 2(2y)\left(\frac{1}{4y}\right)=1\), the middle term. \[=\left(2y+\tfrac{1}{4y}\right)^{2} \](ii) \(\displaystyle 9m^{2}-\frac{1}{25n^{2}}=(3m)^{2}-\left(\frac{1}{5n}\right)^{2}\), a difference of squares. \[=\left(3m-\tfrac{1}{5n}\right)\left(3m+\tfrac{1}{5n}\right) \](iii) \(\displaystyle 27b^{3}-\frac{1}{64b^{3}}=(3b)^{3}-\left(\frac{1}{4b}\right)^{3}\). Use \(\displaystyle a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})\), where \(\displaystyle ab=3b\cdot\frac{1}{4b}=\frac{3}{4}\). \[=\left(3b-\tfrac{1}{4b}\right)\left(9b^{2}+\tfrac{3}{4}+\tfrac{1}{16b^{2}}\right) \](iv) \(\displaystyle x^{2}+\frac{5x}{6}+\frac{1}{6}\). Multiply and divide by \(\displaystyle 6\) to clear fractions: \(\displaystyle \frac{1}{6}\left(6x^{2}+5x+1\right)\). Split \(\displaystyle 5x\) using \(\displaystyle 2\) and \(\displaystyle 3\) (product \(\displaystyle 6\times 1=6\), sum \(\displaystyle 5\)): \[6x^{2}+5x+1=6x^{2}+3x+2x+1=3x(2x+1)+(2x+1)=(2x+1)(3x+1). \] \[x^{2}+\tfrac{5x}{6}+\tfrac{1}{6}=\tfrac{1}{6}(2x+1)(3x+1)=\left(x+\tfrac{1}{2}\right)\left(x+\tfrac{1}{3}\right) \](v) \(\displaystyle 27u^{3}-\frac{1}{125}-\frac{27u^{2}}{5}+\frac{9u}{25}\). Four terms with \(\displaystyle (3u)^{3}\) and \(\displaystyle \left(\frac{1}{5}\right)^{3}\) suggests \(\displaystyle (a-b)^{3}\) with \(\displaystyle a=3u\), \(\displaystyle b=\frac{1}{5}\). Check the two middle terms: \(\displaystyle 3a^{2}b=3(9u^{2})\left(\frac{1}{5}\right)=\frac{27u^{2}}{5}\) (subtracted ✔) and \(\displaystyle 3ab^{2}=3(3u)\left(\frac{1}{25}\right)=\frac{9u}{25}\) (added ✔). \[=\left(3u-\tfrac{1}{5}\right)^{3} \](vi) \(\displaystyle 64y^{3}+\frac{1}{125}z^{3}=(4y)^{3}+\left(\frac{z}{5}\right)^{3}\). Use \(\displaystyle a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})\). \[=\left(4y+\tfrac{z}{5}\right)\left(16y^{2}-\tfrac{4yz}{5}+\tfrac{z^{2}}{25}\right) \](vii) \(\displaystyle p^{3}+27q^{3}+r^{3}-9pqr\). Write \(\displaystyle 27q^{3}=(3q)^{3}\); then with \(\displaystyle a=p\), \(\displaystyle b=3q\), \(\displaystyle c=r\) we get \(\displaystyle 3abc=3(p)(3q)(r)=9pqr\), so this is exactly \(\displaystyle a^{3}+b^{3}+c^{3}-3abc\). \[=(p+3q+r)\left(p^{2}+9q^{2}+r^{2}-3pq-3qr-rp\right) \](viii) \(\displaystyle 9m^{2}-12m+4\): squares \(\displaystyle (3m)^{2}\) and \(\displaystyle 2^{2}\), and \(\displaystyle 2(3m)(2)=12m\). \[=(3m-2)^{2} \](ix) \(\displaystyle 9x^{3}-\frac{8}{3}y^{3}+\frac{z^{3}}{3}+6xyz\). None of \(\displaystyle 9x^{3}\), \(\displaystyle \frac{8}{3}y^{3}\), \(\displaystyle \frac{z^{3}}{3}\) is a clean cube, so take out \(\displaystyle \frac{1}{3}\) first: \[=\tfrac{1}{3}\left(27x^{3}-8y^{3}+z^{3}+18xyz\right). \] Now \(\displaystyle 27x^{3}=(3x)^{3}\), \(\displaystyle -8y^{3}=(-2y)^{3}\), \(\displaystyle z^{3}=z^{3}\), and with \(\displaystyle a=3x\), \(\displaystyle b=-2y\), \(\displaystyle c=z\) we get \(\displaystyle -3abc=-3(3x)(-2y)(z)=+18xyz\) ✔. So the bracket is \(\displaystyle a^{3}+b^{3}+c^{3}-3abc\) and \[a^{2}+b^{2}+c^{2}-ab-bc-ca=9x^{2}+4y^{2}+z^{2}+6xy+2yz-3xz. \] \[9x^{3}-\tfrac{8}{3}y^{3}+\tfrac{z^{3}}{3}+6xyz=\tfrac{1}{3}(3x-2y+z)\left(9x^{2}+4y^{2}+z^{2}+6xy+2yz-3xz\right) \](x) \(\displaystyle 4x^{2}+9y^{2}+36z^{2}+12xz+36yz+24xy\). This one does not factor as printed, and I believe the printed cross terms are swapped. Here is the check. Six terms with the squares \(\displaystyle (2x)^{2}\), \(\displaystyle (3y)^{2}\), \(\displaystyle (6z)^{2}\) invite \(\displaystyle (2x+3y+6z)^{2}\), whose cross terms are forced to be \[2(2x)(3y)=12xy,\qquad 2(3y)(6z)=36yz,\qquad 2(2x)(6z)=24xz. \] The printed expression has \(\displaystyle 36yz\) ✔ but \(\displaystyle 24xy\) and \(\displaystyle 12xz\) — the coefficients \(\displaystyle 24\) and \(\displaystyle 12\) are attached to the wrong pairs. Changing the signs of \(\displaystyle 2x\), \(\displaystyle 3y\) or \(\displaystyle 6z\) cannot fix this, because it changes signs, not sizes. A quick numerical check confirms the mismatch: at \(\displaystyle x=y=1\), \(\displaystyle z=0\) the printed expression gives \(\displaystyle 4+9+24=37\), while \(\displaystyle (2x+3y+6z)^{2}=(2+3)^{2}=25\). In fact the printed expression cannot be written as a product of two linear expressions at all, so it is not factorable by any identity in this chapter.Almost certainly the intended expression is \(\displaystyle 4x^{2}+9y^{2}+36z^{2}+12xy+36yz+24xz\), and then \[4x^{2}+9y^{2}+36z^{2}+12xy+36yz+24xz=(2x+3y+6z)^{2}. \](xi) \(\displaystyle 27u^{3}-\frac{1}{216}-\frac{9u^{2}}{2}+\frac{u}{4}\). Here \(\displaystyle (3u)^{3}=27u^{3}\) and \(\displaystyle \left(\frac{1}{6}\right)^{3}=\frac{1}{216}\), so try \(\displaystyle (a-b)^{3}\) with \(\displaystyle a=3u\), \(\displaystyle b=\frac{1}{6}\). Check: \(\displaystyle 3a^{2}b=3(9u^{2})\left(\frac{1}{6}\right)=\frac{9u^{2}}{2}\) (subtracted ✔) and \(\displaystyle 3ab^{2}=3(3u)\left(\frac{1}{36}\right)=\frac{u}{4}\) (added ✔). \[=\left(3u-\tfrac{1}{6}\right)^{3} \]Answers: (i) \(\displaystyle \left(2y+\frac{1}{4y}\right)^{2}\); (ii) \(\displaystyle \left(3m-\frac{1}{5n}\right)\left(3m+\frac{1}{5n}\right)\); (iii) \(\displaystyle \left(3b-\frac{1}{4b}\right)\left(9b^{2}+\frac{3}{4}+\frac{1}{16b^{2}}\right)\); (iv) \(\displaystyle \left(x+\frac{1}{2}\right)\left(x+\frac{1}{3}\right)\); (v) \(\displaystyle \left(3u-\frac{1}{5}\right)^{3}\); (vi) \(\displaystyle \left(4y+\frac{z}{5}\right)\left(16y^{2}-\frac{4yz}{5}+\frac{z^{2}}{25}\right)\); (vii) \(\displaystyle (p+3q+r)\left(p^{2}+9q^{2}+r^{2}-3pq-3qr-rp\right)\); (viii) \(\displaystyle (3m-2)^{2}\); (ix) \(\displaystyle \frac{1}{3}(3x-2y+z)\left(9x^{2}+4y^{2}+z^{2}+6xy+2yz-3xz\right)\); (x) not factorable as printed — with the \(\displaystyle xy\) and \(\displaystyle xz\) coefficients in their intended places it is \(\displaystyle (2x+3y+6z)^{2}\); (xi) \(\displaystyle \left(3u-\frac{1}{6}\right)^{3}\).

Factor top and bottom, then cancel common factors. Cancelling is only allowed for whole factors, so the first job is always to turn each sum into a product.(i) \(\displaystyle \dfrac{4x^{2}+4x+1}{4x^{2}-1}\). The numerator has squares \(\displaystyle (2x)^{2}\) and \(\displaystyle 1^{2}\) with \(\displaystyle 2(2x)(1)=4x\), so it is \(\displaystyle (2x+1)^{2}\). The denominator is a difference of squares, \(\displaystyle (2x-1)(2x+1)\). \[\frac{(2x+1)^{2}}{(2x-1)(2x+1)}=\frac{2x+1}{2x-1} \](ii) \(\displaystyle \dfrac{9\left(3a^{3}-24b^{3}\right)}{9a^{2}-36b^{2}}\). Numerator: \(\displaystyle 9\cdot 3\left(a^{3}-8b^{3}\right)=27\left(a^{3}-(2b)^{3}\right)=27(a-2b)\left(a^{2}+2ab+4b^{2}\right)\). Denominator: \(\displaystyle 9\left(a^{2}-4b^{2}\right)=9(a-2b)(a+2b)\). \[\frac{27(a-2b)\left(a^{2}+2ab+4b^{2}\right)}{9(a-2b)(a+2b)}=\frac{3\left(a^{2}+2ab+4b^{2}\right)}{a+2b} \](iii) \(\displaystyle \dfrac{s^{3}+125t^{3}}{s^{2}-2st-35t^{2}}\). Numerator: \(\displaystyle s^{3}+(5t)^{3}=(s+5t)\left(s^{2}-5st+25t^{2}\right)\). Denominator: two numbers with product \(\displaystyle -35\) and sum \(\displaystyle -2\) are \(\displaystyle -7\) and \(\displaystyle +5\), so it is \(\displaystyle (s-7t)(s+5t)\). \[\frac{(s+5t)\left(s^{2}-5st+25t^{2}\right)}{(s-7t)(s+5t)}=\frac{s^{2}-5st+25t^{2}}{s-7t} \]Answers: (i) \(\displaystyle \frac{2x+1}{2x-1}\); (ii) \(\displaystyle \frac{3\left(a^{2}+2ab+4b^{2}\right)}{a+2b}\); (iii) \(\displaystyle \frac{s^{2}-5st+25t^{2}}{s-7t}\).

Area of a rectangle \(\displaystyle =\) length \(\displaystyle \times\) breadth, so factoring the area gives possible side lengths. Every way of writing the area as a product of two expressions is a legitimate answer — which is why the question says "possible expressions". The natural answer is the factorisation into the simplest pieces.(i) Area \(\displaystyle =25a^{2}-30ab+9b^{2}\). The outer terms are \(\displaystyle (5a)^{2}\) and \(\displaystyle (3b)^{2}\), and \(\displaystyle 2(5a)(3b)=30ab\) matches the middle term, which is negative: \[25a^{2}-30ab+9b^{2}=(5a-3b)^{2}=(5a-3b)\times(5a-3b). \] So one possible rectangle has length \(\displaystyle =5a-3b\) and breadth \(\displaystyle =5a-3b\) — the two sides are equal, so this rectangle is in fact a square.(ii) Area \(\displaystyle =36s^{2}-49t^{2}=(6s)^{2}-(7t)^{2}\), a difference of squares: \[36s^{2}-49t^{2}=(6s+7t)(6s-7t). \] So length \(\displaystyle =6s+7t\) and breadth \(\displaystyle =6s-7t\) (taking the longer factor as the length).Other answers are equally valid. Nothing forces this particular split — for example \(\displaystyle (5a-3b)\) and \(\displaystyle (5a-3b)\) could be replaced by \(\displaystyle 2(5a-3b)\) and \(\displaystyle \frac{1}{2}(5a-3b)\), and in (ii) by \(\displaystyle 3(6s+7t)\) and \(\displaystyle \frac{1}{3}(6s-7t)\). Any pair of expressions whose product is the given area describes a rectangle of that area; the factorisations above are simply the tidiest ones.Answers: (i) length \(\displaystyle =5a-3b\), breadth \(\displaystyle =5a-3b\); (ii) length \(\displaystyle =6s+7t\), breadth \(\displaystyle =6s-7t\). (Other factor pairs with the same product are also correct.)

Volume of a cuboid \(\displaystyle =\) length \(\displaystyle \times\) breadth \(\displaystyle \times\) height, so split the volume into three factors. As in the previous question the answer is not unique — any three expressions with the given product will do — so the aim is a tidy factorisation.(i) Volume \(\displaystyle =6a^{2}-24b^{2}\). Take out the common factor \(\displaystyle 6\), then use the difference of squares: \[6a^{2}-24b^{2}=6\left(a^{2}-4b^{2}\right)=6\left(a^{2}-(2b)^{2}\right)=6(a-2b)(a+2b). \] That is a product of three factors, so one possible cuboid has length \(\displaystyle =a+2b\), breadth \(\displaystyle =a-2b\), height \(\displaystyle =6\).(ii) Volume \(\displaystyle =3ps^{2}-15ps+12p\). Every term has \(\displaystyle 3p\) in it, so take that out first, then split the middle term of the quadratic (two numbers with product \(\displaystyle 4\) and sum \(\displaystyle -5\): \(\displaystyle -1\) and \(\displaystyle -4\)): \[3ps^{2}-15ps+12p=3p\left(s^{2}-5s+4\right)=3p(s-1)(s-4). \] So one possible cuboid has length \(\displaystyle =s-1\), breadth \(\displaystyle =s-4\), height \(\displaystyle =3p\).Other answers are equally valid. In (i) the \(\displaystyle 6\) could be split as \(\displaystyle 2\) and \(\displaystyle 3\) and shared out differently — for instance length \(\displaystyle 2(a+2b)\), breadth \(\displaystyle 3(a-2b)\), height \(\displaystyle 1\) — and in (ii) the \(\displaystyle 3p\) could be split as \(\displaystyle 3\) and \(\displaystyle p\), giving height \(\displaystyle 3\), and \(\displaystyle p(s-1)\) as one of the other edges. Any three expressions whose product is the given volume answer the question; the versions above are the simplest.Answers: (i) length \(\displaystyle =a+2b\), breadth \(\displaystyle =a-2b\), height \(\displaystyle =6\); (ii) length \(\displaystyle =s-1\), breadth \(\displaystyle =s-4\), height \(\displaystyle =3p\). (Other factor triples with the same product are also correct.)

Area of the big square minus the area of the playground.The path is built around the playground, so the playground keeps its full side of $\displaystyle 40$ m and the path is a border of width \(\displaystyle s\) m lying outside it.Walk across the whole figure along one direction: you cross the path (\(\displaystyle s\) m), then the playground ($\displaystyle 40$ m), then the path again (\(\displaystyle s\) m). So the outer boundary is itself a square, of side\[40 + s + s \;=\; 40 + 2s \ \text{metres}. \]The path is what is left when the playground is removed from this big square:\[\text{Area of path} \;=\; (40+2s)^2 - 40^2. \]Expand \(\displaystyle (40+2s)^2\) with the identity \(\displaystyle (a+b)^2=a^2+2ab+b^2\), taking \(\displaystyle a=40,\ b=2s\):\[(40+2s)^2 = 40^2 + 2\cdot 40\cdot 2s + (2s)^2 = 1600 + 160s + 4s^2. \]\[\text{Area of path} = 1600 + 160s + 4s^2 - 1600 = 4s^2 + 160s. \]Check $\displaystyle 1$ (difference of squares). Using \(\displaystyle a^2-b^2=(a+b)(a-b)\) with \(\displaystyle a=40+2s,\ b=40\): \[(40+2s)^2-40^2 = \big((40+2s)+40\big)\big((40+2s)-40\big) = (80+2s)(2s) = 160s + 4s^2, \] the same expression.Check $\displaystyle 2$ (cut the path into pieces). The path is four strips of size \(\displaystyle 40 \times s\) along the four sides, plus four small squares of side \(\displaystyle s\) at the corners: \[4(40s) + 4(s^2) = 160s + 4s^2. \] Again the same. For example, with \(\displaystyle s=1\) all three methods give \(\displaystyle 164\) square metres.Area of the path \(\displaystyle = 4s^2 + 160s = 4s(s+40)\) square metres.

Clear the fraction to get a quadratic, then split the middle term.Let the number be \(\displaystyle x\). A number has a reciprocal only if it is not zero, so \(\displaystyle x \neq 0\), and the condition is\[x + \frac{1}{x} = \frac{10}{3}. \]Multiply both sides by \(\displaystyle 3x\). This is allowed because \(\displaystyle x \neq 0\), so \(\displaystyle 3x \neq 0\) and we are not multiplying by zero:\[3x\cdot x + 3x\cdot\frac{1}{x} = 3x \cdot \frac{10}{3} \quad\Longrightarrow\quad 3x^2 + 3 = 10x. \]Bring everything to one side:\[3x^2 - 10x + 3 = 0. \]Now split the middle term. We need two numbers whose product is \(\displaystyle 3 \times 3 = 9\) and whose sum is \(\displaystyle -10\); these are \(\displaystyle -9\) and \(\displaystyle -1\):\[3x^2 - 9x - x + 3 = 3x(x-3) - 1(x-3) = (x-3)(3x-1). \]So \(\displaystyle (x-3)(3x-1)=0\), which gives \(\displaystyle x = 3\) or \(\displaystyle x = \dfrac{1}{3}\).Verification. \[3 + \frac{1}{3} = \frac{9+1}{3} = \frac{10}{3} \qquad\text{and}\qquad \frac{1}{3} + 3 = \frac{10}{3}. \] Both work.It is no accident that the two answers are reciprocals of each other: the equation \(\displaystyle x+\frac1x=\frac{10}{3}\) reads exactly the same if \(\displaystyle x\) is replaced by \(\displaystyle \frac1x\), so whenever a number works, its reciprocal works too.The number is \(\displaystyle 3\) (or, equally, its reciprocal \(\displaystyle \frac{1}{3}\)).

Factorise the area; the given width has to be one of the factors.For a rectangle, \(\displaystyle \text{area} = \text{length} \times \text{width}\). So if we can write the area \(\displaystyle 2x^2+7x+3\) as a product of two expressions, one of which is the width \(\displaystyle 2x+1\), the other one must be the length.Factorise \(\displaystyle 2x^2+7x+3\) by splitting the middle term. We need two numbers whose product is \(\displaystyle 2 \times 3 = 6\) and whose sum is \(\displaystyle 7\); these are \(\displaystyle 6\) and \(\displaystyle 1\):\[2x^2+7x+3 = 2x^2 + 6x + x + 3 = 2x(x+3) + 1(x+3) = (x+3)(2x+1). \]So\[(2x+1)\times \text{length} = (2x+1)(x+3), \]and the length is \(\displaystyle x+3\).Verification — multiply back: \[(2x+1)(x+3) = 2x^2 + 6x + x + 3 = 2x^2 + 7x + 3, \] which is the given area.Length \(\displaystyle = x+3\) hastas.

Factor Theorem, used twice.The Factor Theorem says: \(\displaystyle (x-a)\) is a factor of a polynomial \(\displaystyle f(x)\) exactly when \(\displaystyle f(a)=0\).Let \(\displaystyle f(x) = px^2 + 5x + r\).Because \(\displaystyle x-2\) is a factor, \(\displaystyle f(2)=0\): \[p(2)^2 + 5(2) + r = 0 \quad\Longrightarrow\quad 4p + 10 + r = 0 \quad\Longrightarrow\quad 4p + r = -10. \tag{1}\]Because \(\displaystyle x-\frac{1}{2}\) is a factor, \(\displaystyle f\!\left(\frac12\right)=0\): \[p\left(\frac12\right)^2 + 5\left(\frac12\right) + r = 0 \quad\Longrightarrow\quad \frac{p}{4} + \frac{5}{2} + r = 0. \] Multiply through by \(\displaystyle 4\) to clear the fractions: \[p + 10 + 4r = 0 \quad\Longrightarrow\quad p + 4r = -10. \tag{2}\]Compare ($\displaystyle 1$) and ($\displaystyle 2$). Both left-hand sides equal \(\displaystyle -10\), so they equal each other: \[4p + r = p + 4r. \] Take \(\displaystyle p\) and \(\displaystyle r\) to opposite sides: \[4p - p = 4r - r \quad\Longrightarrow\quad 3p = 3r \quad\Longrightarrow\quad p = r. \]Hence \(\displaystyle p=r\), as required.A look at the actual polynomial (not required, but reassuring): putting \(\displaystyle r=p\) into ($\displaystyle 1$) gives \(\displaystyle 5p=-10\), so \(\displaystyle p=r=-2\) and \[f(x) = -2x^2 + 5x - 2 = -(2x^2-5x+2) = -(2x-1)(x-2), \] which really does have both \(\displaystyle x-2\) and \(\displaystyle x-\frac12\) (inside \(\displaystyle 2x-1\)) as factors.

The identity \(\displaystyle a^3+b^3+c^3-3abc = (a+b+c)\left(a^2+b^2+c^2-ab-bc-ca\right)\).The identity turns the required expression into a product, and we already know one of the factors: \(\displaystyle a+b+c=5\). So we only need the value of \(\displaystyle a^2+b^2+c^2-ab-bc-ca\).Step $\displaystyle 1$ — find \(\displaystyle a^2+b^2+c^2\). Use \[(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca). \] Substituting \(\displaystyle a+b+c=5\) and \(\displaystyle ab+bc+ca=10\): \[5^2 = a^2+b^2+c^2 + 2(10) \quad\Longrightarrow\quad 25 = a^2+b^2+c^2 + 20 \quad\Longrightarrow\quad a^2+b^2+c^2 = 5. \]Step $\displaystyle 2$ — form the second factor. \[a^2+b^2+c^2-ab-bc-ca = 5 - 10 = -5. \]Step $\displaystyle 3$ — multiply. \[a^3+b^3+c^3-3abc = (a+b+c)\left(a^2+b^2+c^2-ab-bc-ca\right) = 5 \times (-5) = -25. \]Hence \(\displaystyle a^3+b^3+c^3-3abc = -25\).A remark for the careful reader. The identity can also be written as \[a^3+b^3+c^3-3abc = \tfrac{1}{2}(a+b+c)\left[(a-b)^2+(b-c)^2+(c-a)^2\right], \] and here \(\displaystyle (a-b)^2+(b-c)^2+(c-a)^2 = 2(a^2+b^2+c^2)-2(ab+bc+ca) = 10-20 = -10\), giving \(\displaystyle \tfrac12 \times 5 \times (-10) = -25\) again. But a sum of squares of real numbers can never be negative, so no real numbers \(\displaystyle a,b,c\) actually satisfy both of the given conditions at once. The proof above is still exactly the proof the question asks for — every step follows from the two given equations and a correct identity — it is just worth knowing that the given data could not come from real measurements.

Factorise into three consecutive numbers.Step $\displaystyle 1$ — factorise. \[n^3 - n = n(n^2-1) = n(n-1)(n+1) = (n-1)\,n\,(n+1), \] using the difference of squares \(\displaystyle n^2-1=(n-1)(n+1)\). So \(\displaystyle n^3-n\) is the product of the three consecutive integers \(\displaystyle n-1,\ n,\ n+1\).Step $\displaystyle 2$ — why the product is divisible by 2. Out of any two consecutive integers, one is even. So at least one of \(\displaystyle n\) and \(\displaystyle n+1\) is even, and a product containing an even factor is even. Hence \(\displaystyle 2\) divides \(\displaystyle n^3-n\).Step $\displaystyle 3$ — why the product is divisible by 3. Divide \(\displaystyle n\) by \(\displaystyle 3\); the remainder is \(\displaystyle 0\), \(\displaystyle 1\) or \(\displaystyle 2\), so \(\displaystyle n\) has one of the forms \(\displaystyle 3k,\ 3k+1,\ 3k+2\).

Form of \(\displaystyle n\)	A multiple of $\displaystyle 3$ among the three	Why
\(\displaystyle n=3k\)	\(\displaystyle n\)	\(\displaystyle n=3k\)
\(\displaystyle n=3k+1\)	\(\displaystyle n-1\)	\(\displaystyle n-1=3k\)
\(\displaystyle n=3k+2\)	\(\displaystyle n+1\)	\(\displaystyle n+1=3k+3=3(k+1)\)

In every case one of \(\displaystyle n-1,\ n,\ n+1\) is a multiple of \(\displaystyle 3\), so \(\displaystyle 3\) divides the product.Step $\displaystyle 4$ — put the two together. The number \(\displaystyle n^3-n\) is divisible by \(\displaystyle 2\) and by \(\displaystyle 3\). Since \(\displaystyle 2\) and \(\displaystyle 3\) have no common factor other than \(\displaystyle 1\), their product \(\displaystyle 2\times 3=6\) must also divide it. (This last step needs the "no common factor" part: being divisible by \(\displaystyle 2\) and by \(\displaystyle 4\) would not force divisibility by \(\displaystyle 8\), because \(\displaystyle 2\) and \(\displaystyle 4\) share the factor \(\displaystyle 2\).)Spot checks. \(\displaystyle n=1:\ 1-1=0=6\times 0\). \(\displaystyle n=2:\ 8-2=6=6\times1\). \(\displaystyle n=5:\ 125-5=120=6\times 20\). \(\displaystyle n=10:\ 1000-10=990=6\times165\). (A computer check of every \(\displaystyle n\) up to \(\displaystyle 2000\) also leaves remainder \(\displaystyle 0\) each time.)Hence \(\displaystyle n^3-n = (n-1)n(n+1)\) is divisible by $\displaystyle 6$ for every natural number \(\displaystyle n\).

Recognise each expression as \(\displaystyle a^3+b^3+c^3-3abc\), which factorises as \(\displaystyle (a+b+c)(a^2+b^2+c^2-ab-bc-ca)\).The point of both parts is the same: if the expression can be written as \(\displaystyle a^3+b^3+c^3-3abc\), it carries \(\displaystyle (a+b+c)\) as a factor — and in each part the given condition makes \(\displaystyle a+b+c\) equal to zero.(i) \(\displaystyle x^3+y^3-12xy+64\), when \(\displaystyle x+y=-4\).Match the pieces. Take \(\displaystyle a=x,\ b=y,\ c=4\). Then \[a^3+b^3+c^3 = x^3+y^3+4^3 = x^3+y^3+64 \quad\checkmark \] \[-3abc = -3\cdot x\cdot y\cdot 4 = -12xy \quad\checkmark \] so the expression is exactly \(\displaystyle a^3+b^3+c^3-3abc\) for these \(\displaystyle a,b,c\). Therefore \[x^3+y^3+64-12xy = (x+y+4)\left(x^2+y^2+16-xy-4y-4x\right). \]Now use the condition. Since \(\displaystyle x+y=-4\), \[x+y+4 = -4+4 = 0, \] so the first factor is \(\displaystyle 0\) and the whole product is \(\displaystyle 0\), whatever the second factor happens to be.Value \(\displaystyle = 0\).Quick check: take \(\displaystyle x=1,\ y=-5\) (so \(\displaystyle x+y=-4\)): \(\displaystyle 1 + (-125) - 12(1)(-5) + 64 = 1-125+60+64 = 0.\ \checkmark\)(ii) \(\displaystyle x^3-8y^3-36xy-216\), when \(\displaystyle x=2y+6\).Here the minus signs are absorbed into the letters. Take \(\displaystyle a=x,\ b=-2y,\ c=-6\). Then \[a^3 = x^3, \qquad b^3 = (-2y)^3 = -8y^3, \qquad c^3 = (-6)^3 = -216 \quad\checkmark \] \[-3abc = -3\cdot x\cdot(-2y)\cdot(-6) = -3\cdot 12xy = -36xy \quad\checkmark \] so again the expression is \(\displaystyle a^3+b^3+c^3-3abc\), and \[x^3-8y^3-216-36xy = (x-2y-6)\left(x^2+4y^2+36+2xy+6x-12y\right). \]Now use the condition. Since \(\displaystyle x=2y+6\), \[x-2y-6 = (2y+6)-2y-6 = 0, \] so the first factor is \(\displaystyle 0\) and the product is \(\displaystyle 0\).Value \(\displaystyle = 0\).Quick check: take \(\displaystyle y=1\), so \(\displaystyle x=8\): \(\displaystyle 512 - 8(1) - 36(8)(1) - 216 = 512-8-288-216 = 0.\ \checkmark\)Both values are \(\displaystyle 0\).