NCERT Solutions · Class 9 Mathematics Exploring Algebraic Identities

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) \(\displaystyle (8u-11v-2w)^{2}\).