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NCERT Exemplar · Class 9 Mathematics Polynomials

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EXERCISE 2.4 1–9 (part 8 of 8)

  1. Exercise 1

    If the polynomials az3+4z2+3z−4\displaystyle a z^{3}+4 z^{2}+3 z-4 and z3−4z+a\displaystyle z^{3}-4 z+a leave the same remainder when divided by z−3\displaystyle z-3, find the value of a\displaystyle a.

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    NCERT’s answer
    -$\displaystyle 1$
    By the Remainder Theorem, the remainder on division by \(\displaystyle z-3\) is each polynomial's value at \(\displaystyle z=3\).\[az^{3}+4z^{2}+3z-4 \ \text{at } z=3 = 27a+41 \] \[z^{3}-4z+a \ \text{at } z=3 = a+15 \]Equal remainders:\[27a+41 = a+15 \] \[26a = -26 \] \[a = -1 \]Answer: \(\displaystyle a=-1\).

  2. Exercise 2

    The polynomial p(x)=x4−2x3+3x2−ax+3a−7\displaystyle p(x)=x^{4}-2 x^{3}+3 x^{2}-a x+3 a-7 when divided by x+1\displaystyle x+1 leaves the remainder 19. Find the values of a\displaystyle a. Also find the remainder when p(x)\displaystyle p(x) is divided by x+2\displaystyle x+2.

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    NCERT’s answer
    $\displaystyle a=5 ; 62$
    By the Remainder Theorem, \(\displaystyle p(-1)=19\).\[p(-1) = 1+2+3+a+3a-7 = 4a-1 \] \[4a-1=19 \] \[a=5 \]The remainder on division by \(\displaystyle x+2\) is \(\displaystyle p(-2)\):\[p(x)=x^{4}-2x^{3}+3x^{2}-5x+8 \] \[p(-2)=16+16+12+10+8=62 \]Answer: \(\displaystyle a=5\); remainder \(\displaystyle =62\).

  3. Exercise 3

    If both x−2\displaystyle x-2 and x−12\displaystyle x-\frac{1}{2} are factors of px2+5x+r\displaystyle p x^{2}+5 x+r, show that p=r\displaystyle p=r.

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    Let \(\displaystyle f(x)=px^2+5x+r\). By the Factor Theorem, \(\displaystyle f(2)=0\) and \(\displaystyle f\!\left(\tfrac12\right)=0\).\[f(2)=0 \implies 4p+10+r=0 \implies 4p+r=-10 \] \[f\!\left(\tfrac12\right)=0 \implies \tfrac{p}{4}+\tfrac52+r=0 \implies p+10+4r=0 \implies p+4r=-10 \]Subtracting:\[(4p+r)-(p+4r)=0 \] \[3p-3r=0 \] \[p=r \]Answer: \(\displaystyle p=r\).

  4. Exercise 4

    Without actual division, prove that 2x4−5x3+2x2−x+2\displaystyle 2 x^{4}-5 x^{3}+2 x^{2}-x+2 is divisible by x2−3x+2\displaystyle x^{2}-3 x+2. [Hint: Factorise x2−3x+2\displaystyle x^{2}-3 x+2 ]

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    \[x^{2}-3x+2=(x-1)(x-2) \]By the Factor Theorem, \(\displaystyle x-1\) and \(\displaystyle x-2\) divide \(\displaystyle p(x)=2x^{4}-5x^{3}+2x^{2}-x+2\) iff \(\displaystyle p(1)=0\) and \(\displaystyle p(2)=0\).\[p(1)=2-5+2-1+2=0 \] \[p(2)=32-40+8-2+2=0 \]\(\displaystyle x-1\) and \(\displaystyle x-2\) are coprime factors of \(\displaystyle p(x)\), so their product divides it:\[p(x) \text{ is divisible by } (x-1)(x-2)=x^{2}-3x+2 \]Answer: \(\displaystyle 2x^{4}-5x^{3}+2x^{2}-x+2\) is divisible by \(\displaystyle x^{2}-3x+2\).

  5. Exercise 5

    Simplify (2x−5y)3−(2x+5y)3\displaystyle (2 x-5 y)^{3}-(2 x+5 y)^{3}.

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    NCERT’s answer
    $\displaystyle -120 x^{2} y-250 y^{3}$
    Let \(\displaystyle a=2x-5y,\ b=2x+5y\).\[a^{3}-b^{3}=(a-b)\left(a^{2}+ab+b^{2}\right) \] \[a-b=-10y \] \[a^{2}+ab+b^{2}=12x^{2}+25y^{2} \] \[(2x-5y)^{3}-(2x+5y)^{3}=-10y\left(12x^{2}+25y^{2}\right) \] \[=-120x^{2}y-250y^{3} \]Answer: \(\displaystyle -120x^{2}y-250y^{3}\).

  6. Exercise 6

    Multiply x2+4y2+z2+2xy+xz−2yz\displaystyle x^{2}+4 y^{2}+z^{2}+2 x y+x z-2 y z by (−z+x−2y)\displaystyle (-z+x-2 y).

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    NCERT’s answer
    $\displaystyle x^{3}-8 y^{3}-z^{3}-6 x y z$
    Let \(\displaystyle a=x,\ b=-2y,\ c=-z\).\[a^{2}+b^{2}+c^{2}-ab-bc-ca = x^{2}+4y^{2}+z^{2}+2xy+xz-2yz \] \[a+b+c=x-2y-z \]\[a^{3}+b^{3}+c^{3}-3abc=(a+b+c)\left(a^{2}+b^{2}+c^{2}-ab-bc-ca\right) \]\[\left(x^{2}+4y^{2}+z^{2}+2xy+xz-2yz\right)(x-2y-z)=a^{3}+b^{3}+c^{3}-3abc \] \[=x^{3}-8y^{3}-z^{3}-6xyz \]Answer: \(\displaystyle x^{3}-8y^{3}-z^{3}-6xyz\).

  7. Exercise 7

    If a,b,c\displaystyle a, b, c are all non-zero and a+b+c=0\displaystyle a+b+c=0, prove that a2bc+b2ca+c2ab=3\displaystyle \frac{a^{2}}{b c}+\frac{b^{2}}{c a}+\frac{c^{2}}{a b}=3.

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    \[a+b+c=0 \] \[a^{3}+b^{3}+c^{3}-3abc=(a+b+c)\left(a^{2}+b^{2}+c^{2}-ab-bc-ca\right) \quad \text{(identity)} \] \[\Rightarrow a^{3}+b^{3}+c^{3}=3abc \quad \text{(since } a+b+c=0\text{)} \] \[\frac{a^{2}}{bc}+\frac{b^{2}}{ca}+\frac{c^{2}}{ab}=\frac{a^{3}+b^{3}+c^{3}}{abc} \quad \text{(common denominator } abc\text{)} \] \[=\frac{3abc}{abc}=3 \] Answer: \(\displaystyle 3\)

  8. Exercise 8

    If a+b+c=5\displaystyle a+b+c=5 and ab+bc+ca=10\displaystyle a b+b c+c a=10, then prove that a3+b3+c3−3abc=−25\displaystyle a^{3}+b^{3}+c^{3}-3 a b c=-25.

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    \[(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ca) \] \[5^{2}=a^{2}+b^{2}+c^{2}+2(10) \] \[a^{2}+b^{2}+c^{2}=25-20=5 \] \[a^{3}+b^{3}+c^{3}-3abc=(a+b+c)\left(a^{2}+b^{2}+c^{2}-ab-bc-ca\right) \quad \text{(identity)} \] \[=5(5-10) \] \[=5\times(-5)=-25 \] Answer: \(\displaystyle -25\)

  9. Exercise 9

    Prove that (a+b+c)3−a3−b3−c3=3(a+b)(b+c)(c+a)\displaystyle (a+b+c)^{3}-a^{3}-b^{3}-c^{3}=3(a+b)(b+c)(c+a).

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    Let \(\displaystyle t=b+c\). \[(a+b+c)^{3}=(a+t)^{3}=a^{3}+t^{3}+3at(a+t) \] \[\Rightarrow (a+b+c)^{3}-a^{3}-t^{3}=3at(a+b+c) \] \[t^{3}=(b+c)^{3}=b^{3}+c^{3}+3bc(b+c) \] Substituting: \[(a+b+c)^{3}-a^{3}-b^{3}-c^{3}=3bc(b+c)+3a(b+c)(a+b+c) \] \[=3(b+c)\left[bc+a(a+b+c)\right] \] \[=3(b+c)\left[a^{2}+ab+ac+bc\right] \] \[=3(b+c)\left[a(a+b)+c(a+b)\right] \] \[=3(b+c)(a+b)(a+c) \] \[=3(a+b)(b+c)(c+a) \] Answer: \(\displaystyle 3(a+b)(b+c)(c+a)\)