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NCERT Exemplar · Class 9 Mathematics Polynomials

72 questions · 72 still being checked

EXERCISE 2.3 21–30 (part 6 of 8)

  1. Exercise 21

    Find the value of m\displaystyle m so that 2x−1\displaystyle 2 x-1 be a factor of 8x4+4x3−16x2+10x+m\displaystyle 8 x^{4}+4 x^{3}-16 x^{2}+10 x+m.

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    NCERT’s answer
    -$\displaystyle 2$
    \(\displaystyle 2x-1\) is a factor of \(\displaystyle f(x)=8x^{4}+4x^{3}-16x^{2}+10x+m\) exactly when \(\displaystyle f\!\left(\tfrac{1}{2}\right)=0\). \[f\!\left(\tfrac{1}{2}\right) = 8\cdot\frac{1}{16}+4\cdot\frac{1}{8}-16\cdot\frac{1}{4}+10\cdot\frac{1}{2}+m \] \[= \frac{1}{2}+\frac{1}{2}-4+5+m = 2+m \] \[2+m = 0 \implies m = -2 \] Answer: \(\displaystyle m = -2\).

  2. Exercise 22

    If x+1\displaystyle x+1 is a factor of ax3+x2−2x+4a−9\displaystyle a x^{3}+x^{2}-2 x+4 a-9, find the value of a\displaystyle a.

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    NCERT’s answer
    $\displaystyle 2$
    \(\displaystyle x+1\) is a factor of \(\displaystyle f(x)=ax^{3}+x^{2}-2x+4a-9\) exactly when \(\displaystyle f(-1)=0\). \[f(-1) = a(-1)^{3}+(-1)^{2}-2(-1)+4a-9 = -a+1+2+4a-9 = 3a-6 \] \[3a-6 = 0 \implies a = 2 \] Answer: \(\displaystyle a = 2\).

  3. Exercise 23

    Factorise : (i) x2+9x+18\displaystyle x^{2}+9 x+18 (ii) 6x2+7x−3\displaystyle 6 x^{2}+7 x-3 (iii) 2x2−7x−15\displaystyle 2 x^{2}-7 x-15 (iv) 84−2r−2r2\displaystyle 84-2 r-2 r^{2}

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    NCERT’s answer
    (i)
    $\displaystyle (x+6)(x+3)$ (ii) $\displaystyle (3 x-1)(2 x+3)$ (iii) $\displaystyle (x-5)(2 x+3)$ (iv) $\displaystyle 2(7+r)(6-r)$
    (i) Split the middle term using \(\displaystyle 6\) and \(\displaystyle 3\). \[x^{2}+9x+18 = x^{2}+6x+3x+18 = x(x+6)+3(x+6) = (x+6)(x+3) \] (ii) Split the middle term using \(\displaystyle 9\) and \(\displaystyle -2\). \[6x^{2}+7x-3 = 6x^{2}+9x-2x-3 = 3x(2x+3)-(2x+3) = (3x-1)(2x+3) \] (iii) Split the middle term using \(\displaystyle -10\) and \(\displaystyle 3\). \[2x^{2}-7x-15 = 2x^{2}-10x+3x-15 = 2x(x-5)+3(x-5) = (2x+3)(x-5) \] (iv) Take out \(\displaystyle -2\), then split the middle term using \(\displaystyle 7\) and \(\displaystyle -6\). \[84-2r-2r^{2} = -2\left(r^{2}+r-42\right) = -2\left(r^{2}+7r-6r-42\right) \] \[= -2\big[r(r+7)-6(r+7)\big] = -2(r+7)(r-6) = 2(7+r)(6-r) \] Answer: (i) \(\displaystyle (x+6)(x+3)\) (ii) \(\displaystyle (3x-1)(2x+3)\) (iii) \(\displaystyle (2x+3)(x-5)\) (iv) \(\displaystyle 2(7+r)(6-r)\).

  4. Exercise 24

    Factorise : (i) 2x3−3x2−17x+30\displaystyle 2 x^{3}-3 x^{2}-17 x+30 (ii) x3−6x2+11x−6\displaystyle x^{3}-6 x^{2}+11 x-6 (iii) x3+x2−4x−4\displaystyle x^{3}+x^{2}-4 x-4 (iv) 3x3−x2−3x+1\displaystyle 3 x^{3}-x^{2}-3 x+1

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    NCERT’s answer
    (i)
    $\displaystyle (x-2)(x+3)(2 x-5)$ (ii) $\displaystyle (x-1)(x-2)(x-3)$ (iii) $\displaystyle (x+1)(x-2)(x+2)$ (iv) $\displaystyle (x-1)(x+1)(3 x-1)$
    (i) By the Factor Theorem, test \(\displaystyle x=2\). \[p(2) = 2(2)^{3}-3(2)^{2}-17(2)+30 = 16-12-34+30 = 0 \] \[2x^{3}-3x^{2}-17x+30 = (x-2)\left(2x^{2}+x-15\right) \] \[2x^{2}+x-15 = 2x^{2}+6x-5x-15 = (2x-5)(x+3) \] \[2x^{3}-3x^{2}-17x+30 = (x-2)(2x-5)(x+3) \] (ii) Test \(\displaystyle x=1\). \[p(1) = 1-6+11-6 = 0 \] \[x^{3}-6x^{2}+11x-6 = (x-1)\left(x^{2}-5x+6\right) = (x-1)(x-2)(x-3) \] (iii) Test \(\displaystyle x=-1\). \[p(-1) = -1+1+4-4 = 0 \] \[x^{3}+x^{2}-4x-4 = (x+1)\left(x^{2}-4\right) = (x+1)(x-2)(x+2) \] (iv) Test \(\displaystyle x=1\). \[p(1) = 3-1-3+1 = 0 \] \[3x^{3}-x^{2}-3x+1 = (x-1)\left(3x^{2}+2x-1\right) = (x-1)(3x-1)(x+1) \] Answer: (i) \(\displaystyle (x-2)(2x-5)(x+3)\) (ii) \(\displaystyle (x-1)(x-2)(x-3)\) (iii) \(\displaystyle (x+1)(x-2)(x+2)\) (iv) \(\displaystyle (x-1)(3x-1)(x+1)\)

  5. Exercise 25

    Using suitable identity, evaluate the following: (i) 1033\displaystyle 103^{3} (ii) 101\displaystyle 101 × 102\displaystyle 102 (iii) 9992\displaystyle 999^{2}

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    NCERT’s answer
    (i)
    $\displaystyle 1092727$ (ii) $\displaystyle 10302$ (iii) $\displaystyle 998001$ $\displaystyle 26 \cdot$ (i) $\displaystyle (2 x+5)^{2}$ (ii) $\displaystyle (3 y-11 z)^{2}$ (iii) $\displaystyle \left(3 x-\frac{1}{6}\right)\left(x+\frac{5}{6}\right)$
    \[\text{(i) } 103^{3}=(100+3)^{3}=100^{3}+3(100)^{2}(3)+3(100)(3)^{2}+3^{3}=1092727 \] \[\text{(ii) } 101\times102=(100+1)(100+2)=100^{2}+(1+2)(100)+1\times2=10302 \] \[\text{(iii) } 999^{2}=(1000-1)^{2}=1000^{2}-2(1000)(1)+1^{2}=998001 \]Answer: (i) \(\displaystyle 1092727\) (ii) \(\displaystyle 10302\) (iii) \(\displaystyle 998001\)

  6. Exercise 26

    Factorise the following: (i) 4x2+20x+25\displaystyle 4 x^{2}+20 x+25 (ii) 9y2−66yz+121z2\displaystyle 9 y^{2}-66 y z+121 z^{2} (iii) (2x+13)2−(x−12)2\displaystyle \left(2 x+\frac{1}{3}\right)^{2}-\left(x-\frac{1}{2}\right)^{2}

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    \[\text{(i) } 4x^{2}+20x+25=(2x)^{2}+2(2x)(5)+5^{2}=(2x+5)^{2} \] \[\text{(ii) } 9y^{2}-66yz+121z^{2}=(3y)^{2}-2(3y)(11z)+(11z)^{2}=(3y-11z)^{2} \] \[\text{(iii) } \left(2x+\frac13\right)^{2}-\left(x-\frac12\right)^{2}=\left[\left(2x+\frac13\right)+\left(x-\frac12\right)\right]\left[\left(2x+\frac13\right)-\left(x-\frac12\right)\right] \] \[=\left(3x-\frac16\right)\left(x+\frac56\right) \]Answer: (i) \(\displaystyle (2x+5)^{2}\) (ii) \(\displaystyle (3y-11z)^{2}\) (iii) \(\displaystyle \left(3x-\frac16\right)\left(x+\frac56\right)\)

  7. Exercise 27

    Factorise the following: (i) 9x2−12x+3\displaystyle 9 x^{2}-12 x+3 (ii) 9x2−12x+4\displaystyle 9 x^{2}-12 x+4

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    NCERT’s answer
    (i)
    $\displaystyle 3(x-1)(3 x-1)$ (ii) $\displaystyle (3 x-2)(3 x-2)$
    \[\text{(i) } 9x^{2}-12x+3=3(3x^{2}-4x+1)=3(3x-1)(x-1) \] \[\text{(ii) } 9x^{2}-12x+4=(3x)^{2}-2(3x)(2)+2^{2}=(3x-2)^{2} \]Answer: (i) \(\displaystyle 3(3x-1)(x-1)\) (ii) \(\displaystyle (3x-2)^{2}\)

  8. Exercise 28

    Expand the following: (i) (4a−b+2c)2\displaystyle (4 a-b+2 c)^{2} (ii) (3a−5b−c)2\displaystyle (3 a-5 b-c)^{2} (iii) (−x+2y−3z)2\displaystyle (-x+2 y-3 z)^{2}

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    NCERT’s answer
    (i)
    $\displaystyle 16 a^{2}+b^{2}+4 c^{2}-8 a b-4 b c+16 a c$ (ii) $\displaystyle 9 a^{2}+25 b^{2}+c^{2}-30 a b+10 b c-6 a c$ (iii) $\displaystyle x^{2}+4 y^{2}+9 z^{2}-4 x y-12 y z+6 x z$
    \[(p+q+r)^{2}=p^{2}+q^{2}+r^{2}+2pq+2qr+2rp \] (i) \(\displaystyle p=4a,\ q=-b,\ r=2c\) \[(4a-b+2c)^{2}=(4a)^{2}+(-b)^{2}+(2c)^{2}+2(4a)(-b)+2(-b)(2c)+2(2c)(4a) \] \[=16a^{2}+b^{2}+4c^{2}-8ab-4bc+16ac \] (ii) \(\displaystyle p=3a,\ q=-5b,\ r=-c\) \[(3a-5b-c)^{2}=(3a)^{2}+(-5b)^{2}+(-c)^{2}+2(3a)(-5b)+2(-5b)(-c)+2(-c)(3a) \] \[=9a^{2}+25b^{2}+c^{2}-30ab+10bc-6ac \] (iii) \(\displaystyle p=-x,\ q=2y,\ r=-3z\) \[(-x+2y-3z)^{2}=(-x)^{2}+(2y)^{2}+(-3z)^{2}+2(-x)(2y)+2(2y)(-3z)+2(-3z)(-x) \] \[=x^{2}+4y^{2}+9z^{2}-4xy-12yz+6xz \] Answer: (i) \(\displaystyle 16a^{2}+b^{2}+4c^{2}-8ab-4bc+16ac\) (ii) \(\displaystyle 9a^{2}+25b^{2}+c^{2}-30ab+10bc-6ac\) (iii) \(\displaystyle x^{2}+4y^{2}+9z^{2}-4xy-12yz+6xz\)

  9. Exercise 29

    Factorise the following: (i) 9x2+4y2+16z2+12xy−16yz−24xz\displaystyle 9 x^{2}+4 y^{2}+16 z^{2}+12 x y-16 y z-24 x z (ii) 25x2+16y2+4z2−40xy+16yz−20xz\displaystyle 25 x^{2}+16 y^{2}+4 z^{2}-40 x y+16 y z-20 x z (iii) 16x2+4y2+9z2−16xy−12yz+24xz\displaystyle 16 x^{2}+4 y^{2}+9 z^{2}-16 x y-12 y z+24 x z

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    NCERT’s answer
    (i)
    $\displaystyle (3 x+2 y-4 z)(3 x+2 y-4 z)$ (ii) $\displaystyle (-5 x+4 y+2 z)(-5 x+4 y+2 z)$ (iii) $\displaystyle (4 x-2 y+3 z)(4 x-2 y+3 z)$
    \[\text{(i) } 9x^{2}+4y^{2}+16z^{2}+12xy-16yz-24xz=(3x)^{2}+(2y)^{2}+(-4z)^{2}+2(3x)(2y)+2(2y)(-4z)+2(-4z)(3x)=(3x+2y-4z)^{2} \] \[\text{(ii) } 25x^{2}+16y^{2}+4z^{2}-40xy+16yz-20xz=(5x)^{2}+(-4y)^{2}+(-2z)^{2}+2(5x)(-4y)+2(-4y)(-2z)+2(-2z)(5x)=(5x-4y-2z)^{2} \] \[\text{(iii) } 16x^{2}+4y^{2}+9z^{2}-16xy-12yz+24xz=(4x)^{2}+(-2y)^{2}+(3z)^{2}+2(4x)(-2y)+2(-2y)(3z)+2(3z)(4x)=(4x-2y+3z)^{2} \]Answer: (i) \(\displaystyle (3x+2y-4z)^{2}\) (ii) \(\displaystyle (5x-4y-2z)^{2}\) (iii) \(\displaystyle (4x-2y+3z)^{2}\)

  10. Exercise 30

    If a+b+c=9\displaystyle a+b+c=9 and ab+bc+ca=26\displaystyle a b+b c+c a=26, find a2+b2+c2\displaystyle a^{2}+b^{2}+c^{2}.

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    NCERT’s answer
    $\displaystyle 29$
    \[(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ca) \] \[9^{2}=a^{2}+b^{2}+c^{2}+2(26) \] \[81=a^{2}+b^{2}+c^{2}+52 \] \[a^{2}+b^{2}+c^{2}=29 \]Answer: \(\displaystyle 29\)