# Polynomials : Exercise 2.4 (Mathematics NCERT Class 9th) Q.1В В В В  Determine which of the following polynomials has (x + 1) a factor :
В В В В В В В В В В В В В  (i) ${x^3} + {x^2} + x + 1$
В В В В В В В В В В В В В  (ii) ${x^4} + {x^3} + {x^2} + x + 1$
В В В В В В В В В В В В В  (iii) ${x^4} + 3{x^3} + 3{x^2} + x + 1$
В В В В В В В В В В В В В  (iv) ${x^3} + {x^2} - \left( {2 + \sqrt 2 } \right)x + \sqrt 2$В
Sol.

(i) In order to prove that x + 1 is a factor of $p\left( x \right) = {x^3} + {x^2} + x + 1$, it is sufficient to show that $p\left( { - 1} \right) = 0$
Now, $p\left( { - 1} \right) = {\left( { - 1} \right)^3} + {\left( { - 1} \right)^2} + \left( { - 1} \right) + 1$
$= - 1 + 1 - 1 + 1 = 0$
Hence, (x + 1) is a factor of В $p\left( x \right) = {x^3} + {x^2} + x + 1$

(ii) In order to prove that (x + 1) is a factor of В $p\left( x \right) = {x^4} + {x^3} + {x^2} + x + 1$, it is sufficient to show that p (вЂ“1) = 0 .
Now, В $p\left( { - 1} \right) = {\left( { - 1} \right)^4} + {\left( { - 1} \right)^3} + {\left( { - 1} \right)^2} + \left( { - 1} \right) + 1$
$= 1 - 1 + 1 - 1 + 1 = 1 \ne 0$
Therefore (x + 1) is not a factor of В  ${x^4} + {x^3} + {x^2} + x + 1$

(iii) In order to prove that (x + 1) is a factor of $p\left(x\right) = {x^4} + 3{x^3} + 3{x^2} + x + 1$, it is sufficient to show that p(вЂ“1) = 0.
Now, $p\left({ - 1}\right) = {\left({ - 1}\right)^4} + 3{\left( { - 1}\right)^3} + 3{\left({- 1}\right)^2} +\left({- 1}\right)+1$
$= 1 - 3 + 3 - 1 + 1 = 1 \ne 0$
Therefore (x + 1) is not a factor of В  ${x^4} + 3{x^3} + 3{x^2} + x + 1$.

(iv) In order to prove that (x + 1) is a factor of В $p\left( x \right) = {x^3} - {x^2} - \left( {2 + \sqrt 2 } \right)x + \sqrt 2$, В it is sufficient to show that p(-1)= 0
$p\left( { - 1} \right) = {\left( { - 1} \right)^3} - {\left( { - 1} \right)^2} - \left( {2 + \sqrt 2 } \right)\left( { - 1} \right) + \sqrt 2$
$= - 1 - 1 + 2 + \sqrt 2 + \sqrt 2 = 2\sqrt 2 \ne 0$
Therefore В  (x + 1) is not a factor of В  ${x^3} - {x^2} - \left( {2 + \sqrt 2 } \right)x + \sqrt 2$

Q.2В В В В В  Use the factor theorem to determine whether g(x) is a factor of p (x) in each of the following cases :
В В В В В В В В В В В
(i) $p\left( x \right) = 2{x^3} + {x^2} - 2x - 1,\,g\left( x \right) = x + 1$
В В В В В В В В В В В
(ii) $p\left( x \right) = {x^3} + 3{x^2} + 3x + 1,\,g\left( x \right) = x + 2$
В В В В В В В В В В В
(iii) $p\left( x \right) = {x^3} - 4{x^2} + x + 6,\,g\left( x \right) = x - 3$
Sol.

(i) In order to prove that g(x) = x + 1 is a factor of $p\left( x \right) = 2{x^3} + {x^2} - 2x - 1$, it is sufficient to show that p(вЂ“1) = 0.
Now, $p\left( { - 1} \right) = 2{\left( { - 1} \right)^3} + {\left( { - 1} \right)^2} - 2\left( { - 1} \right) - 1$
$= - 2 + 1 + 2 - 1 = 0$
Therefore g(x) is a factor of p(x).

(ii) In order to prove that g(x) = x + 2 is a factor of В $p\left( x \right) = {x^3} + 3{x^2} + 3x + 1$, it is sufficient to show that p(вЂ“2) = 0
Now, $p\left( { - 2} \right) = {\left( { - 2} \right)^3} + 3{\left( { - 2} \right)^2} + 3\left( { - 2} \right) + 1$
$= - 8 + 12 - 6 + 1$
$= - 1 \ne 0$
Therefore g(x) is not a factor of p(x) .

(iii) In order to prove that g(x) = x вЂ“ 3 is a factor of $p\left( x \right) = {x^3} - 4{x^2} + x + 6$. It is sufficient to show that p(+3) = 0
Now, В $p\left( 3 \right) = {\left( 3 \right)^3} - 4{\left( 3 \right)^2} + 3 + 6$
$= 27 - 36 + 3 + 6$
36 - 36 = 0
Therefore g (x) is В a factor of p(x).

Q.3В В В В В  Find the value of k, if x -1 is a factor of p (x) in each of the following cases :
В В В В В В В В В В В
(i) $p\left( x \right) = {x^2} + x + k$
В В В В В В В В В В В
(ii) $p\left( x \right) = 2{x^2} + kx + \sqrt 2$
В В В В В В В В В В В
(iii) $p\left( x \right) = k{x^2} - \sqrt 2 x + 1$
В В В В В В В В В В В
(iv) $p\left( x \right) = k{x^2} - 3x + k$
Sol.

(i) If (x -1) is a factor of $p\left( x \right) = {x^2} + x + k$, then
$p\left( 1 \right) = 0$
$\Rightarrow$ ${\left( 1 \right)^2} + 1 + k = 0$
$\Rightarrow$ $1 + 1 + k = 0$
$\Rightarrow$ $k = - 2$
Hence, k = -2

(ii) If (x - 1) is a factor of В  $p\left( x \right) = 2{x^2} + kx + \sqrt 2$ , then
$p\left( 1 \right) = 0$
$\Rightarrow$ $2{\left( 1 \right)^2} + k\left( 1 \right) + \sqrt 2 = 0$
$\Rightarrow$ $2 + k + \sqrt 2 = 0$
$\Rightarrow$ $k = - \left( {2 + \sqrt 2 } \right)$

(iii) If (x -1) is a factor of В $p\left( x \right) = k{x^2} - \sqrt 2 x + 1$, В then
$p\left( 1 \right) = 0$
$\Rightarrow$ $k{\left( 1 \right)^2} - \sqrt 2 \left( 1 \right) + 1 = 0$
$\Rightarrow$ $k - \sqrt 2 + 1 = 0$
$\Rightarrow$ $k = \sqrt 2 - 1$

(iv) If (x -1) is a factor of $p\left( x \right) = k{x^2} - 3x + k$, then
$p\left( 1 \right) = 0$
$\Rightarrow$ $k{\left( 1 \right)^2} - 3\left( 1 \right) + k = 0$
$\Rightarrow$ $k - 3 + k = 0$
$\Rightarrow$ $2k = 3$
$\Rightarrow$ $k = {3 \over 2}$

Q.4В В В В В  Factorise :
В В В В В В В В В В В В В В  (i) $12{x^2} - 7x + 1$ В  В  В  В  В  В  В  В  В  В  В  В  В  В
В В В В В В В В В В В  (ii) $2{x^2} + 7x + 3$

В В В В В В В В В В В В В В  (iii) $6{x^2} + 5x - 6$ В  В  В  В  В  В  В  В  В  В  В  В  В
В  В  В  В В В В В  (iv) $3{x^2} - x - 4$

Sol.

(i) Here p + q = coeff. of x = - 7
pq = coeff. of ${x^2} \times$ constant term
= 12 Г— 1 = 12
Therefore p + q = вЂ“ 7 = вЂ“ 4 вЂ“ 3
and pq = 12 = (вЂ“ 4)(вЂ“3)
Therefore $12{x^2} - 7x + 1 = 12{x^2} - 4x - 3x + 1$
$= 4x\left( {3x - 1} \right) - 1\left( {3x - 1} \right)$
$= \left( {3x - 1} \right)\left( {4x - 1} \right)$

(ii) Here p + q = coeff. of x = 7
pq = coeff. of ${x^2} \times$ constant term В
= 2 Г— 3 = 6
Therefore p + q = 7 = 1 + 6
and pq = 6 = 1 Г— 6
Therefore $2{x^2} + 7x + 3 = 2{x^2} + x + 6x + 3$
$= x\left( {2x + 1} \right) + 3\left( {2x + 1} \right)$
$= \left( {2x + 1} \right)\left( {x + 3} \right)$

(iii) Here p + q = coeff. of x = 5
pq = coeff. of ${x^2} \times$ constant term
= 6 Г— (вЂ“ 6)В  = вЂ“ 36
Therefore p + q = 5 = 9 + (вЂ“ 4)
and pq = вЂ“ 36 = 9 Г— (вЂ“ 4)
Therefore $6 {x^2} + 5x - 6 = 6{x^2} + 9x - 4x - 6$
$= 3x\left( {2x + 3} \right) - 2\left( {2x + 3} \right)$
$= \left( {2x + 3} \right)\left( {3x - 2} \right)$

(iv) Here p + q = coeff. of x = вЂ“1
pq = coeff. of ${x^2} \times$ constant term
= 3 Г— (вЂ“ 4) = вЂ“ 12
Therefore p + q = вЂ“1 = 3 + (вЂ“ 4)
and pq = вЂ“ 12 = 3 Г— (вЂ“4)
Therefore $3{x^2} - x - 4 = 3{x^2} + 3x - 4x - 4$
$= 3x\left ({x + 1} \right) - 4\left( {x + 1} \right)$
$= \left( {x + 1} \right)\left( {3x - 4} \right)$

Q.5В В В В В  Factorise :
В В В В В В В В В В В В В В  (i) ${x^3} - 2{x^2} - x + 2$ В  В  В  В  В  В  В  В  В  В  В  В  В  В  В  В
В В В В В В В В В В В  (ii) ${x^3} - 3{x^2} - 9x - 5$
В В В В В В В В В В В
(iii) ${x^3} + 13{x^2} + 32x + 20$ В  В  В  В  В  В  В  В
В В В В В В В В В В В  (iv) $2{y^3} + {y^2} - 2y - 1$

Sol.В

(i) Let $f\left(x \right) = {x^3} - 2{x^2} - x + 2$
The constant term in f(x) is + 2 and factors of + 2 areВ $\pm 1, \pm 2$
Putting x = 1 in f(x),we haveВ
$f\left(1\right) = {\left(1\right)^3} - 2{\left(1\right)^2} - 1 + 2$
$= 1 - 2 - 1 + 2 = 0$
Therefore (xвЂ“1) is a factor of f(x) .В

putting x = вЂ“ 1 in f(x) , we haveВ
$f\left( { - 1} \right) = {\left( { - 1} \right)^3} - 2{\left( { - 1} \right)^2} - \left( { - 1} \right) + 2$В
$= - 1 - 2 + 1 + 2 = 0$В
Therefore x + 1 is a factor of f (x).В
Putting x = 2 in f(x), we haveВ
$f\left( 2 \right) = {\left( 2 \right)^3} - 2{\left( 2 \right)^2} - \left( 2 \right) + 2$
$= 8 - 8 - 2 + 2 = 0$
Therefore (x + 2) is a factor of f(x)

Putting x = вЂ“ 2 in f(x), we have
$f\left( { - 2} \right) = {\left( { - 2} \right)^3} - 2{\left( { - 2} \right)^2} - \left( { - 2} \right) + 2$
$= - 8 - 8 + 2 + 2$
$= - 12 \ne 0$
Therefore x - 2 is not a factor of f (x).
Therefore The factors of f(x) are (x вЂ“ 1), (x + 1) and (x вЂ“ 2).
Let f(x) = k (x вЂ“ 1) (x + 1) (x вЂ“ 2)
$\Rightarrow$ ${x^3} - 2{x^2} - x + 2 = k\left( {x - 1} \right)\left( {x + 1} \right)\left( {x - 2} \right)$
Putting x = 0 on both sides, we have
2 = k (вЂ“1) (1) (вЂ“2) $\Rightarrow$ k = 1
Therefore ${x^3} - 2{x^2} - x + 2 = \left( {x - 1} \right)\left( {x + 1} \right)\left( {x - 2} \right)$

(ii) Let $p\left( x \right) = {x^3} - 3{x^2} - 9x - 5$
We shall look for all factors of вЂ“ 5. Therefore are $\pm 1, \pm 5.$

By trial, we find p (вЂ“1) = вЂ“ 1 вЂ“ 3 + 9 вЂ“ 5 = 0 . So (x + 1) is a factor of p(x) .
Now, divide p (x) by (x + 1) Therefore $p\left( x \right) = \left( {x + 1} \right)\left( {{x^2} - 4x - 5}\right)$
$= \left( {x + 1} \right)\left( {{x^2} + x - 5x - 5} \right)$
$= \left( {x + 1} \right)\left[ {x\left( {x + 1} \right) - 5\left( {x + 1} \right)} \right]$
$= \left( {x + 1} \right)\,\left( {x + 1} \right)\left( {x - 5} \right)$
Therefore ${x^3} - 2{x^2} - x + 2 = \left( {x + 1} \right)\left( {x + 1} \right)\left( {x - 5} \right)$

(iii) Let $p\left( x \right) = {x^3} + 13{x^2} + 32x + 20$
We shall look for all factors of + 20, these are $\pm 1, \pm 2,\, \pm 4, \pm 5, \pm 10\,and\, \pm 20$
By trial , we findВ
$p( - 2) = ( - 2)^3+ 13(2)^2+ 32( - 2) + 20$
p(вЂ“ 2) = вЂ“ 8 + 52 вЂ“ 64 + 20 = 0В
Therefore (x + 2) is a factor of p(x)В
Now, divide p (x) by x + 2 Therefore В $p\left( x \right) = \left( {x + 2} \right)\left( {{x^2} + 11x + 10} \right)$
$= \left( {x + 2} \right)\left( {{x^2} + x + 10x + 10} \right)$
$= \left( {x + 2} \right)\left[ {x\left( {x + 1} \right) + 10\left( {x + 1} \right)} \right]$
$= \left( {x + 2} \right)\,\left( {x + 1} \right)\left( {x + 10} \right)$

(iv) Let $p\left( y \right) = 2{y^3} + {y^2} - 2y - 1$
By trial we find p(1) = 2 + 1 вЂ“ 2 вЂ“ 1 = 0В
So, (y вЂ“ 1) is a factor of p(y)
Now, divide p(y) by y вЂ“ 1 Therefore $p\left( y \right) = \left( {y - 1} \right)\left( {2{y^2} + 3y + 1} \right)$
$= \left( {y - 1} \right)\left( {2{y^2} + 2y + y + 1} \right)$
$= \left( {y - 1} \right)\left[ {2y\left( {y + 1} \right) + 1\left( {y + 1} \right)} \right]$
$= \left( {y - 1} \right)\,\left( {y + 1} \right)\left( {2y + 1} \right)$

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