Factorize each of the following polynomials:

Solution:

1. $x^{3}+13 x^{2}+31 x-45$ given that $x+9$ is a factor

let, $f(x)=x^{3}+13 x^{2}+31 x-45$

given that (x + 9) is the factor

divide f(x) with (x + 9) to get other factors

by , long division

$x^{2}+4 x-5$

$x+9 x^{3}+13 x^{2}+31 x-45$

$x^{3}+9 x^{2}$

(-)       (-)

$4 x^{2}+31 x$

$4 x^{2}+36 x$

(-)        (-)

-5x – 45

-5x – 45

(+)       (+)

0

$=x^{3}+13 x^{2}+31 x-45=(x+9)\left(x^{2}+4 x-5\right)$

Now,

$x^{2}+4 x-5=x^{2}+5 x-x-5$

= x(x + 5) -1(x + 5)

= (x + 5) (x – 1) are the factors

Hence, $x^{3}+13 x^{2}+31 x-45=(x+9)(x+5)(x-1)$

2. $4 x^{3}+20 x^{2}+33 x+18$ given that $2 x+3$ is a factor

let, $f(x)=4 x^{3}+20 x^{2}+33 x+18$

given that 2x + 3 is a factor

divide f(x) with (2x + 3) to get other factors

by, long division

$2 x^{2}+7 x+6$

$2 x+3,4 x^{3}+20 x^{2}+33 x+18$

$4 x^{3}+6 x^{2}$

(-)      (-)

$14 x^{2}-33 x$

$14 x^{2}-21 x$

(-)         (+)

12x + 18

12x + 18

(-)     (-)

0

$=>4 x^{3}+20 x^{2}+33 x+18=(2 x+3)\left(2 x^{2}+7 x+6\right)$

Now,

$2 x^{2}+7 x+6=2 x^{2}+4 x+3 x+6$

= 2x(x + 2) + 3(x + 2)

= (2x + 3)(x + 2) are the factors

Hence, $4 x^{3}+20 x^{2}+33 x+18=(2 x+3)(2 x+3)(x+2)$