Question

Using factor theorem, factorize each of the following polynomial:
x³ +13x²+32x+20

Answer 1

$\large{\mathfrak{Answer:}}$

$\sf{Factors \:of\: 20 = \pm1, \pm2, \pm4, \pm5, \pm10, \pm20}$

Since all the signs in the given polynomial are positive, the factor must be in negative.

When x = -2:

=> (-2)^3 + 13(-2)^2 + 32(-2) + 20

=> -8 + 13(4) - 64 + 20

=> 0

Thus, x = -2

=> x + 2 = 0 is a factor of p(x)

On dividing (x + 2) by p(x), we get:

x^2 + 11x + 10

=> x^2 + 10x + x + 10

=> x(x + 10) + 1(x + 10)

=> (x + 1)(x + 10)

Thus, x + 1 and x + 10 are also the factors of p(x).

x^3 + 13x^2 + 32x + 20 = (x + 2)(x + 1)(x + 10)

_______________

Answer 2

$\underline \mathbb{QUESTION}$

Using factor theorem, factorize each of the following polynomial:

x³ +13x²+32x+20

$\underline \mathbb{SOLUTION}$

$\bold\green{\underline{Let,}}$

$p(x ) = {x}^{3} + 13 {x}^{2} + 32x + 20$

》By trial, we find

$p( - 1) = ( { - 1})^{3} + 13( - 1) {}^{2} + 32( - 1) + 20$

$= > - 1 + 13 - 32 + 20 = 0$

$\bold{\fbox{\color{Red}{By\:factor\:theorem,}}}$

》x-(-1) , (x+1) is a factor of p(x).

${x}^{3} + 13 {x}^{2} + 32 + 20$

$= > {x}^{2} (x + 1) + 12(x)( x + 1) + 20(x + 1)$

$= > (x + 1)( {x}^{2} + 12x + 20)$

$= > (x + 1)( {x}^{2} + 2x + 10x + 20)$

$= > (x + 1){x(x + 2) + 10(x + 2)}$

$= > (x + 1)(x + 2)(x + 10)$

_______________________________________