Question

Factorise : 2x³+7x²-3x-18

Answer 1

2x³ + 7x² - 3x - 18
put x = -2
we get 
2(-2)³ + 7(-2)²-3(-2)-18
= -16+28+6-18
= 34-34 = 0

hence x = -2 is root of 2x + 7x - 3x - 18

hence
2x³ + 7x² - 3x - 18 = (x+2)(2x²+3x-9)
=(x+2)(2x²+6x-3x-9)
=(x+2){2x(x+3)-3(x+3)}
=2x³+7x²-3x-18 = (x+2) (x-3) (2x-3)

Answer 2

Answer:

$2x^3+7x^2-3x-18=(x+2)(x+3)(x-\frac{3}{2})$    

Step-by-step explanation:

Given : Expression  $2x^3+7x^2-3x-18$

To find : Factorise the expression?

Solution :

Expression $2x^3+7x^2-3x-18$

Applying rational root theorem, which state that possible roots are in form $\pm\frac{p}{q}$ where p is the factor of coefficient of higher degree and q is the factor of constant.

Possible roots are $\pm(1,\frac{1}{2},\frac{1}{3},2,\frac{2}{3})$

Substitute x=1 in equation,

$2x^3+7x^2-3x-18=2+7-3-18=-12$

So, x=1 is not a root.

Put x=-2,

$2x^3+7x^2-3x-18=2\times (-8)+7\times 4-3\times (-2)-18=0$

So, x=-2 is one of the root.

Put x=-3

$2x^3+7x^2-3x-18=2\times (-27)+7\times 9-3\times (-3)-18=0$

So, x=-3 is one of the root.

Put $x=\frac{3}{2}$

$2x^3+7x^2-3x-18=2\times (\frac{27}{8})+7\times (\frac{9}{4})-3\times (\frac{3}{2})-18=0$

So, $x=\frac{3}{2}$ is one of the root.

Therefore, The factors of given expression is $(x+2)(x+3)(x-\frac{3}{2})$