Question

factorise and find zeroes of 2x³+7x²-3x-18​

Answer 1

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-182

$Applying \: rational \: root \: \\ theorem, \: which \: state \: that \\ possible \: roots \: are \: in \: form \pm\frac{p}{q}$

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=-122

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.

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

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

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

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