Question

factorisation of
$2x {}^{3} + 7x {}^{2} - 9$
​

Answer 1

•Solution•

=2x³+7x²-9

=2x³-2x²+9x²-9x+9x-9

=2x²(x-1)+9x(x-1)+9(x-1)

=(x-1)(2x²+9x+9)

=(x-1)[2x²+6x+3x+9]

=(x-1)[2x(x+3)+3(x+3)]

=(x-1)(x+3)(2x+3)

clarification[how did I chose

(x-1)

in the above expression

put.x=....-3,-2,-1,0,1,2,3.......

and fir which value of x the expression vanishes or give a value =0

that value of X should be taken..

e.g...

for the above expression...

if we put.x=1,then

2(1)³+7(1)²-9

=2+7-9

=0

so X=1 is taken into consider.

so...X=1 or( x-1)=0

Therefore (x-1) is taken as factor.

Answer 2

Step-by-step explanation:

I think the right question is:

$factorization \: of \: \: \: 2x {}^{3} + 7x {}^{2} - 9x$

$\huge\bf\purple{Solution..}$

$2x {}^{3} + 7x {}^{2} - 9x \\ = 2x {}^{3} + 9x {}^{2} - 2x {}^{2} - 9x \\ = x {}^{2} (2x + 9) - x(2x + 9) \\ = (x {}^{2} - x)(2x + 9) \\ = x(x - 1)(2x + 9)$

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