Question

factorise 2x cube -9x square + x + 12

Answer 1

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Answer 2

All factors of $\bf2 {x}^{3} - 9 {x}^{2} + x + 12$are $\bf \red{(x + 1)(2x - 3)(x - 4) }$.

Given:

• $2 {x}^{3} - 9 {x}^{2} + x + 12$

To find:

• Factors of cubic polynomial.

Solution:

Concept/ Theorem to be used:

Factor theorem: if x-a is a factor of polynomial p(x) than p(a)=0.

Step 1:

Find one of the factor of polynomial by hit and trial using factor theorem.

Let $p(x)=2 {x}^{3} - 9 {x}^{2} + x + 12$

Put x=1

$p(1) = 2 {(1)}^{3} - 9 {(1)}^{2} + 1 + 12$

or

$p(1) = 2 - 9 + 1 + 12$

or

$p(1) = 6 \neq0$

Thus, (x-1) is not a factor of p(x).

Put x= -1.

$p( - 1) = 2 {( - 1)}^{3} - 9 {( - 1)}^{2} - 1 + 12$

or

$p( - 1) = - 2 - 9 - 1 + 12$

or

$p( - 1) = 0$

Thus,

$\bf (x + 1)$is a factor of polynomial.

Step 2:

Divide the polynomial by (x+1).

$x + 1 \: ) \: 2 {x}^{3} - 9 {x}^{2} + x + 12(2 {x}^{2} - 11x + 12 \\ 2 {x}^{3} + 2 {x}^{2} \: \: \: \: \: \: \: \: \: \: \: \\ ( - )( - ) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ - - - - - \: \: \: \: \: \: \: \: \: \: \: \\ - 11 {x}^{2} + x \\ - 11 {x}^{2} - 11x \\ ( + ) \: \: ( + ) \\ - - - - - \\ 12x + 12 \\ 12x + 12 \\ ( - ) \: \: ( - ) \\ - - - - - \\ 0 \\ - - - - -$

Thus,

Quotient polynomial is another factor of p(x).

Step 3:

Factorise quotient polynomial.

$2 {x}^{2} - 11x + 12$

or

$= 2 {x}^{2} - 8x - 3x + 12$

or

$= 2x(x - 4) - 3(x - 4)$

or

$= (2x - 3)(x - 4)$

Thus,

All factors of polynomial are $\bf (x + 1)(2x - 3)(x - 4)$.

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