Question

factorize using factor theorem :2x^3-5x^2+x+2

Answer 1

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Answer 2

All factors of cubic polynomial $\bf 2 {x}^{3} - 5 {x}^{2} + x + 2$ are $\bf \red{(x - 1)(x - 2) \: and \: (2x + 1)}$.

Given:

• $2 {x}^{3} - 5 {x}^{2} + x + 2$

To find:

• Factors using factor theorem.

Solution:

Concept/Theorem to be used:

• A cubic polynomial have three factors, unique or distinct.
• One factor can be find by hit and trial method and using factor theorem.
• Factor theorem: It states that if (x-a) is a factor of polynomial p(x), then p(a)=0.
• Then divide the polynomial by the known factor two find other two factors.

Step 1:

Find one factor using factor theorem.

Let $p(x) = 2 {x}^{3} - 5 {x}^{2} + x + 2$

Put x= 1

$= 2 {(1)}^{3} - 5 {(1)}^{2} + (1) + 2$

or

$= 5 - 5$

or

$\bf p(1) = 0$

So,

(x-1) is one of the factor of p(x).

Step 2:

Divide the polynomial by (x-1).

$x - 1 \: ) \: 2 {x}^{3} - 5 {x}^{2} + x + 2 \: (2 {x}^{2} - 3x - 2 \\ 2 {x}^{3} - 2 {x}^{2} \: \: \: \: \: \: \: \: \: \: \\ ( - ) \: \: \: ( + ) \: \: \: \: \: \: \: \: \: \: \: \\ - - - - - - - \: \: \: \: \: \: \\ - 3 {x}^{2} + x \\ - 3 {x}^{2} + 3x \\ ( + ) \: \: \: ( - ) \\ - - - - - - \\ - 2x + 2 \\ - 2x + 2 \\ ( + ) \: \: ( - ) \\ - - - - - \\ 0 \\ - - - - -$

Thus,

Quotient polynomial is another factor.

But it is not linear, so factorise it further.

Step 3:

Factorise quotient polynomial.

$2 {x}^{2} - 3x - 2$

split the middle term.

$2 {x}^{2} - 4x + x - 2$

or

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

or

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

Thus,

All factors of cubic polynomial $\bf 2 {x}^{3} - 5 {x}^{2} + x + 2$ are $\bf (x - 1)(x - 2) \: and \: (2x + 1)$.

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