Question

2x^3-x^2+x-2 factorize​

Answer 1

Solution :

• 2x³ - x² + x - 2

put x = 1,

• 2•(1³) - 1² + 1 - 2

• 2 - 1 + 1 - 2

• 0

it's value is 0.

hence, it's a solution

divide 2x³ - x² + x - 2 by (x - 1)

• 2x² + x + 2

it can be written as :

• (x - 1)(2x² + x + 2)

Answer 2

Answer:

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

Step-by-step explanation:

Factorize:

$f(x)=2x^3-x^2+x-2$

Factor Theorem

If f(a) = 0 for a polynomial then (x - a) is a factor of the polynomial f(x).

Substitute x = 1 into the function:

$\implies f(1)=2(1)^3-1^2+1-2=0$

Therefore, (x - 1) is a factor.

As the polynomial is cubic:

$\implies f(x)=(x-1)(ax^2+bx+c)$

Expanding the brackets:

$\implies f(x)=ax^3+bx^2+cx-ax^2-bx-c$

$\implies f(x)=ax^3+(b-a)x^2+(c-b)x-c$

Comparing coefficients with the original polynomial:

$\implies ax^3=2x^3 \implies a=2$

$\implies (b-a)x^2=-x^2 \implies b-2=-1 \implies b=1$

$\implies -c=-2 \implies c=2$

Therefore:

$\implies f(x)=(x-1)(2x^2+x+2)$

Cannot be factored any further.