Respuesta :
Answer:
[tex](x-1)(2x^2+x+2)[/tex]
Step-by-step explanation:
Factorize:
[tex]f(x)=2x^3-x^2+x-2[/tex]
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:
[tex]\implies f(1)=2(1)^3-1^2+1-2=0[/tex]
Therefore, (x - 1) is a factor.
As the polynomial is cubic:
[tex]\implies f(x)=(x-1)(ax^2+bx+c)[/tex]
Expanding the brackets:
[tex]\implies f(x)=ax^3+bx^2+cx-ax^2-bx-c[/tex]
[tex]\implies f(x)=ax^3+(b-a)x^2+(c-b)x-c[/tex]
Comparing coefficients with the original polynomial:
[tex]\implies ax^3=2x^3 \implies a=2[/tex]
[tex]\implies (b-a)x^2=-x^2 \implies b-2=-1 \implies b=1[/tex]
[tex]\implies -c=-2 \implies c=2[/tex]
Therefore:
[tex]\implies f(x)=(x-1)(2x^2+x+2)[/tex]
Cannot be factored any further.
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)