Answer:
Factorization of [tex]P(x)=(x+2)(2x+1)(3x-1)[/tex]
Step-by-step explanation:
Given:
The given polynomial is [tex]P(x)=6x^3+13x^2+x-2[/tex]
Also, [tex]P(-2)=0[/tex]
Since, for [tex]x=-2[/tex], [tex]P(x)[/tex] is 0, therefore, [tex]x+2[/tex] is a factor of the polynomial.
Now, let us divide the given polynomial by [tex]x+2[/tex] using long division method. Therefore,
[tex]\frac{6x^3+13x^2+x-2}{x+2}=6x^2+x-1[/tex]
The process of long division is shown below.
Now, factoring the quotient [tex]6x^2+x-1[/tex], we get
[tex]6x^2+x-1=6x^2+3x-2x-1=3x(2x+1)-1(2x+1)=(3x-1)(2x+1)[/tex]
Therefore, the factors of polynomial [tex]P(x)[/tex] are [tex](x+2),(2x+1),\ and\ (3x-1)[/tex]
Hence, the factorization of [tex]P(x)[/tex] is:
[tex]P(x)=(x+2)(2x+1)(3x-1)[/tex]
