How do you factor given that $f(-6)=0$ and $f(x)=2x^3+7x^2-33x-18$ ?

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Answer 1 · 2018-08-08T07:23:33

$f(x)=(x+6)(x-3)(2x+1)$ .

The given cubic polynomial $f$ has $-6$ as its zero.

$:. (x-(-6))=(x+6)$ is a factor of $f(x)$ .

We split the terms of $f(x)$ in such a way that $(x+6)$ may turn

out as a factor, as shown below :-

$f(x)=2x^3+7x^2-33x-18$ ,

$=ul(2x^3+12x^2)-ul(5x^2-30x)-ul(3x-18)$ ,

$=2x^2(x+6)-5x(x+6)-3(x+6)$ ,

$=(x+6){2x^2-5x-3}$ ,

$=(x+6){ul(2x^2-6x)+ul(x-3)}$ ,

$=(x+6){2x(x-3)+1(x-3)}$ .

$rArr f(x)=(x+6)(x-3)(2x+1)$ .

How do you factor given that f(-6)=0f(-6)=0 and f(x)=2x^3+7x^2-33x-18f(x)=2x^3+7x^2-33x-18?