How do you factor given that $f(-3)=0$ and $f(x)=x^3+x^2+2x+24$ ?

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Answer 1 · 2017-05-24T14:39:46

$f(x)=(x+3)(x^2-2x+8).$

We have, for, $f(x)=x^3+x^2+2x+24,$

$f(-3)=-27+9-6+24=0.$

From the Factor Theorem, hence, we conclude that,

$x-(-3)=x+3$ is a factor of $f(x).$

One way, then, is to divide $f(x)$ by $(x+3)$ by Long Division.

Instead, have a look at the following :

$f(x)=x^3+x^2+2x+24,$

$=ul(x^3+3x^2)-ul(2x^2-6x)+ul(8x+24),$

$=x^2(x+3)-2x(x+3)+8(x+3),$

$rArr f(x)=(x+3)(x^2-2x+8).$

For the Quadr. Poly. $x^2-2x+8,$ we have,

$Delta=b^2-4ac=4-4(1)(8)=-28 <0.$

Hence, it can not be factored further in $RR.$

Enjoy Maths.!

How do you factor given that f(-3)=0f(-3)=0 and f(x)=x^3+x^2+2x+24f(x)=x^3+x^2+2x+24?