# Thread: Factoring To Find Absolute Max/Min?

1. ## Factoring To Find Absolute Max/Min?

How do I go about factoring this correctly? (the bold)

f(x)= 2x^3-x^2-4x+8

F^1(x) = 3x^2-2x-4

(3x ) (x ) ?

2. Originally Posted by JumpinJimmy
How do I go about factoring this correctly? (the bold)

f(x)= 2x^3-x^2-4x+8

F^1(x) = 3x^2-2x-4

(3x ) (x ) ?
Dear JumpinJimmy,

Do you mean that you have to factorize, $\displaystyle f(x)=3x^{2}-2x-4$???

3. yeah sorry about the confusion, do not know the short cuts for subscripts and superscripts.

4. Originally Posted by JumpinJimmy
yeah sorry about the confusion, do not know the short cuts for subscripts and superscripts.
Dear JumpinJimmy,

$\displaystyle 3x^{2}-2x-4$

$\displaystyle =3\left(x^{2}-\frac{2}{3}x-\frac{4}{3}\right)$

$\displaystyle =3\left(\left(x-\frac{1}{3}\right)^{2}-\frac{1}{9}-\frac{4}{3}\right)$

$\displaystyle =3\left(\left(x-\frac{1}{3}\right)^{2}-\frac{13}{9}\right)$

$\displaystyle =3\left(\left(x-\frac{1}{3}\right)^{2}-\left(\frac{\sqrt{13}}{3}\right)^{2}\right)$

$\displaystyle =3\left(x-\frac{1}{3}-\frac{\sqrt{13}}{3}\right)\left(x-\frac{1}{3}+\frac{\sqrt{13}}{3}\right)$

$\displaystyle =3\left(x-\frac{(1-\sqrt{13})}{3}\right)\left(x-\frac{(1+\sqrt{13})}{3}\right)$

5. Originally Posted by JumpinJimmy
How do I go about factoring this correctly? (the bold)

f(x)= 2x^3-x^2-4x+8

F^1(x) = 3x^2-2x-4

(3x ) (x ) ?
The derivative is actually

$\displaystyle f'(x) = 6x^2 - 2x - 4$.

So if you are setting it equal to 0 and factorising

$\displaystyle 6x^2 - 2x - 4 = 0$

$\displaystyle 2(3x^2 - x - 2) = 0$

$\displaystyle 2(3x^2 - 3x + 2x -2) = 0$

$\displaystyle 2[3x(x - 1) + 2(x - 1)] = 0$

$\displaystyle 2(x - 1)(3x + 2) = 0$

So $\displaystyle x - 1 = 0$ or $\displaystyle 3x + 2 = 0$

$\displaystyle x = 1$ or $\displaystyle x = -\frac{2}{3}$.