1. ## quadratic function general form to standard form

Ok here is what I did. I'm kinda weak from factoring

$\displaystyle f(x)= 2x^2+8x-2$

group:
$\displaystyle f(x)= (2x^2+8x)-2$

Completing the Square:

$\displaystyle f(x)= (2x^2+8x+16-16)-2$
$\displaystyle f(x)= (2x^2+8x+16)-16-2$
$\displaystyle f(x)= (2x^2+8x+16)-18$

Factoring:

$\displaystyle f(x)= 2(x^2+4x+8)-18$

Where did I go wrong from here?

$\displaystyle f(x)= 2(x+2)-18$

bcoz if I FOIL 2(x+2)(x+2) it will come out:

$\displaystyle 2(x^2+4x+4)$

2. Originally Posted by Anemori
Ok here is what I did. I'm kinda weak from factoring

$\displaystyle f(x)= 2x^2+8x-2$

group:
$\displaystyle f(x)= (2x^2+8x)-2$

Completing the Square:

$\displaystyle f(x)= (2x^2+8x+16-16)-2$
$\displaystyle f(x)= (2x^2+8x+16)-16-2$
$\displaystyle f(x)= (2x^2+8x+16)-18$

Factoring:

$\displaystyle f(x)= 2(x^2+4x+8)-18$

Where did I go wrong from here?

$\displaystyle f(x)= 2(x+2)-18$

bcoz if I FOIL 2(x+2)(x+2) it will come out:

$\displaystyle 2(x^2+4x+4)$
You can't complete the square unless the coefficient of $\displaystyle x^2$ is $\displaystyle 1$.

$\displaystyle f(x)= 2x^2+8x-2$

$\displaystyle = 2(x^2 + 4x - 1)$

$\displaystyle = 2(x^2 + 4x + 2^2 - 2^2 - 1)$

$\displaystyle = 2[(x + 2)^2 - 5]$

$\displaystyle = 2(x + 2)^2 - 10$.

3. Originally Posted by Prove It
You can't complete the square unless the coefficient of $\displaystyle x^2$ is $\displaystyle 1$.

$\displaystyle f(x)= 2x^2+8x-2$

$\displaystyle = 2(x^2 + 4x - 1)$

$\displaystyle = 2(x^2 + 4x + 2^2 - 2^2 - 1)$

$\displaystyle = 2[(x + 2)^2 - 5]$

$\displaystyle = 2(x + 2)^2 - 10$.

yea i look back my answer and i made a mistake after the grouping.
thanks!