I don't get how we're supposed to FOIL this quadratic equation...

I know how to FOIL, of course, but I don't get how they got the -10x here. Can anyone help me out? These are the steps it showed:

f(x) = 2(x – 5)2

f(x) = 2(x^2 – 10x 25)

f(x) = 2x^2 – 20x 50

Update:

Sorry, the equation got messed up a bit. It should read like this:

f(x) = 2(x – 5)2

f(x) = 2(x^2 – 10x + 25)

f(x) = 2x^2 – 20x + 50)

Relevance

When you expand (x-5)^2, the first step should look like this:

x*x+x*-5+-5*x+-5*-5

Then:

x^2-5x-5x+25

x^2-10x+25

It's -10x because there are two -5x's that add together.

A good rule to remember is that:

(a-b)^2=a^2-2ab+b^2

In this case, "-2ab" is -2*x*5, or -10x

2(x^2-10x+25)

2x^2-20x+50

You can rewrite the (x - 5)² as (x - 5)(x - 5)

f(x) = 2(x - 5)²

f(x) = 2(x - 5)(x - 5) --> Now you can FOIL the (x - 5)(x - 5)

f(x) = 2(x² - 5x - 5x + 25) --> Combine the like terms: -5x - 5x

f(x) = 2(x² - 10x + 25) --> Then distribute the 2

f(x) = 2x² - 20x + 50

i think you meant 2(x-5)^2

so cuz it is to the power of 2 it can be expanded into

2(x-5)(x-5) and then you foil it

then you get 2(x^2-10x+25)

if you time the whole equation by 2

you get 2x^2-20x+50

hope this helpz!

• Anonymous

f(x)=2(x-5)^2

f(x)=2(x - 5)(x - 5)

FOIL = First, Outside, Inside, Last

First = x * x = x^2

Outside = x * -5 = -5x

Inside = -5 * x = -5x

Last = -5 * -5 = 25

x^2 - 5x - 5x + 25 = x^2 - 10x + 25

The equation should readf(x) = 2(x-5)^2

(x-5)^2 is (x-5)*(x-5)

F x*x = x^2

O x*-5 = -5x

I x*-5 =-5x

L -5*-5 =25

x^2 -5x -5x + 25

x^2 -10x +25

then 2(x^2 -10x +25) = 2x^2 -20x +50

Hope this helps