- anonymous

How to factor these two functions?

- chestercat

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- anonymous

f(x)=-(x-3)^2+2
f(x)=-4x^2+16-15

- anonymous

I don't really need the answer. I just need a quick review of how to factor.

- anonymous

I know the first one can't be factored.

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## More answers

- jim_thompson5910

is the second one supposed to be f(x)=-4x^2+16x-15
I put an x after the 16

- anonymous

Yes.

- anonymous

I apologize, that was careless of me.

- jim_thompson5910

that's fine

- jim_thompson5910

have you learned the quadratic formula?

- anonymous

I know the formula.

- jim_thompson5910

you can use the formula to solve -4x^2+16x-15 = 0 for x
you'll get two roots x = p and x = q
you can use those two solutions to find the factorization in the form
k*(x-p)*(x-q) = 0
the k is some fixed number that scales the graph and determines whether the graph opens up or down

- anonymous

Oh my gosh. Thank you!

- jim_thompson5910

let me know what you get

- anonymous

x= 1.5
x=2.5

- anonymous

is this correct?

- jim_thompson5910

that's the same as x = 3/2 and x = 5/2

- anonymous

Oh!

- jim_thompson5910

what we can do is multiply both sides of each equation by 2
x = 3/2 ----> 3x = 2
x = 5/2 ----> 5x = 2

- jim_thompson5910

then move the 2's over through subtraction
3x = 2 ----> 3x-2 = 0
5x = 2 ----> 5x-2 = 0

- jim_thompson5910

what would come next?

- anonymous

I'm not sure. I've never seen this done before.

- jim_thompson5910

oh wow I made a big typo I'm just noticing now

- jim_thompson5910

x = 3/2 should turn into 2x = 3

- jim_thompson5910

and x = 5/2 should turn into 2x = 5

- anonymous

Oh. Would you multiply them from there to get the original function?

- jim_thompson5910

ok so through the quadratic formula, we get these 2 solutions
x = 3/2, x = 5/2
multiply both sides by 2
2x = 3, 2x = 5
then move everything to one side
2x-3=0, 2x-5=0

- anonymous

Oh.

- jim_thompson5910

we have 2x-3=0, 2x-5=0
they would turn into (2x-3)*(2x-5) = 0
I think that's what you had in mind?

- anonymous

Yup.

- jim_thompson5910

what would (2x-3)*(2x-5) expand out into?

- anonymous

4x^2+16x-15

- jim_thompson5910

that 16 should be negative
try again

- anonymous

Oops.
4x^2-16x+15

- jim_thompson5910

now compare 4x^2-16x+15 (what you just got when you expanded) with -4x^2+16x-15 (the original function). Are they the same? If not, what can we do to make them the same?

- anonymous

I don't think they are the same. I'm not sure. :[

- anonymous

Would you need a negative one in front of the one I got?

- jim_thompson5910

they aren't the same
notice how the 4x^2 is positive in 4x^2-16x+15
then we have -4x^2 in -4x^2+16x-15

- jim_thompson5910

same for the 16x terms and the 15 terms too

- anonymous

Right. So...what did I do wrong?

- anonymous

Why aren't they the same?

- jim_thompson5910

well to easily fix this, we can stick a -1 in front of the factorization

- jim_thompson5910

since that will make +4x^2 turn into -4x^2
the -16x turn into +16x
and the +15 turn into -15

- jim_thompson5910

so k = -1 is that constant I was talking about

- anonymous

Oh! I think I get it. I understand evrything except the constant. Why is that -1?

- anonymous

Where did the -1 come from?

- jim_thompson5910

if we expanded out (2x-3)*(2x-5) we end up with 4x^2 as one of the terms
we want -4x^2
so we can make (2x-3)*(2x-5) negative and say -1*(2x-3)*(2x-5) instead to fix that issue

- jim_thompson5910

expanding out (2x-3)*(2x-5) will have -16x
but we want +16x
that -1 out front fixes the issue

- anonymous

Understood.
Thank you so much, Jim!

- jim_thompson5910

you're welcome

- anonymous

You're the best! Have a good night!

- jim_thompson5910

you have a good night as well

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