Question

How to factor these two functions?

Answer 1

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

Answer 2

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

Answer 3

I know the first one can't be factored.

Answer 4

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

I put an x after the 16

Answer 5

Yes.

Answer 6

I apologize, that was careless of me.

Answer 7

that's fine

Answer 8

have you learned the quadratic formula?

Answer 9

I know the formula.

Answer 10

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

Answer 11

Oh my gosh. Thank you!

Answer 12

let me know what you get

Answer 13

x= 1.5
x=2.5

Answer 14

is this correct?

Answer 15

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

Answer 16

Oh!

Answer 17

what we can do is multiply both sides of each equation by 2

x = 3/2 ----> 3x = 2
x = 5/2 ----> 5x = 2

Answer 18

then move the 2's over through subtraction

3x = 2 ----> 3x-2 = 0
5x = 2 ----> 5x-2 = 0

Answer 19

what would come next?

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

Oh.

Answer 27

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?

Answer 28

Yup.

Answer 29

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

Answer 30

4x^2+16x-15

Answer 31

that 16 should be negative

try again

Answer 32

Oops.
4x^2-16x+15

Answer 33

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?

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

same for the 16x terms and the 15 terms too

Answer 38

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

Answer 39

Why aren't they the same?

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

Where did the -1 come from?

Answer 45

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

Answer 46

expanding out (2x-3)*(2x-5) will have -16x
but we want +16x

that -1 out front fixes the issue

Answer 47

Understood.
Thank you so much, Jim!

Answer 48

you're welcome

Answer 49

You're the best! Have a good night!

Answer 50

you have a good night as well