anonymous
  • anonymous
How to factor these two functions?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
f(x)=-(x-3)^2+2 f(x)=-4x^2+16-15
anonymous
  • anonymous
I don't really need the answer. I just need a quick review of how to factor.
anonymous
  • anonymous
I know the first one can't be factored.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

jim_thompson5910
  • jim_thompson5910
is the second one supposed to be f(x)=-4x^2+16x-15 I put an x after the 16
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
I apologize, that was careless of me.
jim_thompson5910
  • jim_thompson5910
that's fine
jim_thompson5910
  • jim_thompson5910
have you learned the quadratic formula?
anonymous
  • anonymous
I know the formula.
jim_thompson5910
  • 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
  • anonymous
Oh my gosh. Thank you!
jim_thompson5910
  • jim_thompson5910
let me know what you get
anonymous
  • anonymous
x= 1.5 x=2.5
anonymous
  • anonymous
is this correct?
jim_thompson5910
  • jim_thompson5910
that's the same as x = 3/2 and x = 5/2
anonymous
  • anonymous
Oh!
jim_thompson5910
  • 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
  • jim_thompson5910
then move the 2's over through subtraction 3x = 2 ----> 3x-2 = 0 5x = 2 ----> 5x-2 = 0
jim_thompson5910
  • jim_thompson5910
what would come next?
anonymous
  • anonymous
I'm not sure. I've never seen this done before.
jim_thompson5910
  • jim_thompson5910
oh wow I made a big typo I'm just noticing now
jim_thompson5910
  • jim_thompson5910
x = 3/2 should turn into 2x = 3
jim_thompson5910
  • jim_thompson5910
and x = 5/2 should turn into 2x = 5
anonymous
  • anonymous
Oh. Would you multiply them from there to get the original function?
jim_thompson5910
  • 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
  • anonymous
Oh.
jim_thompson5910
  • 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
  • anonymous
Yup.
jim_thompson5910
  • jim_thompson5910
what would (2x-3)*(2x-5) expand out into?
anonymous
  • anonymous
4x^2+16x-15
jim_thompson5910
  • jim_thompson5910
that 16 should be negative try again
anonymous
  • anonymous
Oops. 4x^2-16x+15
jim_thompson5910
  • 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
  • anonymous
I don't think they are the same. I'm not sure. :[
anonymous
  • anonymous
Would you need a negative one in front of the one I got?
jim_thompson5910
  • 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
  • jim_thompson5910
same for the 16x terms and the 15 terms too
anonymous
  • anonymous
Right. So...what did I do wrong?
anonymous
  • anonymous
Why aren't they the same?
jim_thompson5910
  • jim_thompson5910
well to easily fix this, we can stick a -1 in front of the factorization
jim_thompson5910
  • 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
  • jim_thompson5910
so k = -1 is that constant I was talking about
anonymous
  • anonymous
Oh! I think I get it. I understand evrything except the constant. Why is that -1?
anonymous
  • anonymous
Where did the -1 come from?
jim_thompson5910
  • 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
  • jim_thompson5910
expanding out (2x-3)*(2x-5) will have -16x but we want +16x that -1 out front fixes the issue
anonymous
  • anonymous
Understood. Thank you so much, Jim!
jim_thompson5910
  • jim_thompson5910
you're welcome
anonymous
  • anonymous
You're the best! Have a good night!
jim_thompson5910
  • jim_thompson5910
you have a good night as well

Looking for something else?

Not the answer you are looking for? Search for more explanations.