Question

How do you solve 8x^2+20x-48?

Answer 1

By "solve" do you mean find the values of \(x\) such that \[8x^2+20x-48 = 0\]

Answer 2

Or are you trying to factor the polynomial?

Answer 3

To find the x's, you have to factor.

Answer 4

No, you don't have to factor, but that is one way you can find them.

Answer 5

In fact, you can find the values of \(x\) even if the polynomial is irreducible (cannot be factored).

Answer 6

You have to factor it. How do you do that?

Answer 7

Is your assignment to factor it, or find the values of \(x\) that make it equal to 0?

Answer 8

@Thefaceless uh, that's not correct...

Answer 9

\[8x^2+20x - 48 = 2*2*2*x^2+2*2*5x-2*2*2*2*3\]so you can factor out 2*2 = 4 giving you\[4(2x^2+5x-12)\]

Answer 10

Oh whoops forgot + and 2

Answer 11

@j2lie one last time: are you supposed to factor that polynomial, or solve it for the values of \(x\) that make it equal to 0?

Answer 12

She said factor

Answer 13

use the quadratic formula
\[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]
@j2lie

Answer 14

Okay, if you are factoring it, we already got out the common factor of 4:
\[4(2x^2+5x-12)\]
Now we are going to have two binomials, one of which will be \((x+a)\) for some value of \(a\) to be determined, and the other of which will be \((2x + b\), for some value of \(b\) to be determined.

We know that one of them must have a \(2\) because we need to end up with \(2x^2\) as our highest order term.

Let's try an experiment:
\[(x+a)(2x+b) = 2x^2 + bx + 2ax + ab\]\[=2x^2 + (2a+b)x + ab\]
Compare with our polynomial to be factored:\[2x^2+5x-12\]

To make those the same, we need to find values of \(a,b\) such that\[2a+b=5\]\[a*b=-12\]

Can you think of two numbers that satisfy those conditions?

Answer 15

Obviously, one will be positive and the other negative. I'll list the factors of 12:

1*12
2*6
3*4
4*3
6*2
12*1

If you put a - sign in front of one number in the pair, can you find the pair that works?

Answer 16

quadratic formula is much faster :3

Answer 17

i figured it out. but I got stuck.

Answer 18

Yes, it is, but it is important to know how to factor, and that's what she says she is supposed to do in this problem.

Answer 19

oh. if she doesn't know that, guess she's ought to do it the hard way.

Answer 20

you're supposed to split the middle but it doesn't work.

Answer 21

how did you split the middle?

Answer 22

-3x and 8x

Answer 23

try & use this once if you still can't figure it out:
\[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

Answer 24

ah, but you have to divide one by 2 because of the multiplier...

Answer 25

@sayakdbz countless millions of algebra students have learned how to factor trinomials. j2lie can do it too :-)

Answer 26

@whpalmer4 i'll leave this one to u :)

Answer 27

i am confused

Answer 28

Okay. do you agree that the factoring will look something like
\[(2x + a)(x+b)\]?

Answer 29

yes

Answer 30

Good. So if you "split the middle", don't you have to divide one of the pieces in half? It's going to be multiplied by 2, not 1.

Answer 31

u are guiding it in the wrong way. its supposed to be:
\[x^2+(a+b)x+ab=(x+a)*(x+b)\]
@whpalmer4

Answer 32

@sayakdbz how many polynomials do you think I've factored in the 35 years I've been doing this?

Answer 33

1

Answer 34

@thefaceless 2, actually, this will be the 3rd :-)

Answer 35

lol

Answer 36

XD

Answer 37

if you're factoring by grouping:

\[ax^2+bx+c\]you multiply a*c, and choose two factors of the product a*c that sum to b.
-8 and 3 are such a pair.

\[2x^2 -8x + 3x - 12\]Group terms:\[(2x^2-8x) + (3x-12)\]Now factor each group. what do you get?

Answer 38

\[(2x^2-8x)+(3x-12) = 2x(x-4) + 3(x-4)\]Notice anything interesting about the right hand side?

Answer 39

Sorry, we're not doing \(2x^2-5x+12\) we're doing \(2x^2+5x-12\)!

-3 and 8 are our split of the middle:

\[2x^2 + 8x -3x -12\]\[(2x^2+8x)-(3x+12)\]\[2x(x+4)-3(x+4)\]

What do we get when we factor \((x+4)\) out of both of those terms? Essentially using the distributive property in reverse...

\[2x(x+4)+3(x+4) = (x+4)(2x+3)\]

Combine that with the common factor of \(4\) that we pulled out earlier:
\[8x^2+20x-48 = 4(x+4)(2x-3)\]

If we want to "solve" that (find the values of \(x\) that make the value of that be \(0\), we can set
\[4(x+4)(2x-3) = 0\]\[(x+4)=0,\,x=-4\]\[2x-3=0,\,2x=3,\,x=3/2\]
So \(x=-4\) and \(x = 3/2\) are the solutions.

Answer 40

the solution is (x+4)(8x-12)

Answer 41

What do you think \(4(2x-3)\) = ?

Answer 42

8x-12

Answer 43

As I said several lines earlier:

\[8x^2+20x-48 = 4(x+4)(2x-3)\]That is the complete factoring. \[(x+4)(8x-12)\]is an incomplete factoring, because there is still a factor of \(4\) which can be taken out of the \((8x-12)\) product term.

"Solving" a polynomial is commonly understood to mean finding the roots, aka zeros, aka values that make the polynomial = 0.

Answer 44

I understand what you mean. but it's from 8x^2+32x_12x-48.

Answer 45

\[8x^2+32x-12x-48 = 8x^2 + 20x - 48\]What's your point? If you're factoring that by grouping:
\[(8x^2+32x) - (12x+48) = 8x(x+4)-12(x+4)\]\[ =(8x-12) (x+4)\]\[ = 4(2x-3)(x+4)\]

Answer 46

I did it this way
8x^2+20x-48.
48*8=384
8x^2+32x-12x-48
8x(x+4)-12(x+4)=
8x^2+20x-48

Answer 47

That's all good except for the last line. You need to factor out the \((x+4)\) from both terms — that's why you did the splitting/grouping.

Answer 48

whoops I forgot but I did that.

Answer 49

Do you recognize that \(8x(x+4) - 12(x+4)\) is simply a case of
\[b*a + c*a\]where \(a = (x+4)\)?

Answer 50

and then we can rewrite it as \[a(b+c)\]

Answer 51

yes it's like a foiling problem.

Answer 52

so \[8x(x+4) - 12(x+4)\]can be written as \[(x+4)*8x+(x+4)*(-12) = (x+4)(8x-12)\]which can be further factored to \[4(x+4)(2x-3)\]

Answer 53

to me it's easier leaving it that way because the 4 is confusing.

Answer 54

okay, but it isn't fully factored without pulling out the 4. And in many applications, you'll want it fully factored so you can cancel common factors (say this is in the numerator of a fraction, and the denominator contains \((2x-3)\) — will you realize that you can cancel it out?)

Answer 55

yes

Answer 56

Anything else you need to know about this problem?

Answer 57

i did know how to factor the 4 with the terms in the parentheses.

Answer 58

Yes, I'm sure you do — the question is whether you will notice it in the middle of a large problem when it isn't an obvious number to factor out. You want to stack the deck in your favor. Trusting that you'll spot a common factor like the 4 in this case isn't doing that.

Answer 59

I did it you're way and the answer is the same.

Answer 60

@whpalmer4 thank you for helping me.

Answer 61

You're welcome! Sorry if I seemed a bit grumpy — my pizza was late :-)

Answer 62

it's because I was arguing with you, that's why you're grumpy.