Question

Multiply. State any restrictions on the variable.
12-x-x^2/x^2-8x+15
Can you guys show me the steps to this? I don't get it

Answer 1

\[\frac{12-x-x^2}{x^2-8x+15}\]Factor the numerator and denominator. Before doing any cancellation, find the values of \(x\) that make the denominator = 0. Those values are the restricted values — \(x\) cannot be any of them as you cannot meaningfully divide by 0.

Answer 2

You have to do this check before you cancel common terms from numerator and denominator because they are still restrictions on the simplified fraction.

Answer 3

I know it's (x,-3) and (x,-5). What do I do next?

Answer 4

Show me your factored denominator...

Answer 5

What is the factored denominator?

Answer 6

I asked you to factor the denominator of that fraction. What does it look like after you have factored it?

Answer 7

I didn't factor it. I don't know how.

Answer 8

You don't know how to factor \[x^2-8x+15\]?

Answer 9

:l Sorry, I'm stupid.

Answer 10

No, not stupid, but maybe you haven't learned it for some reason. Has it been covered in class? You really can't do this problem without knowing how to factor...

Answer 11

I dunno

Answer 12

can you show me the steps? c:

Answer 13

please?

Answer 14

yes, sorry, had to take care of something

Answer 15

Okay, we have a trinomial (fancy word for 3 terms):
\[x^2-8x+15\]
We want to split it into the product of two binomials (fancy word for 2 terms):
\[x^2-8x+15 = (x+a)(x+b)\]But how?

Answer 16

it's okay:p

Answer 17

Let's multiply \((x+a)(x+b)\):
\[(x+a)(x+b) = x(x+b) + a(x+b) = x*x + x*b + a*x + a*b\]\[=x^2 + (a+b)x + a*b\]Agreed so far?

Answer 18

hello...

Answer 19

sorrry. This thing wasn't letting me type.

Answer 20

no problem...okay with what I've done so far?

Answer 21

yes. And if I don't respond, just keep showing me the steps. Cause this thing is getting stupid.

Answer 22

the typing

Answer 23

so let's compare what we've got:
\[x^2+(a+b)x + a*b\]\[x^2-8x+15\]To make those equivalent, we need to pick \(a,b\) such that \[(a+b) = -8\]and\[a*b=15\]Your mission is to find a pair of numbers that make that work.

Answer 24

Hopefully it is obvious that they both have to be negative in this case...

Answer 25

yes

Answer 26

how can we get a*b = 15, with both a and b negative?

-1*-15 = 15
-3*-5 = 15
-5*-3 = 15
-15*-1 = 15

We also need them to add up to -8
That leaves a=-3, b=-5 (or vice versa)

So our factored version is \[(x-3)(x-5)\]Let's check by multiplying that out:
\[(x-3)(x-5) = x*x -5*x -3*x-3*(-5) = x^2-5x-3x+15\]\[(x-3)(x-5)=x^2-8x+15\]

That means our fraction can be written\[\frac{12-x-x^2}{(x-3)(x-5)}\]

Now, to find the restrictions on \(x\), we set the denominator = 0 and solve for the values of \(x\) that make it = 0. We have a product as the denominator, so that makes this task very easy:
\[(x-3)(x-5) = 0\]\[x-3=0\]\[x-5=0\]\[x=3,\,x=5\]

So our restrictions are that \(x \ne 3, x\ne 5\)

Answer 27

is that it?

Answer 28

We should factor the numerator (even though we don't need to for this problem) for practice.

\[-x^2-x+12\](I reversed the order of the terms, but of course the value does not change)
We can factor out a common \(-1\) from each term:\[-1(x^2) -1(x) -1(-12) = -1(x^2+x-12)\]We'll set that \(-1\) aside and just factor the quantity in the parentheses. We'll put the \(-1\) factor back when we are done.

\[x^2+x-12 = x^2+1x-12\]Again, we want to find two numbers \(a,b\) such that \(a+b = 1\) and \(a*b = -12\)

Let's factor -12:
-1*12
-2*6
-3*4
-4*3
-6*2
-12*1

which of those pairs adds up to +1? That will be our factoring.
-3+4 = +1
\[x^2+x-12 = (x-3)(x+4)\]Checking:
\[(x-3)(x+4) = x*x + 4*x -3*x -3(4) = x^2 + x - 12\]
We had a \(-1\) that we factored out, so our fully factored numerator would be:
\[-x^2-x+12 = -1(x-3)(x+4)\]

Giving us a fully factored fraction of \[\frac{-1(x-3)(x+4)}{(x-3)(x-5)}\] We could then simplify that:
\[\frac{-1(x-3)(x+4)}{(x-3)(x-5)}= \frac{-1\cancel{(x-3)}(x+4)}{\cancel{(x-3)}(x-5)} = -\frac{(x+4)}{(x-5)}\] But our restrictions remain the same, even though we no longer have the \((x-3)\) term in the denominator!

Answer 29

That final fraction could also be written as \[-\frac{x+4}{x-5} = \frac{x+4}{5-x}\]

Answer 30

or \[\frac{4-x}{x-5}\]

Answer 31

Now, I only covered factoring trinomials where the coefficient of the \(x^2\) term is 1. Unfortunately, it doesn't have to be 1, and it's more complicated to factor when it isn't. The procedure is sort of similar, and I'm happy to show it to you right now, or give you a rain check and explain it at some other time. Which do you want?

Answer 32

well do I have everything I need already?

Answer 33

you have the answer to this problem, and I've given you a brief tutorial on how to factor trinomials where the leading coefficient is 1. That may be all you need for now.

Answer 34

By the way, the first word of your problem statement is "multiply", but then you showed a division...we only have the restriction on variable business if it is a division, did you make a mistake when you typed "multiply"?

Answer 35

no. :/

Answer 36

And the fraction as I typed it, that's what appears in your problem?

Answer 37

i copied what it said in the book

Answer 38

That's not answering my question...

Answer 39

Does the fraction as I formatted it look like the problem in the book?

Answer 40

yes

Answer 41

Okay, well, must be an error in the book. Wouldn't be the first time that has happened, or likely the last.

Any questions or things that you'd like cleared up?

Answer 42

nope. thanks do

Answer 43

Did you block me for some reason?