Question

Give the complete factorization for the polynomial. 10x - 3 - 3x2
can somebody please explain how to do this???

Answer 1

So the polynomial can be rearranged as:

\[-3x^2+10x-3=0\]
note that the polynomial is in the form:
\[ax^2+bx+c=0\]

To start factoring, find two numbers that multiply to (C*A) So find two numbers that multiply to 9 and add to +10. What are they?

Answer 2

Could I include negatives?

Answer 3

You could, but you don't need them here, since both numbers are positive.

Answer 4

can't 9 only be divided by 1,3 and itself?

Answer 5

Good, going on that in reverse, what are the only two sets of numbers that multiply to 9? (Two sets of numbers)

Answer 6

9, 1 and 3, 3. So that would be 1,9 because they equal 10

Answer 7

Yup, you got it!

Now, you know how we have 10x? We're gonna split that up into 9x+1x (it's the same thing)

\[-3x^2+9x+1x-3=0\]

Now we factor out the negative in the front to make stuff easier for us.

\[-(3x^2-9x-1x+3)=0\]

Now we want to factor this by groups, the first group is:

\[3x^2-9x\] and second group is:
\[-1x+3\]

So now factor these two grouping individually please.

Answer 8

@craftshark98, so any ideas how to do it?

Answer 9

Yeah I'm trying to figure it out right now. Sorry it's taking so long.

Answer 10

Here's a hint, take out 3x for the first one, and take out -1 for the second one ;D

Answer 11

what do you mean? :/

Answer 12

yeah I think I'm pretty stuck lol

Answer 13

So basically what we do for the first group is take out 3x, since that's the common factor:

\[3x^2-9x \rightarrow 3x(x-3)\]

And the second group we take out -1, which is the common factor to get:

\[-1x+3\rightarrow -1(x-3)\]

Answer 14

So understand what I did there?

Answer 15

yeah....but wha I don't understand is how in the second one +3 becomes -3

Answer 16

So what happens when we divide a + by a negative? It becomes negative right? So that's what I did here.

Answer 17

Essentially the main idea is to get the same thing in the parenthesis for both groups.

Answer 18

what did you divide it by?

Answer 19

-1, so +3/-1 = -3 right?

Answer 20

yes it is :D
but why do we want (x - 3) in the parenthesis of both?

Answer 21

Since that is one of our factors :D Better move on:

So first we have this:
\[−(3x^2−9x−1x+3)=0\]Now sub for each group:
\[−(3x(x-3)−1(x-3))=0\]

Now you see that x-3 is in both sets? That's 1 group, and we factor that out now!

So we end up being left with this when factoring:

\[-((3x-1)(x-3))=0\]

Once we have that, we can get rid of outer parenthesis and we are left with factored form:

\[-(3x-1)(x-3)=0\]

Answer 22

Get it?

Answer 23

yeah. thank you! :D
though I might need help in applying this to other problems.

Answer 24

It's okay, hope this helped.