Question

How would I factor this???

8c2 - 26c + 15

Answer 1

8c^2 - 26c + 15

Answer 2

1. Try to factor out a common factor out of all terms if there is one.

Answer 3

I don't think there is one.

Answer 4

At least not one small enough to use in this equation.

Answer 5

Correct.
2. This is a trinomial of the form
ax^2 + bx + c
Multiply a and c together. In this case it's 8 * 15

Answer 6

ac = 8 * 15 =

Answer 7

120

Answer 8

Good.
3. Come up with two factors of ac that add to b.
In this case you need two factors of 120 that add up to -26
Since 120 is positive and -26 is negative, the numbers you need are both negative.

Answer 9

So I need negative factors of 120?

Answer 10

Yes. You need two factors of 120; they must both be negative; they need to multiply to 120 and add to -26.

Answer 11

-4 and -30!

Answer 12

No.

Answer 13

-6 and -20

Answer 14

Right

Answer 15

4. Call these factors p and q. Now rewrite the trinomial with the middle term (bx) broken up into two terms (px + qx), ax^2 + px + qx + c
In this case you get
8c^2 - 20c - 6c + 15
If you wrote
8c^2 - 6c - 20c + 15, it also works.

Answer 16

5. Now factor by parts. Factor a common factor out of the first two terms, and factor a common factor out of the last two terms.
8c^2 - 20c - 6c + 15
=4c(2c - 5) - 3(2c - 5)

Answer 17

6. Factor out the common term.
4c(2c - 5) - 3(2c - 5)
Here, the common term is (2c - 5), so we factor it out:
(2c - 5)(4c - 3)
Now it's factored.

Answer 18

This is called factoring by parts.

Answer 19

for part 5, where did (2c-5) come from?

Answer 20

Ohhh I think I know. 2 comes from 8 (because 4*2 = 8) and the c come from the 8. Then the 5 comes from the 20. Right?

Answer 21

Right.
Then there is another 2c - 5 that comes from -6c + 15

Answer 22

Ohh. Makes sense.

Answer 23

Now I get it. I guess I just need to get used to this. I haven't fully caught on to factoring stuff yet.

Answer 24

Thank you for all your patience and help!!!

Answer 25

This method of factoring is good for a trinomial like yours which has a number other than 1 with the x^2 term. In your case you had 8x^2. If you have a simpler trinomial which starts with just x^2, there is a simpler way of doing it.

Answer 26

You're welcome.