Question

help! i don't get factoring at all!

Answer 1

What kind of factoring do you mean?

Answer 2

Factoring gcd from a sum? Or factoring quadratics, etc

Answer 3

(x-3)(x^2+7x+4)

Answer 4

It's a similar process, but there are subtle differences.

Answer 5

if you can multiply, you can factor (or unmultiply)

Answer 6

That looks like it's factored already.

Answer 7

I mean find the product

Answer 8

Oh, that's unfactoring. Also called distribution.

Do you know how to do this?

\[a(b+c - d)\]

Answer 9

ab+ac-ad ??

Answer 10

That's right!

Answer 11

This works exactly the same way

Answer 12

\[(x-3)(x^2+7x+4)\]\[= x^2(x-3) + 7x(x-3) + 4(x-3)\]

Answer 13

x-3 is just like the 'a' from the previous one

Answer 14

Except that then you have to distribute again with each term

Answer 15

???
wouldn't x * x^2 be x^3 for the first step?

Answer 16

Yes

Answer 17

But you also have to distribute it to the -3

Answer 18

so confused...

Answer 19

Why? You did it with a, now do it with other things

Answer 20

but there was only a single term. now there are two terms.

Answer 21

There was 3 terms in your a example

Answer 22

i mean to the left

Answer 23

Ok watch this:
Let a = x-3
\[(x-3)(x^2+7x+4)\]\[=a(x^2 + 7x + 4)\]

Answer 24

Now do it from there with a.

Answer 25

x^3+7x^2+4x-x^2-21x-12

Answer 26

Close, but not quite

Answer 27

Can you do what I said with the a?

Answer 28

ax^2+7ax+4a

Answer 29

Right. Now just turn each of those a's back one at a time and multiply it out:

\[= x^2(x-3) + 7xa + 4a\]

Answer 30

so distribute the x^2

Answer 31

x^3-3x^2+7x^3a+4ax^2

Answer 32

no no. the 7xa and 4a terms don't change. They aren't being multiplied.

Answer 33

The only term that you distribute to is the one being multiplied by x^2

Answer 34

x^3-3x^2+7xa+4a

Answer 35

Right. Now lets do the same thing to the 7xa term:

\[=x^3 - 3x^2 + 7x(x-3) + 4a\]

Answer 36

x^3-3x^2+7x^2-21x+4a

Answer 37

Yep. Now combine those two x^2 terms to simplify

Answer 38

x^3+4x^2-21x+4a

Answer 39

it really takes that many steps?

Answer 40

It doesn't have to. This is just to get you comfortable with the process

Answer 41

I'll recap at the end and you can see

Answer 42

Now convert the 4a into 4(x-3) and distribute

Answer 43

then combine any like terms you can find

Answer 44

x^3+4^2-21x+4(x-3)
+4x-12
x^3+4^2-17x-12

Answer 45

am i close?

Answer 46

Yep. Exactly right.
Now here's the short version:

\[(x-3)(x^2 + 7x + 4)\]\[=x^2(x-3) + 7x(x-3) + 4(x-3)\]\[=x^3 - 3x^2 + 7x^2 - 21x + 4x - 12\]\[=x^3 + 4x^2 - 17x - 12\]

Answer 47

Try to follow it, step by step and see how it's similar to what we did before but without all the 'a' stuff.