Question

I'm no good at factoring. Suggestions?

Answer 1

http://www.purplemath.com/modules/factnumb.htm
Read the whole page, all topics will help you. :)

Answer 2

What shall I do with the equation 24x^2-8x?

Answer 3

take 8x out:
8x(3x-1)

Answer 4

Could you help me with this one? x^3-5x^2-9x+45

Answer 5

first take a look at what can u factor out.. first part looks like x^2 can be taken out and in the 2nd, 9 can be taken out...

Answer 6

x^2(x-5) + 9(-x+5), now if you observe that once you take -1 out in 9(-x+5) you will get an (x-5) there.. so

Answer 7

x^2(x-5) + 9(-x+5)
x^2(x-5) -9(x-5)
Now take x-5 out:
(x-5)[x^2-9]

Answer 8

now x^2 - 9 is in the form a^2 - b^2 = (a-b)(a+b)
x^2-9 = x^2 - 3^2 = (x-3)(x+3)
(x-5)[x^2-9] = (x-5)(x-3)(x+3)

Answer 9

Yeah now I'm lost.

Answer 10

where exactly are you lost? which step?

Answer 11

Where did the x^2-3^2 come from?

Answer 12

9 = 3^2

Answer 13

Wouldn't the two (x-5)s cancel out? I figured you'd be left with x^2-9 and take the square root of the nine to end up with x^2(x+3)(x-3).

Answer 14

U can't just drop it, I took it out..

Answer 15

x^2(x-5) -9(x-5)=(x-5)[x^2-9] = (x-5)(x-3)(x+3)

Answer 16

So one of the x-5s goes away, then? And you're left with (x-5)(x-3)(x+3) <- that bit?

Answer 17

yes and the x-5 doesnt go away anywhere, I just take it outside of big parenthesis, just like you would do:
3x-3y = 3(x-y), see the same term comes out,
x^2(x-5) -9(x-5) = (x-5)(x^2 * 1 - 9 * 1)

Answer 18

x^2(x-5) -9(x-5) = (x-5)(x^2 * 1 - 9 * 1) = (x-5)(x^2-9)

Answer 19

I see.

Answer 20

any more questions?