anonymous
  • anonymous
I'm no good at factoring. Suggestions?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
http://www.purplemath.com/modules/factnumb.htm Read the whole page, all topics will help you. :)
anonymous
  • anonymous
What shall I do with the equation 24x^2-8x?
bahrom7893
  • bahrom7893
take 8x out: 8x(3x-1)

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anonymous
  • anonymous
Could you help me with this one? x^3-5x^2-9x+45
bahrom7893
  • bahrom7893
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...
bahrom7893
  • bahrom7893
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
bahrom7893
  • bahrom7893
x^2(x-5) + 9(-x+5) x^2(x-5) -9(x-5) Now take x-5 out: (x-5)[x^2-9]
bahrom7893
  • bahrom7893
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)
anonymous
  • anonymous
Yeah now I'm lost.
bahrom7893
  • bahrom7893
where exactly are you lost? which step?
anonymous
  • anonymous
Where did the x^2-3^2 come from?
bahrom7893
  • bahrom7893
9 = 3^2
anonymous
  • anonymous
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).
bahrom7893
  • bahrom7893
U can't just drop it, I took it out..
bahrom7893
  • bahrom7893
x^2(x-5) -9(x-5)=(x-5)[x^2-9] = (x-5)(x-3)(x+3)
anonymous
  • anonymous
So one of the x-5s goes away, then? And you're left with (x-5)(x-3)(x+3) <- that bit?
bahrom7893
  • bahrom7893
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)
bahrom7893
  • bahrom7893
x^2(x-5) -9(x-5) = (x-5)(x^2 * 1 - 9 * 1) = (x-5)(x^2-9)
anonymous
  • anonymous
I see.
bahrom7893
  • bahrom7893
any more questions?

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