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October 4th, 2014, 05:40 AM   #1
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How do I solve x^3-5x^2-500 = 0?

Hi, I need help understanding how x^3 - 5x^2 - 500 = 0 reduces to (x-10)(x^2 + 5x + 50). Can someone please explain how they achieved this factorization??

Last edited by sainikbiswas; October 4th, 2014 at 05:44 AM.
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October 4th, 2014, 06:50 AM   #2
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Do you know the formulas for a difference of cubes and a difference of squares? If we have

$\displaystyle a^3-b^3$

then

$\displaystyle a^3-b^3=(a-b)(a^2+ab+b^2)$

and if we have

$\displaystyle a^2-b^2$

then

$\displaystyle a^2-b^2 = (a-b)(a+b)$

For your problem: Rewrite as

$\displaystyle x^3-1000-5x^2+500$

$\displaystyle =(x-10)(x^2+10x+100)-5(x-10)(x+10)$

$\displaystyle =(x-10)(x^2+5x+50)$
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October 4th, 2014, 07:19 AM   #3
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Hi greg1313, thanks for explaining this, it was really helpful. After understanding this, I feel so dumb. Thanks a bunch

Last edited by skipjack; October 4th, 2014 at 06:11 PM.
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October 4th, 2014, 07:21 AM   #4
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$\displaystyle x^3-5x^2-500=x^3-10x^2+5x^2-500=x^2(x-10)+5(x^2-100)=x^2(x-10)+5(x^2-10^2)=
\\\;\\=x^2(x-10)+5(x-10)(x+10)=(x-10)[x^2+5(x+10)]=(x-10)(x^2+5x+50)$
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October 4th, 2014, 07:35 AM   #5
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hi then how can we solve this factorization 12q^3 - 20q^2 -144 = 0
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October 4th, 2014, 07:47 AM   #6
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hi thanks to your teachings solved it
12q^3 - 20q^2 - 144 = 0
3q^3 - 5q^2 - 36 = 0 [Divided by 4 on both sides]
3q^3 - 9q^2 + 4q^2 - 36 = 0
3q^2(q-3) + 4(q^2 - 9) = 0
3q^2(q-3) + 4(q-3)(q+3) = 0
(q-3)(3q^2 + 4q + 12) = 0
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