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 October 4th, 2014, 05:40 AM #1 Newbie   Joined: Oct 2014 From: Bangalore Posts: 4 Thanks: 4 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.
 October 4th, 2014, 06:50 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond 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 $\displaystyle x^3-1000-5x^2+500$ $\displaystyle =(x-10)(x^2+10x+100)-5(x-10)(x+10)$ $\displaystyle =(x-10)(x^2+5x+50)$ Thanks from MarkFL, topsquark and sainikbiswas
 October 4th, 2014, 07:19 AM #3 Newbie   Joined: Oct 2014 From: Bangalore Posts: 4 Thanks: 4 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.
 October 4th, 2014, 07:21 AM #4 Senior Member     Joined: Apr 2014 From: Europa Posts: 584 Thanks: 177 $\displaystyle x^3-5x^2-500=x^3-10x^2+5x^2-500=x^2(x-10)+5(x^2-100)=x^2(x-10)+5(x^2-10^2)= \\\;\\=x^2(x-10)+5(x-10)(x+10)=(x-10)[x^2+5(x+10)]=(x-10)(x^2+5x+50)$ Thanks from sainikbiswas
 October 4th, 2014, 07:35 AM #5 Newbie   Joined: Oct 2014 From: Bangalore Posts: 4 Thanks: 4 hi then how can we solve this factorization 12q^3 - 20q^2 -144 = 0
 October 4th, 2014, 07:47 AM #6 Newbie   Joined: Oct 2014 From: Bangalore Posts: 4 Thanks: 4 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 Thanks from greg1313, Hoempa, MarkFL and 1 others

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