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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^35x^2500 = 0?
Hi, I need help understanding how x^3  5x^2  500 = 0 reduces to (x10)(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 $\displaystyle a^3b^3$ then $\displaystyle a^3b^3=(ab)(a^2+ab+b^2)$ and if we have $\displaystyle a^2b^2$ then $\displaystyle a^2b^2 = (ab)(a+b)$ For your problem: Rewrite as $\displaystyle x^310005x^2+500$ $\displaystyle =(x10)(x^2+10x+100)5(x10)(x+10)$ $\displaystyle =(x10)(x^2+5x+50)$ 
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  $\displaystyle x^35x^2500=x^310x^2+5x^2500=x^2(x10)+5(x^2100)=x^2(x10)+5(x^210^2)= \\\;\\=x^2(x10)+5(x10)(x+10)=(x10)[x^2+5(x+10)]=(x10)(x^2+5x+50)$ 
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(q3) + 4(q^2  9) = 0 3q^2(q3) + 4(q3)(q+3) = 0 (q3)(3q^2 + 4q + 12) = 0 