My Math Forum Factorization help

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 March 9th, 2010, 10:20 AM #1 Newbie   Joined: Mar 2010 Posts: 20 Thanks: 0 Factorization help I know that: x(32-2x)(40-2x) - 2016 = 0 can be factorised into: 4(x-2)(x^2 - 34x +252) but can someone please show me the steps to how you get that exactly? thank you .. very appreciated.
 March 9th, 2010, 01:53 PM #2 Senior Member   Joined: Mar 2007 Posts: 428 Thanks: 0 Re: Factorization help Are you not asking someone to complete the problem you just stated and were helped with? You are asking how to factor a cubic, x(16 - x)(20 - x) - 504 . I'd expand that to standard form Ax^3 + Bx^2 + Cx + D, then there are several methods for factoring the cubic, the easiest of which, when it applies [and it does here] is The Factor Theorem. When you have studied that method, and used it to find the first factor, you will use Polynomial Long Division to find the second quadratic factor. Again, Google, and a little study, is you friend. Since the method is standard, there's no sense anyone spending the time rewriting the book, so why not let Google help you if it is not available to you in a textbook?
 March 9th, 2010, 03:31 PM #3 Senior Member   Joined: Dec 2006 Posts: 166 Thanks: 3 Re: Factorization help When you expand the equation, you get $x^3-36x^2+320x-504=0$ With integer coefficient cubic equation, you may try to find the rational roots of the form a/b with a factor of constant term as the numerator and a factor of the coefficient of $x^3$ as the denominator. In this case we need only to consider the factors of $504= 2^3\cdot 3^2 \cdot 7$ Since $2^3-36\cdot 2^2 + 320\cdot 2 - 504= 0$, we find a root $x= 2$ at the first try. Now $\frac{x^3-36x^2+320x-504}{x-2}= x^2 - 34 x + 252$ and your result follows
 March 9th, 2010, 07:31 PM #4 Newbie   Joined: Mar 2010 Posts: 20 Thanks: 0 Re: Factorization help thanks so much guys

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