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 March 2nd, 2014, 06:56 AM #1 Senior Member     Joined: Feb 2014 From: Louisiana Posts: 156 Thanks: 6 Math Focus: algebra and the calculus How to use the factor theorem for a cubic polynomial How exactly is the factor theorem used for a cubic polynomial 2x^3 + x^2 + x - 1, where the leading coefficient is not 1.
 March 2nd, 2014, 08:15 AM #2 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: How to use the factor theorem for a cubic polynomial I am assuming you want the complete factorization. Here is one way to do it, $P(x)= 2x^3 + x^2 + x - 1 = a(x - r_1)(x - r_2)(x -r_3)$ Using the Rational Root Theorem and trial and error, we see $\ \ r_1= x= \frac{ 1}{2} \ $/extract_itex] is the only rational root. Next, do long division or synthetic division to get the quotient polynomial. $\frac{ 2x^3 + x^2 + x - 1 }{x - \frac{ 1}{2}} \= \ 2(x^2 + x + 1)$ So far, you have this $2x^3 + x^2 + x - 1 \= \ 2(x - \frac{ 1}{2}) (x^2 + x + 1)$ You can stop now if you don't need the complete factoring, otherwise continue. Using the quadratic formula to get the other two roots, you should get, $r_2= \frac{ -1 + i \sqrt{3}}{2}$ $r_3= \frac{ -1 - i \sqrt{3}} {2}$ Putting it all together gives, $2x^3 + x^2 + x - 1 \= \ 2 $$x \ - \ \frac{ 1}{2}$$ \[x \ - \ $$\frac{ -1 + i \sqrt{3}}{2}$$$ $x \ - \ $$\frac{ -1 - i \sqrt{3}} {2}$$$$ I leave to you the pleasure of multiplying out the RHS and confirming it is equal to the LHS.

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