I have a question that tells me to use synthetic division on a polynomial to get the fully factored form... the polynomial is P(x)=2x^3+7x^2+4x-4 but thats it what do I divide it by?
@Directrix
@steve816
@mathmate
@inkyvoyd
Hey, check out this website. It should help. https://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/Polynomials-and-rational-expressions.faq.question.199776.html
thanks @steve816
\(\color{#0cbb34}{\text{Originally Posted by}}\) @steve816 Hey, check out this website. It should help. https://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/Polynomials-and-rational-expressions.faq.question.199776.html \(\color{#0cbb34}{\text{End of Quote}}\)
@bonnieisflash1.0 What was that for?
\(\color{blue}{\text{Originally Posted by}}\) @steve816 @bonnieisflash1.0 What was that for? \(\color{blue}{\text{End of Quote}}\) nothing
\(P(x)=2x^3+7x^2+4x-4\) When you have a cubic, and a relatively small number as the constant term, it would be worthwhile to try +1 or -1 as a root. To check if x-1 is a factor, add up all the coefficients (2+7+4-4)=9 \(\ne\)0, so x-1 is not a factor, i.e. x=1 is not a root. Similarly, try (x+1), but this time, add up all the even powered coefficients, and \(subtract\) the odd powered coefficients. -2+7-4-4=-3\(\ne\)0, so (x+1) is not a factor (i.e. x=-1 is not a root). Since the constant term is only 4, the possible factors are +1,-1,+2, -2, +4, -4. We have already eliminated two of them. So try put x=2, or calculate f(2)=2*2^3+7*2^2+4*2-4, by inspection, we know that f(x) > 0, or =16+28+8-4 >0 Now try f(-2)=2*2^3+7*2^2+4*2-4 we only need to reverse the sign of the odd powered terms as before, so f(-2)=-16+28-8-4=0, so (x+2) is a factor (remember x+2=0 => x=-2) Now you can proceed to do the synthetic division and get the quadratic. It would be useful as well for you to get familiar with the factor theorem and the Descartes rule of signs. http://www.purplemath.com/modules/factrthm.htm http://www.purplemath.com/modules/drofsign.htm
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