Question

help please (:

determine whether binomial is a factor of x^3+x^2-10x+8

1. x-2

Answer 1

okay, here are a couple of ways to determine that.

First: can you divide \[x^3+x^2-10x+8\]by \((x-2)\) and get a result which has no remainder? If so, the binomial \((x-2)\) is a factor.

Second: if a binomial \((x-a)\) is a factor of a polynomial \(P(x)\), then \(P(a) = 0\). Does \[(2)^3+(2)^2-10(2)+8=0\]?

Answer 2

let me do a different polynomial as an example.

I will construct it first by multiplying three binomials:
\[(x-1)(x-2)(x+3)=(x-1)(x*x+3x -2x-2*3) = (x-1)(x^2+x-6)\]\[=x*x^2+x*x+x(-6)-x^2-x+6\]\[=x^3-7x+6\]

we can write any polynomial in this fashion when we know its roots. This polynomial has roots of \(x=1\), \(x=2\) and \(x=-3\). Roots are values of the variable that make the polynomial equal \(0\), and they have a one-to-one relationship with the factors.

if you look at the factored form of our polynomial, putting in any of the roots will cause one or more of the product terms to be equal to \(0\), and any number of things multiplied with \(0\) always gives \(0\) as a product.. That means if we know \[P(a)=0\]then \[(x-a)\]is a factor of \[P(x)\]
It is also the case that the opposite is true: if \((x-a)\) is a factor of \(P(x)\) then \(P(a)=0\) .

Let"s test my polynomial with \(x=1\), which corresponds to \((x-1)\):

\[x^3-7x+6\]\[(1)^3-7(1)+6=1-7+6=7-7=0\checkmark\]
\(x=2\), which corresponds to \((x-2)\):
\[(2)^3-7(2)+6=8-14+6=14-14=0\checkmark\]
and finally \(x=-3\) which corresponds to \((x-(-3)) = (x+3)\):
\[(-3)^3-7(-3)+6=-27+21+6=-27+27=0\checkmark\]

I will not try to illustrate polynomial long division here, but if you took my polynomial and divided it by any of the factors, you would get a simpler polynomial which is equal to the remaining factors multiplied together.

I hope that helps eliminate some of the confusion, but if not, ask a question here and I will try to answer it.