Factoring Polynomial functions

Question

Determine if f(x)=(x-1)(x-6)(x+3) is the correct factorization of f(x)=x^3+7x^2+4x-12. Explain.

Answer 1

Just plug in thesuspected roots, 1, 6 and -3 into the cubic given polynomial, set equal to zero. 1 works, 6 and -3 don't 1 is a root, 6 and -3 are not, so (x-6) and (x+3) are not factors.

Since you know 1 is a root, (x-1) is a factor. Just change the sign of the root to get the factor

divide x-1 into the cubic and you'll get a quadratic with a constant term of 12, because the cubic has a constant term of 12, and you're dividing by x-1, the 1 will go into 12 exactly 12 times.

If there are integer roots, they will multiply to 12. They must be either 6 and 2 or 4 and 3 or 12 and 1, those are factors of 12.

plug those numbers into the quadratic or the cubic equation. -6 and- 2 work, 4 and 3 don't, 12 doesn't.

the other two factors are (x+6) and (x+2) for root of -2, -8+28-8-12 = 0 for the original cubic polynomial

for -6, -216+252-24-12=0.for the original cubic polynomial set equal to zero. A factor just changes the sign of the root. -6 corresponds to (x+6) -2 corresponds to (x+2)

(x-1)(x-6)(x-2) multiplied together to give you the original cubic equation.

Answer 2

There are a few ways you can verify whether the two functions are the same.

Method 1: If you are allowed to use a graphic calculator, you may enter both functions to the calculator to see if the resulting graphs are identical.

Method 2: Algebraically, you can multiple the polynomials in each parentheses such as the demonstration below:

f(x)=[x*x+x*(-6)+(-1)*x+(-1)*(-6)](x+3)

=[x²-6x-x+6](x+3) (combine the like terms)

=[x²-7x+6](x+3)

=x²*x+(-7x)*x+6*x+3*x²+(-7x)*3+6*3

=x³-7x²+6x+3x²-21x+18

=x³-4x²-15x+18

As you can see, the two polynomials aren't the same. Therefore, f(x)=(x-1)(x-6)(x+3) is NOT the correct factorization of f(x)=x³+7x²+4x-12

Method 3: Instead of distributing the factors, you can also try to factor the polynomial f(x)=x³+7x²+4x-12.

Sometimes, it's easier to distribute than factor, other times it might be faster to factor.