Question

Jamison graphs the function ƒ(x) = x4 − x3 − 19x2 − x − 20 and sees two zeros: −4 and 5. Since this is a polynomial of degree 4 and he only sees two zeros, he determines that the Fundamental Theorem of Algebra does not apply to this equation. Is Jamison correct? Why or why not?
A) No, the root 3 has multiplicity of 3.
B) No, there are two imaginary solutions.
C) No, the root −2 has multiplicity of 3.
D) Yes, the Fundamental Theorem of Algebra does not apply to this equation.

Answer 1

We are given polynomial function f(x)=$x^4-x^3-19x^2-x-20$

We are given two zeros -4 and 5.

Therefore, x=-4 and x=5 is given.

So, the two factors of the polynomial will be (x+4)(x-5).

Let us write some steps to factor the given polynomial.

Let us take above factor (x+4) first.

On factoring (x+4) we get factors

$x^4-x^3-19x^2-x-20=\left(x+4\right)\left(x^3-5x^2+x-5\right)$

$\mathrm{:Let \ us \ factor}\:x^3-5x^2+x-5 \ now.$

Grouping

$=\left(x^3-5x^2\right)+\left(x-5\right)$

Factoring out gcf from each group, we get

$=x^2\left(x-5\right)+\left(x-5\right)$

$=\left(x-5\right)\left(x^2+1\right)$

So, the final factored form of given polynomial will be

$x^4-x^3-19x^2-x-20=\left(x+4\right)\left(x-5\right)\left(x^2+1\right)$

For first to factors (x+4) and (x-5) we have given roots: -4 and 5.

Let us find the root of third factor we got.

$x^2+1=0$

Subtracting both sides by 1.

x^2 = - 1.

On taking square root on both sides, we get a square root(-1)

$x=\sqrt{-1}=$+ i and -i.

Those are imaginary solutions.

Therefore, correct option is B) No, there are two imaginary solutions.