Question

Which answer best describes the complex zeros of the polynomial function? f(x)=x3+x2+10x+10 The function has three real zeros. The graph of the function intersects the x-axis at exactly three locations. The function has one real zero and two nonreal zeros. The graph of the function intersects the x-axis at exactly two locations. The function has two real zeros and one nonreal zero. The graph of the function intersects the x-axis at exactly one location. The function has one real zero and two nonreal zeros. The graph of the function intersects the x-axis at exactly one location.

Answer 1

ANSWER

The function has one real zero and two non real zeros

EXPLANATION

GRAPHICAL APPROACH

Since the graph of

$f(x)=x^3+x^2+10x+10$

intersects the x-axis at only one point, it has only one real root.

But the polynomial has degree 3, and hence has three roots. Due to the fact that only one of the roots are real, then the remaining two are non real. Therefore the non real roots are imaginary that is why they don't intersect the x-axis.

NON-GRAPHICAL APPROACH

$f(x)=x^3+x^2+10x+10$

$f(-1)=(-1)^3+(-1)^2+10(-1)+10$

$f(-1)=-1+1-10+10$

$f(-1)=0$

Since $f(-1)=0$, it means $(x+1)$ is a factor.

We now use long division to obtain the other factor.

$f(x)=x^3+x^2+10x+10=(x+1)(x^2+10)$

We can see that the discriminant of the quadratic factor will be less than zero. This gives two non real roots.

$B^2-4AC=0^2-4(1)(10)=-40<\:0$

Also by directly equating the quadratic factor to zero, you see clearly that they are two non real roots.

$x=\pm\sqrt{-10}$

See diagram.

Answer 2

Just took the quiz and can confirm that the answer is indeed:

"The function has one real zero and two nonreal zeros. The graph of the function intersects the x-axis at exactly one location."

Hope I could help :)