Question

Write a polynomial function that has the given zero

3/4 multiplicity of 2, -4

Answer 1

Answer:

$x^{3}+\frac{5}{2}x^{2}-\frac{87}{16}x+\frac{9}{4}$

Step-by-step explanation:

First, let's set up our given zeros.

We are told that $\frac{3}{4}$ is a zero with a multiplicity of 2, meaning it is a squared term. So,

$(x-\frac{3}{4} )(x-\frac{3}{4} )$

Obviously, this can be simplified to $(x-\frac{3}{4} )^{2}$.

Now, let's add in the zero of -4.

$(x-\frac{3}{4})^{2}(x+4)$

Now, all we need to do is FOIL. Let's start by FOIL-ing the first two terms.

$(x-\frac{3}{4} )(x-\frac{3}{4} )(x+4)$

$(x^{2}-\frac{3}{4}x-\frac{3}{4}x+\frac{9}{16} )(x+4)$

$(x^{2}-\frac{3}{2}x+\frac{9}{16} )(x+4)$

Now we're going to FOIL what we have left.

$x^{3}-\frac{3}{2} x^{2}+\frac{9}{16}x+4x^{2}-6x+\frac{9}{4}$

Simplify:

$x^{3}+\frac{5}{2}x^{2}-\frac{87}{16}x+\frac{9}{4}$

I hope this helps!