Answer:
[tex]x^{3}+\frac{5}{2}x^{2}-\frac{87}{16}x+\frac{9}{4}[/tex]
Step-by-step explanation:
First, let's set up our given zeros.
We are told that [tex]\frac{3}{4}[/tex] is a zero with a multiplicity of 2, meaning it is a squared term. So,
[tex](x-\frac{3}{4} )(x-\frac{3}{4} )[/tex]
Obviously, this can be simplified to [tex](x-\frac{3}{4} )^{2}[/tex].
Now, let's add in the zero of -4.
[tex](x-\frac{3}{4})^{2}(x+4)[/tex]
Now, all we need to do is FOIL. Let's start by FOIL-ing the first two terms.
[tex](x-\frac{3}{4} )(x-\frac{3}{4} )(x+4)[/tex]
[tex](x^{2}-\frac{3}{4}x-\frac{3}{4}x+\frac{9}{16} )(x+4)[/tex]
[tex](x^{2}-\frac{3}{2}x+\frac{9}{16} )(x+4)[/tex]
Now we're going to FOIL what we have left.
[tex]x^{3}-\frac{3}{2} x^{2}+\frac{9}{16}x+4x^{2}-6x+\frac{9}{4}[/tex]
Simplify:
[tex]x^{3}+\frac{5}{2}x^{2}-\frac{87}{16}x+\frac{9}{4}[/tex]
I hope this helps!