Answer:
See below.
Step-by-step explanation:
So, we have the zeros -4 with a multiplicity of 1, zeros 2 with a multiplicity of 3, and f(0)=64.
Recall that if something is a zero, then the equation must contain (x - n), where n is that something. In other words, for a polynomial with a zero of -4 with a multiplicity of 1, then (x+4)^1 must be a factor.
Therefore, (x-2)^3 (multiplicity of 3) must also be a factor.
Lastly, f(0)=64 tells that when x=0, f(x)=64. Don't simply add 64 (like what I did, horribly wrong). Instead, to keep the zeros constant, we need to multiply like this:
In other words, we will have:
[tex]f(x)=(x+4)(x-2)^3\cdot n[/tex], where n is some value.
Let's determine n first. We know that f(0)=64, thus:
[tex]f(0)=64=4(-2)^3\cdot n[/tex]
[tex]64=-32n, n=-2[/tex]
Now, let's expand:
Expand:
[tex]f(x)=(x+4)(x^2-4x+4)(x-2)(-2)[/tex]
[tex]f(x)=(x^2+2x-8)(x^2-4x+4)(-2)[/tex]
[tex]f(x)=(x^4-4x^3+4x^2+2x^3-8x^2+8x-8x^2+32x-32)(-2)[/tex]
[tex]f(x)=-2x^4+4x^3+24x^2-80x+64[/tex]
This is the simplest it can get.
