Question

Find the equation for a polynomial f(x) that satisfies the following:Degree 5• Root of multiplicity 1 at x = 1• Root of multiplicity 2 at x = • Root of multiplicity 2 at x = -2y-intercept of (0, -32)f(x) =

Answer 1

`2x^5-2x^4-16x^3+16x^2+32x-32=0`

1) Since the degree is the 5th and we have multiplicities of 1 and 2, it means that we have identical roots (multiplicity 2), so let's start by writing out a factored form for that:

`y=a(x-1)(x-2)(x-2)(x+2)(x+2)`

Each parentheses has one root following the pattern (x-x_1)(x-x_2)...

Note that we were told about another point (0, -32).

2) So let's plug that point to find the leading coefficient:

`\begin{gathered} y=a(x-1)(x-2)(x-2)(x+2)(x+2) \\ -32=a(-16) \\ -16a=-32 \\ -(16)/(-16)a=(-32)/(-16) \\ a=2 \end{gathered}`

So we can plug into that a=2, then expand that:

`\begin{gathered} 0=a(x-1)(x-2)(x-2)(x+2)(x+2) \\ 0=2(x-1)(x-2)(x-2)(x+2)(x+2) \\ 2\mleft(x-2\mright)^2\mleft(x+2\mright)^2\mleft(x-1\mright)=0 \\ 2(x^2-4x+4)(x^2+4x+4)(x-1)= \\ 2x^5-2x^4-16x^3+16x^2+32x-32 \end{gathered}`

3) Hence, the answer is:

`2x^5-2x^4-16x^3+16x^2+32x-32=0`