[tex]2x^5-2x^4-16x^3+16x^2+32x-32=0[/tex]
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:
[tex]y=a(x-1)(x-2)(x-2)(x+2)(x+2)[/tex]
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:
[tex]\begin{gathered} y=a(x-1)(x-2)(x-2)(x+2)(x+2) \\ -32=a(-16) \\ -16a=-32 \\ -\frac{16}{-16}a=\frac{-32}{-16} \\ a=2 \end{gathered}[/tex]
So we can plug into that a=2, then expand that:
[tex]\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}[/tex]
3) Hence, the answer is:
[tex]2x^5-2x^4-16x^3+16x^2+32x-32=0[/tex]
