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(C) Let $f(x) = x^4+4x^3-16x-16$. To find the multiple roots,we find the derivative $f'(x) = 4x^3+12x^2-16$. Setting $f'(x) = 0$,we get $4(x^3+3x^2-4)

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$x^2+x-2=0$

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(C) Let $f(x) = x^4+4x^3-16x-16$. To find the multiple roots,we find the derivative $f'(x) = 4x^3+12x^2-16$. Setting $f'(x) = 0$,we get $4(x^3+3x^2-4) = 0$. By inspection,$x=1$ is a root of $f'(x)$,so $(x-1)$ is a factor. Dividing $x^3+3x^2-4$ by $(x-1)$,we get $(x-1)(x^2+4x+4) = (x-1)(x+2)^2 = 0$. The roots of $f'(x)=0$ are $x=1$ and $x=-2$. Checking these in $f(x)$: $f(1) = 1+4-16-16 = -27 
eq 0$. $f(-2) = (-2)^4+4(-2)^3-16(-2)-16 = 16-32+32-16 = 0$. Since $f(-2)=0$ and $f'(-2)=0$,$x=-2$ is a multiple root. For $x=-2$ to be a root of a quadratic equation $x^2+ax+b=0$,we check the options: $A: (-2)^2+2(-2)-3 = 4-4-3 = -3 
eq 0$. $B: (-2)^2-3(-2)+2 = 4+6+2 = 12 
eq 0$. $C: (-2)^2+(-2)-2 = 4-2-2 = 0$. $D: (-2)^2-4(-2)+3 = 4+8+3 = 15 
eq 0$. Thus,the equation is $x^2+x-2=0$.

The equation having the multiple root of the equation $x^4+4x^3-16x-16=0$ as its root is

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