1953 AHSME Problems/Problem 4

Revision as of 20:18, 1 April 2017 by Xiej (talk | contribs) (Solution)

The roots of $x(x^2+8x+16)(4-x)=0$ are:

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 0,4 \qquad \textbf{(C)}\ 0,4,-4 \qquad \textbf{(D)}\ 0,4,-4,-4 \qquad \textbf{(E)}\ \text{none of these}$

Solution

We can divide both sides by $-1$ to get the equation:

$x(x^2+8x+16)(x-4)=0$

We factor the middle part into $(x+4)^2$.

The equation now becomes $x(x+4)^2(x-4)=0$

The solutions are then $0, -4, 4$. So the answer is $\boxed{\text{C}}$

See Also

1953 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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