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Consider a cubic polynomial function $y=f(x)$ with the following properties:

$f(x) \ge 0$ only for $x=-1$ and $x\ge3$

when $f(x)$ is divided by $(x-4)$ the remainder is $50$.

Find the equation for $y=f(x)$.

I'm not entirely certain on how to begin this, but I can identify that it is a remainder theorem question. With some of the information I think I can make out that $f(x)=50$. And I know that $\dfrac{f(x)}{ x-4}$ will give me the quotient $q(x)$ with a remainder of $50$ and vice versa $f(x)=(x-4)q(x) + 50$.

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  • $\begingroup$ Check the question, and I think you may have meant $f(4)=50$. $\endgroup$ – Macavity Oct 22 '14 at 17:53
  • $\begingroup$ I suppose you could put it that way, but this was how the question was written. It has a remainder of 50 when it is divided by x-4. $\endgroup$ – Ashton Oct 22 '14 at 19:01
  • $\begingroup$ You have written $f(x)=50$. That is what I refe to. $\endgroup$ – Macavity Oct 23 '14 at 2:34
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Hint: the first condition gives $(x+1)^2$ as a factor, while the second gives $x-3$. So we must have $f(x)=a(x+1)^2(x-3)$ for some non-negative constant $a$. Use the remainder theorem to find $a$.

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  • $\begingroup$ *Why does it become (x+1)^2? I understand where the factor comes from, but I simply am not sure why it is to the power of two. $\endgroup$ – Ashton Oct 22 '14 at 19:30
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    $\begingroup$ Observe that $f(x) < 0$ if $x < -1$ and if $-1 < x < 3$. Hence, $f(x) = 0$ at $x = -1$, but is negative to both the left and right of $x = -1$. Since $f(x)$ is a cubic polynomial, its roots have multiplicity $1$, $2$ or $3$. If the multiplicity of the root at $x = -1$ were $1$ or $3$, the graph would change sign at $x = -1$. It does not. Therefore, $x = -1$ is a root of multiplicity $2$. Since $3$ is also a root, the polynomial has the form $f(x) = a(x + 1)^2(x - 3)$. $\endgroup$ – N. F. Taussig Oct 22 '14 at 21:25

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