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I'm having trouble with this homework problem:

Find an​ $n^\text{th}$-degree polynomial function with real coefficients satisfying the given conditions. If you are using a graphing​ utility, use it to graph the function and verify the real zeros and the given function value.

  • $n = 3$
  • $-3$ and $6+4i$ are zeros
  • $f(-1) = 130$

Here's my work so far:

$$\begin{align} f(x)&=a_n(x+3)(x-6+4i)(x-6-4i)\\ &=a_n(x+3)(x(x-6-4i)-6(x-6-4i)+4i(x-6-4i)) \\ &=a_n(x+3)(x^2-6x-4ix)-(6+36+24i)+(4ix-24i-4i^2) \\ &=a_n(x+3)(x^2-12x+4+36) \\ &=a_n(x+3)(x^2-12x+40) \\ \end{align}$$

I can't seem to figure out this calculation. The book shows the correct calculation is $x^2-12x+52$. It doesn't show the work, just the answer. What am I doing wrong?

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You made a mistake in this step:

$$f(x)=a_n(x+3)\left(x^2-6x-4ix\right)-(6+36+24i)+\left(4ix-24i-4i^2\right)$$

You've distributed wrong and neglected some parentheses. Backing up to the previous step:

$$f(x)=a_n(x+3)\big[x(x-6-4i)-6(x-6-4i)+4i(x-6-4i)\big]$$

Once you distribute you should get:

$$f(x)=a_n(x+3)\color{red}{\big[}\left(x^2-6x-4ix\right)\color{blue}{-(6x-36-24i)}+\left(4ix-24i\color{green}{-16i^2}\right)\color{red}{\big]}$$

Notice the parentheses in red, they are important! Also notice the blue expression. Here, $6$ was distributed to each term, and then the subtraction applied to the whole quantity. Finally, you forgot to distribute the $4i$ to the final $-4i$ to create $-16i^2$ in green. Thus:

$$\begin{align} f(x)&=a_n(x+3)\left(x^2-6x-4ix-6x+36+24i+4ix-24i-16i^2\right) \\ &=a_n(x+3)\left(x^2-12x+36-16i^2\right) \\ &=a_n(x+3)\left(x^2-12x+52\right) \end{align}$$

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  • $\begingroup$ Oh!! I see... I also didn't multiply $4i*4i$ properly either. It's been a few years since I've taken any math classes. Thanks for clearing it up! $\endgroup$ – Haruku Mar 24 '18 at 23:41
  • $\begingroup$ @Haruku Yes, that was also something you missed. Glad I could help. $\endgroup$ – Andrew Li Mar 24 '18 at 23:41

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