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Find a third degree polynomial with real coefficients which has $6$ and $-5+2i$ as zeros and $f(2)=-63$

I tried to substitute in the equation $$ f(x)=ax^3 +bx^2 + cx +d = 0$$ by the two given zeros and use $f(2)= -63$. But I think we need another equation to determine all the coefficients.

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    $\begingroup$ Note that putting $-5+2i$ in the equation is worth two equations here because you can equate real and imaginary parts and $a,b,c,d$ are all real. $\endgroup$ – Mark Bennet Dec 30 '17 at 22:19
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$$\begin{align}f(x)&=a(x-\alpha)(x-\beta)(x-\gamma)\end{align}$$ Using complex conjugate theorem $$x=-5+2i\text{ is a root}\implies x=-5-2i\text{ is a root}\\(x+5)^2=-4\\x^2+10x+29=0$$ The other factor is $(x-6)$

So far, we have

$$\begin{align}a(x-6)(x^2+10x+29)&=0\end{align}$$

Use the fact that $f(2)=-63$

$$\begin{align}a(-4)(4+20+29)&=-63\\212a&=63\\a&=\dfrac{63}{212}\end{align}$$

$$f(x)=\dfrac{63}{212}(x-6)(x^2+10x+29)$$

is the required polynomial.

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    $\begingroup$ it should be $x-6$ not $x+6$ so $a$ becomes positive $\endgroup$ – daulomb Dec 30 '17 at 22:21
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    $\begingroup$ Notice that $f(2)=-63$, so $a=\frac{63}{212}$. $\endgroup$ – daulomb Dec 30 '17 at 22:27
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Since for a polynomial with real coefficients the complex roots must be conjugate, we have:

$$p(x)=a(x-6)(x+5-2i)(x+5+2i)$$

The value for "a" is given by the condition $p(2)=-63$.

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Hint : $-5-2i$ must be the third root of the polynomial

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Hint: If $p $ is a polynomial with real coefficients, and $p (z)=0$ for some $z\in\mathbb C $, what is $p (\overline z) $?

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