# Thread: Complex Numbers, Find The Unknown - Help

1. ## Complex Numbers, Find The Unknown - Help

I am stuck on question 11, it is below:

If anyone is able to do this it would be great, but if you can somehow show me the working out/steps involved. Though, any tips/help would also be appreciated.

thanks,
Matt.

EDIT: Sorry! I posted in the wrong section!

2. Hello,

\begin{aligned}P(x)&=ax^4+bx^3+cx^2+dx+e \\
&=(5x^2+6x+5)(fx^2+gx+8) \end{aligned}

Let $z=-2+2i=2(-1+i)$.

----------------

We know that $P(z)=0$.

So $(5z^2+6z+5)(fz^2+gz+8)=0$.

Because you know all the coefficients of $5z^2+6x+5$, calculate if it is equal to 0.
After some calculations (that you will do), it seems that it is not equal to 0.

Therefore, $fz^2+gz+8=0$.

Substituting z by its value, and after some calculations (that you will do), we get :

$2ig-2g-8if+8=0$

$(-2g+8)+i(2g-8f)=0$

This is a complex number. But a complex number equals to 0 if and only if the real and the imaginary parts are both equal to 0, that is to say :

$\left\{ \begin{array}{ccc}-2g+8&=&0 \\ 2g-8f&=&0 \end{array}\right.$

This gives f and g

--------------

I have to go out right now, so if no one else does, I'll help you find a,b,c,d, and e

3. Thanks, Moo! That helped me a lot

I have solved a,b,c,d,e my self, thanks again for the help!

- Matt.

4. Originally Posted by Phatmat
I am stuck on question 11, it is below:

If anyone is able to do this it would be great, but if you can somehow show me the working out/steps involved. Though, any tips/help would also be appreciated.

thanks,
Matt.

EDIT: Sorry! I posted in the wrong section!
Since the coefficients of p(x) are real, z = 2 - 2i is also a root by the conjugate root theorem.

Hence a quadratic factor of p(x) is $(x - [-2 + 2i])(x - [-2 - 2i]) ([x + 2] - 2i)([x + 2] + 2i) = (x+2)^2 + 4 = x^2 + 4x + 8$.

So you now have f = 1 and g = 4 in the second expression. Now expand the second expression to get the coefficients a, b, c, d, e in the first.

5. Originally Posted by Phatmat
I am stuck on question 11, it is below:
We have

$P(x) = ax^4 + bx^3 + cx^2 + dx + e = \left(5x^2 + 6x + 5\right)\left(fx^2 + gx + 8\right),\;a,b,c,d,e,f,g\in\mathbb{R}$

First, note that a polynomial of degree 4 has at most four distinct roots (this follows from the fundamental theorem of algebra), so we should get at most four distinct solutions to the equation $P(x) = 0$. Also note that the nonreal roots of a polynomial with real coefficients always come in conjugate pairs. So if $-2 + 2{\rm i}$ is a root, then so must $-2 - 2{\rm i}$.

We may easily find the other two roots thus:

$5x^2 + 6x + 5 = 0\Rightarrow x = \frac{-3\pm4{\rm i}}5$

This means that the first two roots we identified must come from the $\left(fx^2 + gx + 8\right)$ factor. So we have

$\left(fx^2 + gx + 8\right) = C\left(x + 2 - 2{\rm i}\right)\left(x + 2 + 2{\rm i}\right)$

$= C\left(x^2 + 4x + 8\right)$ for some $C\in\mathbb{R}$

Equating coefficients, we find that $C = 1$, so

$f = 1, g = 4$

Then we have

$ax^4 + bx^3 + cx^2 + dx + e = \left(5x^2 + 6x + 5\right)\left(x^2 + 4x + 8\right)$

$=5x^4+26x^3+69x^2+68x+40$

$\Rightarrow\left\{\begin{array}{rcl}
a & = & 5\\
b & = & 26\\
c & = & 69\\
d & = & 68\\
e & = & 40
\end{array}\right.$

If you like, you can do the substitutions and verify that these are the correct values.

6. Thanks for the help, those are the correct answers

,

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