Finding a Quadratic Factor of z⁴+16

[XtremePhysX]

Yes you're right, but I don't understand why they'd make you stop there.

It's a 2 mark question in an MX2 paper, I think you know what MX2 is ;)

 

[ehild]

The second one is obvious: The product of the other two linear factors: z²-2√2 z+4. Check if multiplying these two factors you really get z⁴ +16.

ehild

 

[XtremePhysX]

The second one is obvious: The product of the other two linear factors: z²-2√2 z+4. Check if multiplying these two factors you really get z⁴ +16.

ehild

I completely understand it now, thanks to everyone, much appreciated.

 

[Mentallic]

It's a 2 mark question in an MX2 paper, I think you know what MX2 is ;)

Oh, well then a lot of time would've been wasted finding the other complex roots as well if only one quadratic factor is necessary. Look at how quick and easy the solution can be:

$$z^4+16=0$$
$$z^4=-16$$
$$z^4=16cis(\pi+2k\pi)$$
$$z=2cis(\frac{\pi(1+2k)}{4})$$ for $k=-2,-1,0,1$

Now let's just consider one of the complex conjugate pairs:

$$z_1=2cis(\pi/4)$$
$$z_2=2cis(-\pi/4)=\bar{z_1}$$

Therefore the quadratic factor is,

$$(z-z_1)(z-\bar{z_1})$$
$$=z^2-(z_1+\bar{z_1})z+z_1\bar{z_1}$$
Where $$z_1+\bar{z_1}=2Re(z_1)=4cos(\pi/4)=2\sqrt{2}$$
and $$z_1\bar{z_1}=4|z_1|^2=4$$

Therefore we have the quadratic factor
$$z^2-2\sqrt{2}z+4$$

 

[XtremePhysX]

Oh, well then a lot of time would've been wasted finding the other complex roots as well if only one quadratic factor is necessary. Look at how quick and easy the solution can be:

$$z^4+16=0$$
$$z^4=-16$$
$$z^4=16cis(\pi+2k\pi)$$
$$z=2cis(\frac{\pi(1+2k)}{4})$$ for $k=-2,-1,0,1$

Now let's just consider one of the complex conjugate pairs:

$$z_1=2cis(\pi/4)$$
$$z_2=2cis(-\pi/4)=\bar{z_1}$$

Therefore the quadratic factor is,

$$(z-z_1)(z-\bar{z_1})$$
$$=z^2-(z_1+\bar{z_1})z+z_1\bar{z_1}$$
Where $$z_1+\bar{z_1}=2Re(z_1)=4cos(\pi/4)=2\sqrt{2}$$
and $$z_1\bar{z_1}=4|z_1|^2=4$$

Therefore we have the quadratic factor
$$z^2-2\sqrt{2}z+4$$

Neat solution, thank you.

 

[Bohrok]

I'm surprised no one mentioned the other way without using Euler!

$z^4 + 16 = z^4 + 8z^2 + 16 - 8z^2 = (z^2 + 4)^2 - 8z^2$

This is a difference of squares that will factor and give you not just one but both real quadratic factors.