Factoring...help?~Will reward
4a^2=9/64

x^n+6 +x^n+2 +x^n

Question

Factoring...help?~Will reward
4a^2=9/64

x^n+6 +x^n+2 +x^n

zepdrix · Answer

$$\large 4a^2=\frac{9}{64}$$Subtracting 9/64 from each side gives us,$$\large 4a^2-\frac{9}{64}=0$$

9/64 is actually a perfect square divided by a perfect square, let's write it like that. And we'll do the same with the 4.$$\large 2^2a^2-\frac{3^2}{8^2}$$Which we can write as,$$\large (2a)^2-\left(\frac{3}{8}\right)^2$$

We haven't factored yet. That part won't be too bad.
But so far, do you understand what I did?
I rewrote some stuff as squares so we can get through the next part easily.

anonymous · Answer

Yes

zepdrix · Answer

What we have here is the DIFFERENCE OF SQUARES.
There is a handy dandy formula for factoring the difference of squares.$$\large x^2-y^2=(x-y)(x+y)$$

In our case, $2a$ is our X that's being squared,
while $\huge \frac{3}{8}$ is our Y.

$$\large (2a)^2-\left(\frac{3}{8}\right)^2 \qquad \rightarrow \qquad \left(2a-\frac{3}{8}\right)\left(2a+\frac{3}{8}\right)$$

zepdrix · Answer

Are you suppose to solve for $a$? Or did they just want it factored?

anonymous · Answer

It said to solve each equation by factoring and the answers they had were -3/16 and 3/16

anonymous · Answer

I was wondering how they got there...

zepdrix · Answer

From here,$$\large \left(2a-\frac{3}{8}\right)\left(2a+\frac{3}{8}\right)=0$$We can apply the Zero Factor Property. We'll set each individual factor equal to 0 and solve for $a$.$$\large 2a-\frac{3}{8}=0, \qquad \qquad 2a+\frac{3}{8}=0$$

zepdrix · Answer

Looking at the first one, we'll add 3/8 to each side.$$\large 2a=\frac{3}{8}$$We'll divide both sides by 2, giving us,$$\large a=\frac{3}{16}$$

zepdrix · Answer

We can do the same with the other factor.

If you're having trouble with the factoring, you can actually solve this problem without factoring.

As a first step, take the square root of both sides.$$\large \large 4a^2=\frac{9}{64} \qquad \rightarrow \qquad 2a= \pm \frac{3}{8}$$Then divide both sides by 2,$$\large a=\pm \frac{3}{16}$$

anonymous · Answer

^-^ I get it now ;-;

zepdrix · Answer

That certainly isn't the way they wanted you to do it, but it's much easier :D heh