1. ## It's a shame but I have to ask this

First attempt:

$4a^2+2a+\frac{1}{4}$ → Original Problem

$2[2a^2+a+\frac{1}{8}]$

...but I can't keep on.

So I tried again:

$4a^2+2a+\frac{1}{4}$

$4a^2+2a+\frac{1}{2^2}$

$4a^2+2a+1^{-2}$

...but then again, this is quite likely to be mistaken.

2. Originally Posted by Alienis Back
First attempt:

$4a^2+2a+\frac{1}{4}$ → Original Problem

$2[2a^2+a+\frac{1}{8}]$

...but I can't keep on.

So I tried again:

$4a^2+2a+\frac{1}{4}$

$4a^2+2a+\frac{1}{2^2}$

$4a^2+2a+1^{-2}$

...but then again, this is quite likely to be mistaken.
Let's stick to the first method you have choosen:

$4a^2+2a+\frac{1}{4}$ → Original Problem

$4[a^2+\frac12 a+\frac{1}{16}] = 4\left(a+\frac14\right)^2$

3. Originally Posted by Alienis Back
First attempt:

$4a^2+2a+\frac{1}{4}$ → Original Problem

$2[2a^2+a+\frac{1}{8}]$

...but I can't keep on.

So I tried again:

$4a^2+2a+\frac{1}{4}$

$4a^2+2a+\frac{1}{2^2}$

$4a^2+2a+1^{-2}$

...but then again, this is quite likely to be mistaken.
Are you trying to factorise it? Complete the square:

$4a^2+2a+\frac{1}{4}$

$= 4 \left(a^2 + \frac{1}{2} a+\frac{1}{16}\right)$

$= 4 \left( \left[a + \frac{1}{4}\right]^2 - \frac{1}{16} +\frac{1}{16}\right)$

$= 4 \left(a + \frac{1}{4}\right)^2$

$= \left(2a + \frac{1}{2}\right)^2$.

In hindsight you should recognise that you had a perfect square.

4. Originally Posted by Alienis Back
First attempt:

$4a^2+2a+\frac{1}{4}$ → Original Problem

$2[2a^2+a+\frac{1}{8}]$

...but I can't keep on.

So I tried again:

$4a^2+2a+\frac{1}{4}$

$4a^2+2a+\frac{1}{2^2}$

$4a^2+2a+1^{-2}$

...but then again, this is quite likely to be mistaken.

the best way i think is to multiply and divide a 4

So it becomes $\frac{16a^2+8a+1}{4}$
$=\left(\frac{4a+1}{2}\right)^2$
$=\left(2a+\frac{1}{2}\right)^2$

DONE!!!

5. ## Completed square??

Ok. Somebody here told me that I should learn to factor polynomials by the method of "completing the square" which I had no idea of, so I resourced to wikipedia, which was quite usefull. Now, as my book doesn't have answers for pair numbered excercises I want you to check this for me and tell me if I'm going the right way.

$4a^2+2a+\frac{1}{4}$ Original problem

$4\left[a^2+\frac{2}{4}a+\frac{1}{16}\right]$

We'll leave that $4$ out for a while...

$a^2+\frac{1}{2}a=-\frac{1}{16}$

$a^2+\frac{1}{4}a+\left(\frac{1}{4}\right)^2=-\frac{1}{16}+\left(\frac{1}{4}\right)^2$

$\left(a+\frac{1}{4}\right)^2=-\frac{1}{16}+\frac{1}{16}$

$4\left(a+\frac{1}{4}\right)^2$

...and then we bring it ( $4$) back in place again.

$2^2\left(a+\frac{1}{4}\right)^2$

$\left(2a+\frac{2}{4}\right)^2$

$\left(2a+\frac{1}{2}\right)^2$

$\left(2a+\frac{1}{2}\right)\left(2a+\frac{1}{2}\ri ght)$

6. Instead of going through the motions each time, just plug a,b,c into the general formula for completing the square.

$a(x+\frac{b}{2a})^{2}+\underbrace{c-\frac{b^{2}}{4a}}_{\text{constant}}$

In your case, a=4, b=2, c=1/4.

7. In my textbook and class we use perfecting the square to solve quadratic equations, so I have no idea why they told you that you should factor by perfecting the square. To me it is not as efffective for factoring with fractions. Well, it is effective, but just a pain in the butt. It is easier to divide by the co-effecient of a, to make it 1, perfect the square and solve for a, but that is just my opinion.

If we use perfecting the square to find a, the answer is -1/4

BTW this is already a perfect square trinomial, so you if you identified it in the beginning, there would be no need to perfect the square. My professor and textbook told us to only perfect the square, when a trinomial can no longer be factored, because it is faster.