# Thread: Simplifying a complex fraction

1. ## Simplifying a complex fraction

Hi, recently I was given a question by a teacher:

Simplify: $\frac{2a^4 +a^3 + 4a^2 + a + 2}{4a^4 + 4a^3 + a^2 - 4}$

After the whole class struggled over the week, our teacher revealed the answer: $\frac{a^2 + 1}{2a^2 + a - 2}$

How do I get to this answer??

2. Hello BG
Originally Posted by BG5965
Hi, recently I was given a question by a teacher:

Simplify: $\frac{2a^4 +a^3 + 4a^2 + a + 2}{4a^4 + 4a^3 + a^2 - 4}$

After the whole class struggled over the week, our teacher revealed the answer: $\frac{a^2 + 1}{2a^2 + a - 2}$

How do I get to this answer??

Your only hope would be to find the factors of the numerator and denominator - and then 'cancel' the common factor, which turns out to be $(2a^2 + a + 2)$; i.e.

$\frac{2a^4 +a^3 + 4a^2 + a + 2}{4a^4 + 4a^3 + a^2 - 4}=\frac{\color{red}(2a^2+a+2)\color{black}(a^2+1) }{\color{red}(2a^2+a+2)\color{black}(2a^2+a-2)}$

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

The only way I can suggest you might have done this is to use a 'trial and error' process on each of the quartic expressions. Given that, for example, $2a^4$ cannot be broken down in many ways -

• $2a^3 \times a$

• $2a^2 \times a^2$; and

• $2a \times a^3$

- this shouldn't have taken too long.

But I must admit: I cheated and worked back from the answer you gave me using algebraic long division.

Does that help to shed some light?

3. Originally Posted by BG5965
Hi, recently I was given a question by a teacher:

Simplify: $\frac{2a^4 +a^3 + 4a^2 + a + 2}{4a^4 + 4a^3 + a^2 - 4}$

After the whole class struggled over the week, our teacher revealed the answer: $\frac{a^2 + 1}{2a^2 + a - 2}$

How do I get to this answer??

$\frac{2a^4 +a^3 + 4a^2 + a + 2}{4a^4 + 4a^3 + a^2 - 4}$

$=\frac{(a^2+1)(2a^2+a+2)}{(2a^2+a-2)(2a^2+a+2)}$ it's just some tricky factoring you have to play with

Then we cancel the $2a^2+2a$ in the numerator and denominator to get the answer

4. Originally Posted by artvandalay11
$\frac{2a^4 +a^3 + 4a^2 + a + 2}{4a^4 + 4a^3 + a^2 - 4}$

$=\frac{(a^2+1)(2a^2+2a)}{(2a^2+a-2)(2a^2+2a)}$ it's just some tricky factoring you have to play with

Then we cancel the $2a^2+2a$ in the numerator and denominator to get the answer
hi, can u explain how you did the factoring.

5. Hello reiward
Originally Posted by reiward
hi, can u explain how you did the factoring.
You haven't quoted artvandalay11's posting quite accurately. The expressions factorise as follows:

$2a^4 +a^3 + 4a^2 + a + 2 = (a^2+1)(2a^2+a+2)$

$4a^4 + 4a^3 + a^2 - 4=(2a^2+a-2)(2a^2 +a+2)$

and then the common factor $(2a^2+a+2)$ is 'cancelled'.

So how do we find these factors? If you look back to my first posting, you'll see that I've given some clues, by looking at the possible factors of $2a^4$. The process is not dissimilar to factorising quadratics - just a bit more complicated!

Any factors (if they exist at all) of a quartic expression in $x$ will be either in the form:

• A linear factor and a cubic factor, like $(ax+b)(cx^3+dx^2+ex+f)$, where the cubic may or may not factorise further; or

• Two quadratic factors, like $(ax^2+bx+c)(dx^2+ex+f)$, where neither quadratic factorises further.

Looking back to my posting, then, you have three possibilities for $2a^4 +a^3 + 4a^2 + a + 2$:

• Linear and cubic: $(2a + ...)(a^3 + ...)$

• Linear and cubic: $(a+...)(2a^3 + ...)$

• Two quadratics: $(2a^2 + ...)(a^2 + ...)$

Then you'd look at the factors of the constant term, $2$, which of course are simply $2\times 1$, and, by trial and error eliminate the first two of these three possibilities, by considering the coefficients of $a^3, a^2$ and $a$.

This would leave you with either $(2a^2 + pa + 2)(a^2 + qa +1)$ or $(2a^2 + pa+1)(a^2+qa+2)$, where $p$ and $q$ are constants still to be found. Again, by looking at the coefficients of $a^3, a^2$ and $a$ you should fairly quickly see that the factors are $(2a^2+a+2)(a^2+1)$.

The denominator can be factorised in a similar way, except that by now you will have found the factors of the numerator, and you'll be hoping that one of these will also be a factor of the denominator in order to allow some simplification to take place. So this gives quite a helpful clue as to what the factors are likely to be!

Not easy, but definitely achievable!