# How do you factor 2n^4 - 9n^2 +4 = 0?

Jun 30, 2018

$\left(2 {n}^{2} - 1\right) \left(n - 2\right) \left(n + 2\right) = 0$

#### Explanation:

Start by noticing that there are no ${n}^{3}$ or $n$ terms in the expression on the LHS to be factored. Thus we can attempt factorisation as a quadratic in ${n}^{2}$ of the form $\left(a {n}^{2} + b\right) \left(c {n}^{2} + d\right)$=0

If we are looking for integer factorisations, we immediately know $a$ and $c$ - one must be 2, the other 1 in order to get the correct ${n}^{4}$ coefficient. So our desired form is $\left(2 {n}^{2} + b\right) \left({n}^{2} + d\right)$=0

Multiply this out:
$2 {n}^{4} + \left(b + 2 d\right) {n}^{2} + b d = 0$
where
$b + 2 d = - 9$ and $b d = 4$.

We now known that $b$ and $d$ must have the same sign and that sign must be negative. So, assuming integer solutions, the second equation tells us that $b$ and $d$ must either be $\left(- 1 , - 4\right)$, $\left(- 4 , - 1\right)$ or $\left(- 2 , - 2\right)$. Trialling these in the first equation, we see that $b = - 1$, $d = - 4$.

$\left(2 {n}^{2} - 1\right) \left({n}^{2} - 4\right) = 0$

Can we factor this further? Yes. The second bracket is a difference of two squares, ${n}^{2} - 4 = \left(n - 2\right) \left(n + 2\right)$. The first requires a square root to express the same, so we leave it as a quadratic. Thus our full integer factorisation is

$\left(2 {n}^{2} - 1\right) \left(n - 2\right) \left(n + 2\right) = 0$

The question doesn't ask it, but if we wish to solve this equation, then the four quartic roots are immediately available:
$n = \left(- \frac{1}{\sqrt{2}} , + \frac{1}{\sqrt{2}} , - 2 , + 2\right)$
It is good practice to verify that each of these is in fact a solution to the original equation.

Jun 30, 2018

$\left({n}^{2} - 4\right) \left(2 {n}^{2} - 1\right)$

#### Explanation:

Re-Write the given equation as follows:

(2.(n^4)) - 3^2n^2)+4=0

$\left(2 {n}^{4} - {3}^{2.} {n}^{2}\right) + 4 = 0$

Step 1: Trying to factor by splitting the middle term

Find two factors of $8$ whose sum equals the coefficient of the middle term, which is $- 9$ .

$- 8 + - 1 = - 9$ and $- 8 \times - 1 = 8$

Step 2: Rewrite the polynomial splitting the middle term using the two factors found in step 1 above, $- 8$ and $- 1$

$2 {n}^{4} - 8 {n}^{2} - {n}^{2} + 4$

Pulling out the common factors

$2 {n}^{2} \left({n}^{2} - 4\right) - 1 \left({n}^{2} - 4\right)$

$\left({n}^{2} - 4\right) \left(2 {n}^{2} - 1\right)$ ----> these are the factors!