Question

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

Answer 1

`(2n^2-1)(n-2)(n+2)=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 `(an^2+b)(cn^2+d)`=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 `(2n^2+b)(n^2+d)`=0

Multiply this out:
`2n^4+(b+2d)n^2+bd=0`
where
`b+2d=-9` and `bd=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 `(-1,-4)`, `(-4,-1)` or `(-2,-2)`. Trialling these in the first equation, we see that `b=-1`, `d=-4`.

So our quadratic factorisation is
`(2n^2-1)(n^2-4)=0`

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

`(2n^2-1)(n-2)(n+2)=0`

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

Answer 2

`(n^2-4)(2n^2-1)`

Explanation:

Re-Write the given equation as follows:

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

`(2n^4 - 3^2.n^2)+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 xx -1 = 8`

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

`2n^4-8n^2-n^2+4`

Pulling out the common factors

`2n^2(n^2-4) - 1(n^2-4)`

`(n^2-4)(2n^2-1)` ----> these are the factors!