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

2 Answers
Jun 30, 2018

(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.

Jun 30, 2018

(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!