How do you factor 2n49n2+4=0?

2 Answers
Jun 30, 2018

(2n21)(n2)(n+2)=0

Explanation:

Start by noticing that there are no n3 or n terms in the expression on the LHS to be factored. Thus we can attempt factorisation as a quadratic in n2 of the form (an2+b)(cn2+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 n4 coefficient. So our desired form is (2n2+b)(n2+d)=0

Multiply this out:
2n4+(b+2d)n2+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
(2n21)(n24)=0

Can we factor this further? Yes. The second bracket is a difference of two squares, n24=(n2)(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

(2n21)(n2)(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=(12,+12,2,+2)
It is good practice to verify that each of these is in fact a solution to the original equation.

Jun 30, 2018

(n24)(2n21)

Explanation:

Re-Write the given equation as follows:

(2.(n4))32n2)+4=0

(2n432.n2)+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 8×1=8

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

2n48n2n2+4

Pulling out the common factors

2n2(n24)1(n24)

(n24)(2n21) ----> these are the factors!