Question

I need some help factoring.
Find the real solutions of the equation 21x^4-4x^2-1=0

Answer 1

I understand the problem and there is an explanation on how to do the problem in the book but there is one step in the factoring that I dont understand

Answer 2

first, you express it to quadratic factors
\[21x^{4}-4x ^{2}-1 = (7x ^{2}+1)(3x ^{2}-1)\]

Answer 3

I don't understand where the 1 and the 3 come from. I know they come from the 4x^2 but how/why?

Answer 4

in finding the factors, you need to find the factors of 21 such that when added will have the second term. but since we also have to consider the third term, we also have to have the factors of -1

Answer 5

7*3=21
1*-1=-1
the arrangement of the signs is for you to identify by inspection

Answer 6

ok I think that makes sense. What always throws me of is if the first term is multiplied by more than 1

Answer 7

*off

Answer 8

for cases like that, you just have to consider it's factors

Answer 9

the factors for the numerical coefficients

Answer 10

ok thank you :)

Answer 11

yw :)