Answer:
(2x-1)(2x+1)(x^2+2) = 0
Step-by-step explanation:
Here's a trick: Use a temporary substitution for x^2. Let p = x^2. Then 4x^4+7x^2-2=0 becomes 4p^2 + 7p - 2 = 0.
Find p using the quadratic formula: a = 4, b = 7 and c = -2. Then the discriminant is b^2-4ac, or (7)^2-4(4)(-2), or 49+32, or 81.
Then the roots are:
-7 plus or minus √81
p= --------------------------------
8
p = 2/8 = 1/4 and p = -16/8 = -2.
Recalling that p = x^2, we let p = x^2 = 1/4, finding that x = plus or minus 1/2. We cannot do quite the same thing with the factor p= -2 because the roots would be complex.
If x = 1/2 is a root, then 2x - 1 is a factor. If x = -1/2 is a root, then 2x+1 is a factor.
Let's multiply these two factors, (2x-1) and (2x+1), together, obtaining 4x^2 - 1. Let's divide this 4x^2 - 1 into 4x^4+7x^2-2=0. We get x^2+2 as quotient.
Then, 4x^4+7x^2-2=0 in factored form, is (2x-1)(2x+1)(x^2+2) = 0.
