2x^4+x^2-3 = 2(x^2)^2 + (x^2) - 32x4+x2−3=2(x2)2+(x2)−3
is quadratic in x^2x2
Notice that the sum of the coefficients is 00
2+1-3 = 02+1−3=0
So x^2 = 1x2=1 is a root and (x^2 - 1)(x2−1) is a factor.
x^2 - 1 = x^2 - 1^2 = (x-1)(x+1)x2−1=x2−12=(x−1)(x+1)
using the difference of squares identity:
a^2-b^2 = (a-b)(a+b)a2−b2=(a−b)(a+b)
The other factor of 2x^4+x^2-32x4+x2−3 must be (2x^2+3)(2x2+3) in order that when multiplied by (x^2-1)(x2−1) it results in the leading 2x^42x4 term and the trailing -3−3 term.
2x^2+3 = 02x2+3=0 has no real roots, since x^2 >= 0x2≥0, hence 2x^2+3 >= 32x2+3≥3
So the only real roots of the original quartic equation are x = +-1x=±1
