how many real solutions of equation x^4 + 4x^3 + 2x^2 - x + 6 = 0

Question

find the real solutions of this equation x^4 + 4x^3 + 2x^2 - x + 6 = 0

Answer 1

x^4+4x^3+2x^2-x+6 = 0

By the rational root therom

Since the leading coofficient is 1 and constant term is 6.

So the possible real solutions are ±1,±2,±3,±6.

Now we test the each of the possibilities.

x^4+4x^3+2x^2-x+6 = 0

P(1) = 1^4+4(1)^3+2(1)^2-1+6 = 1+4+2-1+6 = 12

P(-1) = (-1)^4+4(-1)^3+2(-1)^2-(-1)+6 = 1-4+2+1+6 = 6

P(2) = (2)^4+4(2)^3+2(2)^2-2+6 = 16+32+8-2+6 = 60

P(-2) = (-2)^4+4(-2)^3+2(-2)^2-(-2)+6 = 16-32+8+2+6 = 0

P(3) = (3)^4+4(3)^3+2(3)^2-3+6 = 81+108+18-3+6 = 210

P(-3) = (-3)^4+4(-3)^3+2(-3)^2-(-3)+6 = 81-108+18+3+6 = 0

P(6) = (6)^4+4(6)^3+2(6)^2-6+6 = 1296+864+72 = 2232

P(-6) = (-6)^4+4(-6)^3+2(-6)^2-(-6)+6 = 1296-864+72+6+6 = 516

So the real solutions of x^4+4x^3+2x^2-x+6 is -2,-3.