Question

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

Answer 1

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

Since the leading coefficiant is 1, any rational zero must be  a divisor of the constant term 6.

So the possible rational zeros are ±1,±2,±3,±6.

now we test each of these possibilities.

P(1) = 1+4+2-1+6 is not equals to 0.

P(-1) = 1-4+2+1+6 is not equls to 0.

P(2) = 16+32+8-2+6  is not equls to 0.

P(-2) = 16-32+8+2+6 = 0.

P(3) = 81+108+18-3+6 is not equls to 0.

P(-3) =81-108+18+3+6 = 0.

P(6) = 1296-864+72-6+6  is not equals to 0.

P(-6) = 1296+864+72+6+6 is not equals to 0.

Real solution of given polynomial is x = -2,-3