P(x) = x^4 − 5x^3 + 6x^2 + 4x − 8

Question

 Find all the real zeros of P. (Enter your answers as a comma-separated list. Enter all answers including repetitions.)

P(x) = x^4 − 5x^3 + 6x^2 + 4x − 8

x =

Answer 1

The Rational Root Theorem states that possible rational zeroes are ±8/1, ±4/1, ±2/1, ±1/1.

Descartes' Rules of Signs states that there are three positive roots because there are 3 sign changes for f(x) (so most likely a repeated root) and one negative root because there is 1 sign change for f(-x).

Now, use synthetic division which can help determine which rational zeroes work, if any:

-1 | 1 -5 6 4 -8

__-1_6 -12_8_

1 -6 12 -8 | 0 <-- (x+1) is a factor

Hence, x^4 − 5x^3 + 6x^2 + 4x − 8 = (x+1)(x³-6x²+12x-8)

1 | 1 -6 12 -8

___ 1 -6_ 6_

1 -6 6 | -2 <-- (x-1) is not a factor

2 | 1 -6 12 -8

___ 2 -8_8_

1 -4 4 | 0 <-- (x-2) is a factor

Hence, (x+1)(x³-6x²+12x-8) = (x+1)(x-2)(x²-4x+4) = (x+1)(x-2)(x-2)² = (x+1)(x-2)³.

Hence, x=-1 and x=2 (with multiplicity 3) are the roots!

Answer 2

Use Rational Root Theorem to determine possible rational roots.

Use DesCartes Rule of signs to determine number of positive and negative roots.

Use synthetics substituttion to test possible roots.