Solve the equation in the real number system: x^5-4x^4+4x^3+2x^2-5x+2=0 (MULTIPLE CHOICE)

Question

Solve the equation in the real number system: x⁵-4x⁴+4x³+2x²-5x+2=0

 

a. x=-5, -4, 1, 2, 4

b. x=-1, 1, 2

c. x=-5, -4, 1, 2

d. x=-2, -1, 1

Answer 1

This equation doesn't appear to factor, and seems kind of tricky, sonce it's a 5th order equation. We are looking for real numbers, but we don't necessarily know if we are dealing with rational or irrational roots.

 

Descartes' Rule of Signs shows 4 sign changes, indicating 4 or 2 or 0 positive real roots, and also there is one negative real root.

 

The Rational Root Theorem says (here) that if there are any rational roots, they will be factors of 2, so that could be ±2, ±1. Let's just try a negative root, say -1, and we can use synthetic division.

 

This leaves no remainder, so -1 is a root, and what we have left as the quotient is x⁴ - 5x³ + 9x² - 7x + 2

 

Now let's try -2, because -1*-2 = 2, which is the constant term at the end of the original equation. However, using synthetic division leaves a remainder of 108, and so it is not a root.

 

So we'll go for 2, and synthetic division shows that it is a root. What's left as the quotient is  x³ - 3x² + 3x - 1

 

This cannot be factored by grouping, so we could try +1 as a root. Synthetic division leaves no remainder, so it is a root. The quotient is

x² - 2x + 1, which factors into (x - 1)² . Setting this = 0 gives the final 2 roots of 1 (twice).

 

SO THE ROOTS ARE -1,1(3 TIMES),2, SO THE ANSWER IS (b)