Question

F(x) = x3 + x2 -8x - 6

According to the Fundamental Theorem of Algebra, how many solutions/roots will there be?
According to Descartes' Rule of Signs, what are the possible combinations of positive, negative, and/or complex roots will there be?
Using the Rational Root Theorem, list all the possible rational roots.
Use a combination of Synthetic Division, Factoring, and/or the Quadratic Formula to find all the roots. PLEASE SHOW ALL WORK!
This is my 4th time posting this and no ones helping. Please someone who is smart help me out lol

Answer 1

It has highest degree 3 so 3 roots

• 1 positive and 2 negative roots

Lets find

• x³+x²-8x-6=0
• x²(x+3)-2x(x+3)-2(x+3)=0
• (x+3)(x²-2x-2)=0
• (x+3)(x-2.732)(x+0.732)=0

Roots are

• -3,2.732,-0.732

Answer 2

Given function:

• f(x) = x³ + x² - 8x - 6

This is the third degree polynomial, so it has total 3 roots.

Lets factor it and find the roots:

• x³ + x² - 8x - 6 =
• x³ + 3x² - 2x² - 6x - 2x - 6 =
• x²(x + 3) - 2x(x + 3) - 2(x + 3) =
• (x + 3)(x² - 2x - 2) =
• (x + 3)(x² - 2x + 1 - 3) =
• (x + 3)((x - 1)² - 3) =
• (x + 3)(x - 1 + √3)(x - 1 - √3)

The roots are:

• x = -3
• x = 1 - √3
• x = 1 + √3