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!

The fundamental theorem of algebra can help you find imaginary roots. Imaginary roots appear in a quadratic equation when the discriminant of the quadratic equation — the part under the square root sign (b2 – 4ac) — is negative. If this value is negative, you can’t actually take the square root, and the answers are not real. In other words, there is no real solution; therefore, the graph won’t cross the x-axis.

Using the quadratic formula always gives you two solutions, because the plus/minus sign means you’re both adding and subtracting and getting two completely different answers. When the number underneath the square-root sign in the quadratic formula is negative, the answers are called complex conjugates. One is r + si and the othe

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gimmee one of those gold crowns!

~coco

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