Question

Using the tools you learned in this lesson, find all solutions (real and non-real) for the polynomial below. Explain how we can use Descartes' rule of signs to find the number of positive and negative real zeros of this polynomial. 3x^3-4x^2+11x+10=0To earn full credit please share all work, calculations and thinking. If you prefer you can do the work by hand on a piece of paper, take a picture of that work and upload it.

Answer 1

Step 1:

Write the polynomial equation

`3x^3-4x^2+11x+10=0`

Step 2:

First, find the first factor of the polynomial.

The first factor is 3x + 2

`\begin{gathered} \text{Divide }3x^3-4x^2+11x+10=0\text{ by 3x + 2 using long division} \\ \text{ x}^2\text{ - 2x + 5} \\ 3x\text{ + 2 }\sqrt[]{3x^3-4x^2+11x+10} \\ \text{ -(3x}^3+2x^2) \\ --------------------------- \\ \text{ -6x}^2+11x\text{ + 10} \\ \text{ -}(-6x^2\text{ - 4x)} \\ --------------------------- \\ \text{ 15x + 10} \\ \text{ -(15x + 10)} \\ ---------------------------------- \\ \text{ 0} \end{gathered}`

Step 3

The polynomial

`3x^3-4x^2+11x+10canbeexpressionas(3x+2)((x^2-2x+5)`

Step 4:

`\begin{gathered} \text{Use the quadratic formula to find the solution to } \\ x^2\text{ - 2x + 5} \\ x\text{ = }\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{a = 1, b = -2 and c = 5} \\ \text{x = }\frac{2\pm\sqrt[]{(-2)^2-4*1*5}}{2*1} \\ \text{x = }\frac{2\pm\sqrt[]{4-20}}{2} \\ \text{x = }\frac{2\pm\sqrt[]{-16}}{2} \\ \text{x = }\frac{2\text{ + 4i}}{2}\text{ , x = }\frac{2\text{ - 4i}}{2} \\ \text{x = 1 + 2i , x = 1 - 2i} \end{gathered}`

Descartes rule of sign

It tells us that the number of positive real zeros in a polynomial function f(x) is the same or less than by an even number as the number of changes in the sign of the coefficients. The number of negative real zeros of the f(x) is the same as the number of changes in sign of the coefficients of the terms of f(-x) or less than this by an even number.

`There\text{ are 1 negative real root x = }(-2)/(3)`