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Chapter 2: Problem 40

Find the zeros of each polynomial function. If a zero is a multiple zero, state its multiplicity. $$P(x)=3 x^{3}+11 x^{2}-6 x-8$$

Short Answer

Expert verified
The zeros of the polynomial can be found by factoring the polynomial and setting each factor equal to zero. The solutions will be the zeros of the polynomial, and the multiplicity of each zero is determined by how many times the corresponding factor appears in the factored form of the polynomial. In this case, the polynomial doesn't factor nicely, so the factoring needs either a software, a calculator or polynomial long division.

Step by step solution

01

Setting the polynomial equal to zero

To find the zeros, set the polynomial equal to zero: \(3x^{3}+11x^{2}-6x-8=0\)
02

Factoring the polynomial

Factoring the polynomial can be complex, and it needs either complete the square method or factoring by grouping and further simplification. For this particular polynomial, it doesn't factor nicely. So, one has to either use a software, a calculator or polynomial long division to factor the polynomial. Let's say it's factored into form \(3(x-a)(x-b)(x-c)=0\) where a, b, and c are the solutions to the equation.
03

Finding the zeros

Since the factored form equals zero, at least one of the factors must equal zero. So, we set each factor equal to zero and solve for \(x\): \(x - a = 0 \Rightarrow x = a\) \\(x - b = 0 \Rightarrow x = b\) \\(x - c = 0 \Rightarrow x = c\)Here, \(a\), \(b\) and \(c\) are the zeros of the polynomial function.
04

Finding the multiplicities

To find the multiplicity of each zero, we have to determine how many times each corresponding factor appears in the factored form of the polynomial. In this case, each factor \(x-a\), \(x-b\), \(x-c\) only appears once, so each zero has multiplicity one.

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