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Chapter 2: Problem 4

Find the zeros of the polynomial function and state the multiplicity of each zero. $$P(x)=x^{3}(2 x+1)(3 x-12)^{2}$$

Short Answer

Expert verified
The zeros of the given polynomial function are \(x=0\), \(x=-1/2\), \(x=4\) with multiplicities of 3, 1, 1 respectively.

Step by step solution

01

Set Polynomial Equal to Zero

First, set the polynomial function \(P(x)\) equal to zero. This means we have the equation: \[0 = x^{3}(2 x+1)(3 x-12)^{2}\]
02

Solve for Each Factor

Next, equate each factor in the polynomial to zero and solve to get the zeros of the polynomial. We have the three factors: \(x^3\), \(2x+1\), and \((3x-12)^2\). Equate each factor to zero and solve for \(x\). \[x^3 = 0, 2x+1 = 0, (3x-12)^2 = 0\] Solving above equations, we get \(x = 0\), \(x=-1/2\), \(x=4\).
03

Find the Multiplicities

The multiplicities of the zeros can be identified from the powers of the factors in the polynomial. \(x=0\) repeats 3 times as it is raised to power 3. This implies the multiplicity is 3. Both \(x=-1/2\) and \(x=4\) appear once (count the powers, not the multiplication '2' in \(2x+1\)). Hence both have a multiplicity of 1.

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