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Chapter 3: Problem 64

If you know that \(-2\) is a zero of $$f(x)=x^{3}+7 x^{2}+4 x-12$$ explain how to solve the equation $$x^{3}+7 x^{2}+4 x-12=0$$

Short Answer

Expert verified
The solutions to the equation \(x^{3}+7 x^{2}+4 x-12=0\) are \(x= -2, -2, -3\)

Step by step solution

01

Apply the Factor Theorem

The factor theorem states that a polynomial has a factor \(x-k\) if and only if \(f(k) = 0\). Since \(-2\) is a root of the polynomial \(f(x)=x^{3}+7 x^{2}+4 x-12\), it means \(x+2\) is a factor of this polynomial
02

Perform Polynomial Division

Divide the polynomial \(f(x)=x^{3}+7 x^{2}+4 x-12\) by the factor \(x+2\) to get the quotient. Performing the division we get: \[\frac{{x^{3}+7 x^{2}+4 x-12}}{{x+2}} = x^{2}+5x+6\]So, we can write the original polynomial as: \[f(x)= (x+2)(x^{2}+5x+6)\]
03

Solve for \(x\)

To solve the equation \(x^{3}+7 x^{2}+4 x-12=0\) now, we set the factored form: \((x+2)(x^{2}+5x+6)=0\) to equal zero. A product can be zero only if at least one of the factors is zero. So, either \((x+2) = 0\) or \((x^{2}+5x+6) = 0\).We already know the solution for the first equation \(x=-2\). For the equation \((x^{2}+5x+6) = 0\), we can further factorize \(x^{2}+5x+6\) to \((x+2)(x+3)\). So, the roots also include \(x=-2\) and \(x=-3\).
04

List the Solutions

Our solutions include all the roots of the equation. Therefore, the solutions to the equation \(x^{3}+7 x^{2}+4 x-12=0\) are \(x= -2, -2, -3\) corresponding to the factors \((x+2), (x+2), (x+3)\) respectively.

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