/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 45 Solve the equation \(12 x^{3}+16... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 45

Solve the equation \(12 x^{3}+16 x^{2}-5 x-3=0\) given that \(-\frac{3}{2}\) is a root.

Short Answer

Expert verified
The roots of the equation \(12 x^{3}+16 x^{2}-5 x-3=0\) are \(-\frac{3}{2}\), \( \frac{1}{3}\) and \(-\frac{1}{2}\).

Step by step solution

01

Verify the root

First step is to ensure that \(-\frac{3}{2}\) is indeed a root of the equation. Substitute the given root into the equation, if it satisfies the equation then it's indeed a root. Hence, substituting \(-\frac{3}{2}\) into the equation we get \( 12 (-\frac{3}{2})^3 + 16 (-\frac{3}{2})^2 - 5 (-\frac{3}{2}) -3 = 0 \). Thus, verifying that \(-\frac{3}{2}\) is indeed a root of the given equation.
02

Factorization

Now that we have confirmed that \(-\frac{3}{2}\) is a root of the equation, we can factorize the equation. Since, if r is a root of the equation, (x - r) is a factor of the equation. As a root is given, we can sketch out a factor. For the root \(-\frac{3}{2}\), the corresponding factor is \(2x + 3\). Now, perform polynomial division or use synthetic division to obtain the other factor. Therefore, the cubic equation \(12x^3 + 16x^2 -5x - 3 =0\) can be factored to \( (2x + 3)(6x^2 - 2x -1) = 0 \).
03

Solving the quadratic

Now that we have factored the equation, we need to find the roots from the two factors. Since, we already know that root from the factor \(2x + 3\) is \(-\frac{3}{2}\). The other roots are obtained by solving the quadratic equation \(6x^2 - 2x -1 = 0\). This can be solved by using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Hence the remaining roots of the equation are obtained as \( x = \frac{1}{3}, -\frac{1}{2} \).

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