/*! 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 29 Find the zeros of the function a... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 29

Find the zeros of the function algebraically. $$ f(x)=4 x^{3}-24 x^{2}-x+6 $$

Short Answer

Expert verified
The zeros of the function \(f(x)=4x^3-24x^2-x+6\) are \(x=1\), \(x=1.5\), and \(x=-1\).

Step by step solution

01

Rewrite the Equation

Create an equation by setting the function equal to zero: \(4x^3-24x^2-x+6=0\).
02

Factor the Polynomial

Because it is a cubic polynomial, one approach is to use trial and error to discover a factor. Substitute different integers into the equation until finding one that makes it equal zero. Start with small numbers. For instance, try \(x=1\). You'll find that \(4(1)^3 - 24(1)^2 - (1) + 6 = 0\). So, \(x=1\) is a zero of the function. This implies that \((x-1)\) is a factor. Divide \((x-1)\) into the cubic polynomial, finding another factor \((4x^2 - 20x - 6)\). The equation becomes \((x-1)(4x^2 - 20x - 6) = 0\). Here, you should then factor the quadratic \(4x^2 - 20x - 6\) into \((2x-3)(2x+2)\). Thus, the factored form of the equation is \((x-1)(2x-3)(2x+2)=0\).
03

Solve for Zeros

Solve for \(x\) by setting each factor equal to zero and solving. In this case, the zeros are found when \(x-1=0\), \(2x-3=0\), and \(2x+2=0\). Solving for \(x\) in each of these equations yield \(x=1\), \(x=1.5\), and \(x=-1\) respectively. These are the zeros of the function.

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Most popular questions from this chapter

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