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Chapter 12: Problem 31

You are given a polynomial equation and one or more roots. Check that the given numbers really are roots, then solve each equation using the method shown in Examples 5 and \(6 .\) To help you decide whether you have found all the roots in each case, you may rely on the following theorem, discussed in the next section: A polynomial equation of degree n has at most \(n\) (not necessarily distinct) roots. $$3 x^{3}-5 x^{2}-16 x+12=0 ;-2 \text { is a root. }$$

Short Answer

Expert verified
The roots of the polynomial are \(-2, 3, \frac{2}{3}\).

Step by step solution

01

Verify Given Root

We need to check whether \(-2\) is indeed a root of the polynomial equation \(3x^3 - 5x^2 - 16x + 12 = 0\). To do this, substitute \(x = -2\) into the polynomial:\[3(-2)^3 - 5(-2)^2 - 16(-2) + 12 = 0\]Calculate each term:- \(3(-2)^3 = 3(-8) = -24\)- \(-5(-2)^2 = -5(4) = -20\)- \(-16(-2) = 32\)- The constant \(12\)Adding these: \(-24 - 20 + 32 + 12 = 0\). Since the result is 0, \(-2\) is indeed a root of the polynomial.
02

Polynomial Division

Now, use polynomial division to divide the polynomial \(3x^3 - 5x^2 - 16x + 12\) by \(x + 2\) (since \(-2\) is a root, that corresponds to the factor \(x + 2\)).Using synthetic division, multiply \(-2\) successively:- Bring down the 3. - Multiply by \(-2\), place the result under the next coefficient: \(-2 \times 3 = -6\).- Add \(-5\) and \(-6\): \(-11\).- Multiply \(-11 \times -2 = 22\), add it to \(-16\): \(6\).- Multiply \(6 \times -2 = -12\), add it to \(12\): \(0\).The remainder is 0, confirming \(-2\) is a root. The quotient is \(3x^2 - 11x + 6\).
03

Solve Reduced Quadratic

We need to solve the quadratic equation \(3x^2 - 11x + 6 = 0\) obtained from polynomial division.Use the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where \(a = 3\), \(b = -11\), and \(c = 6\):1. Calculate the discriminant: \((-11)^2 - 4 \times 3 \times 6 = 121 - 72 = 49\).2. The square root of 49 is 7.3. Plug values into the formula: - \(x = \frac{11 \pm 7}{6}\) - Solutions are \(x = \frac{18}{6} = 3\) and \(x = \frac{4}{6} = \frac{2}{3}\).
04

List All Roots

We now compile the roots of the equation. The polynomial equation \(3x^3 - 5x^2 - 16x + 12 = 0\) has a degree of 3, meaning it can have at most 3 roots.We have found the roots as \(-2\), \(3\), and \(\frac{2}{3}\). These are all the roots since the polynomial is cubic, and we've accounted for all possible roots.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Polynomial Division
To understand polynomial division, think of dividing a polynomial, like you do numbers. Consider dividing a cubic polynomial by a linear polynomial. The goal is to rewrite your original polynomial into the form of another polynomial (the quotient) times your divisor plus a remainder, ideally zero if the divisor is a root.

In this case, we are working with:
  • The cubic polynomial: \(3x^3 - 5x^2 - 16x + 12\).
  • The linear divisor: \(x + 2\), since \(-2\) is a given root.
Conducting polynomial division helps reduce the degree of the polynomial. It's much like how long division reduces a large number to a simpler one. In this example, after dividing, you retrieve a new quadratic expression: \(3x^2 - 11x + 6\). This is easier to work with and solve for additional roots.
Synthetic Division
Synthetic division is an efficient shortcut to divide a polynomial when dealing with linear divisors of the form \(x - r\). It鈥檚 particularly useful for confirming root existence and simplifying calculations.

To execute synthetic division:
  • Write the coefficients of your polynomial: \(3, -5, -16, 12\).
  • Use the root provided, \(-2\), to substitute as the divisor.
You then go through a series of multiplication and addition steps, which somewhat mimic long division. The result yields the coefficients of the resulting polynomial, and a remainder.

When the remainder is zero, as it was here, the divisor \(-2\) is indeed a root. The new polynomial we obtain is \(3x^2 - 11x + 6\). This new equation is easier to solve and find other roots.
Quadratic Formula
When the polynomial is reduced to a quadratic form, as is obtained after synthetic division in our example, the next step is to solve using the quadratic formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

This formula allows you to find the roots of any quadratic equation. In the example given, \(a = 3\), \(b = -11\), and \(c = 6\). Plugging these into the formula, you calculate:
  • The discriminant, \(b^2 - 4ac\), determining the nature of the roots.
  • Calculate exact roots using the solution to the formula.
In this case, the discriminant is positive (49), indicating two real roots. Thus, the roots are \(x = 3\) and \(x = \frac{2}{3}\). This method is invaluable as it provides a guaranteed way to solve quadratics.
Cubic Polynomial
A cubic polynomial is a polynomial of degree three of the form \(ax^3 + bx^2 + cx + d\). Polynomial degrees indicate the highest power of the variable in the expression. Cubic polynomials can have up to three real roots.

Understanding cubic polynomials involves:
  • Realizing there can be multiple roots: depending on the polynomial, they can all be real, or some can be complex.
  • Identifying potential rational roots using methods like the Rational Root Theorem.
  • Verifying roots through substitution and calculations like synthetic division.
Cubic polynomials often need to be broken down or factorized to solve for the roots step by step. First, check for obvious roots, followed by polynomial division to reduce the degree to a quadratic polynomial to find further roots.
Root Verification
Root verification is crucial to determine whether a proposed solution actually solves the polynomial equation. It involves substituting the potential root back into the original polynomial.

In our example, we were given \(-2\) as a root.
  • By plugging \(-2\) into the polynomial \(3x^3 - 5x^2 - 16x + 12\), and showing it results in zero confirms it's a true root.
  • If a proposed value, when substituted, results in zero, it's verified as a genuine root.
Root verification is an essential step especially before performing synthetic division, ensuring accuracy in following steps. This step consolidates understanding and confirms calculations are on the right track, leading to the discovery of all roots in a systematic manner.

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Most popular questions from this chapter

(a) Use a calculator to verify that the number \(\tan 9^{\circ}\) appears to be a root of the following equation: $$ x^{4}-4 x^{3}-14 x^{2}-4 x+1=0 $$ In parts (b) through (d) of this exercise, you will prove that \(\tan 9^{\circ}\) is indeed a root and that \(\tan 9^{\circ}\) is irrational. (b) Use the trigonometric identity $$ \tan 5 \theta=\frac{\tan ^{5} \theta-10 \tan ^{3} \theta+5 \tan \theta}{5 \tan ^{4} \theta-10 \tan ^{2} \theta+1} $$ to show that the number \(x=\tan 9^{\circ}\) is a root of the fifth-degree equation $$ x^{5}-5 x^{4}-10 x^{3}+10 x^{2}+5 x-1=0 $$Hint: In the given trigonometric identity, substitute \(\theta=9^{\circ}\) (c) List the possibilities for the rational roots of equation (2). Then use synthetic division and the remainder theorem to show that there is only one rational root. What is the reduced equation in this case? (d) Use your work in parts (b) and (c) to explain (in complete sentences) why the number \(\tan 9^{\circ}\) is an irrational root of equation (1).

Determine values for \(a\) and \(b\) such that \(x-1\) is a factor of both \(x^{3}+x^{2}+a x+b\) and \(x^{3}-x^{2}-a x+b\).

Find all roots of each equation. Hints: First, factor by grouping. In Exercises 71 and 72 each equation has three roots; in Exercise 73 the equation has six roots; in Exercise 74 there are five roots. $$x^{3}-3 x^{2}+4 x-12=0$$

(a) Let \(r_{1}, r_{2},\) and \(r_{3}\) be three distinct numbers that are roots of the equation \(f(x)=A x^{2}+B x+C=0 .\) Show that \(f(x)=0\) for all values of \(x .\) Hint: You need to show that \(A=B=C=0 .\) First show that both \(A\) and \(B\) are zero as follows. If either \(A\) or \(B\) were nonzero, then the equation \(f(x)=0\) would be a polynomial equation of degree at most 2 with three distinct roots. Why is that impossible? (b) Use the result in part (a) to prove the following identity: $$\begin{aligned}\frac{a^{2}-x^{2}}{(a-b)(a-c)} &+\frac{b^{2}-x^{2}}{(b-c)(b-a)} \\\&+\frac{c^{2}-x^{2}}{(c-a)(c-b)}-1=0 \end{aligned}$$ Hint: Let \(f(x)\) denote the quadratic expression on the left- hand side of the equation. Compute \(f(a), f(b)\) and \(f(c)\)

First graph the two functions. Then use the method of successive approximations to locate, between successive thousandths, the \(x\)-coordinate of the point where the graphs intersect.Use a graphing utility to draw the graphs as well as to check your final answer. $$y=x^{3}-5 ; y=2 x-3$$
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