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Chapter 3: Problem 13

For each polynomial, one or more zeros are given. Find all remaining zeros. \(P(x)=x^{3}-x^{2}-4 x-6 ; \quad 3\) is a zero.

Short Answer

Expert verified
The zeros are 3, \(-1 + i\), and \(-1 - i\).

Step by step solution

01

Use the Given Zero

It is given that 3 is a zero of the polynomial \(P(x) = x^3 - x^2 - 4x - 6\). This means that \(P(3) = 0\). We will use synthetic division with the zero 3 to help factor the polynomial.
02

Perform Synthetic Division

Set up the synthetic division using 3 as the zero on the left and the coefficients of the polynomial \([1, -1, -4, -6]\) on the top row. 1. Bring down the first coefficient 1. 2. Multiply 3 by 1 (the number just brought down), giving 3, and write it under the next coefficient. 3. Add -1 and 3, resulting in 2. 4. Multiply 3 by 2, giving 6, and write it under the next coefficient. 5. Add -4 and 6, resulting in 2. 6. Multiply 3 by 2, giving 6, and write it under -6. 7. Add -6 and 6 to get 0. The remainder is 0, confirming that 3 is indeed a zero, and the reduced polynomial (quotient) is \(x^2 + 2x + 2\).
03

Find Remaining Zeros from Quotient

Now solve the quadratic equation \(x^2 + 2x + 2 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 2\), and \(c = 2\).
04

Calculate the Discriminant

Calculate the discriminant \(b^2 - 4ac\):\[2^2 - 4(1)(2) = 4 - 8 = -4\]The discriminant is -4, which indicates two complex zeros.
05

Solve for Complex Zeros

Use the quadratic formula to find the complex zeros:\[x = \frac{-2 \pm \sqrt{-4}}{2(1)} = \frac{-2 \pm 2i}{2} = -1 \pm i\]Thus, the other two zeros are \(-1 + i\) and \(-1 - i\).

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Key Concepts

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

Synthetic Division
Did you know that synthetic division is a quick way to divide a polynomial by a linear factor of the form \(x - c\)? It鈥檚 super efficient because you only need the coefficients of the polynomial and the zero of the linear factor. Let鈥檚 see how this works with our example polynomial \(P(x) = x^3 - x^2 - 4x - 6\).

We already have a known zero, \(3\), meaning \(P(3) = 0\). To divide the polynomial by \(x - 3\), we follow these steps:
  • Start with the coefficients: \([1, -1, -4, -6]\).
  • Place \(3\) to the left (this is our divisor).
  • Bring down the first coefficient, \(1\).
  • Multiply \(3\) by the number just brought down, \(1\), and write the product item below the next coefficient.
  • Add -1 and 3, then repeat this process for all coefficients.
After this neat side-by-side arithmetic, you end up with \(0\) as a remainder, confirming \(3\) is indeed a zero, leaving us with a simpler quadratic: \(x^2 + 2x + 2\). This new polynomial we get is called the quotient.
Quadratic Formula
Faced with a quadratic equation like \(x^2 + 2x + 2 = 0\), you can use the quadratic formula to quickly find the zeros. This formula is super handy, especially when the quadratic equation does not easily factor. The quadratic formula is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]In our problem, the coefficients are \(a = 1\), \(b = 2\), and \(c = 2\).

Before plugging into the formula, calculate the discriminant \(b^2 - 4ac\):
  • \(2^2 = 4\)
  • \(4(1)(2) = 8\)
  • So, \(4 - 8 = -4\)
The discriminant tells us if the roots will be real or complex. A negative value indicates that the zeros are complex. Learn to embrace the quadratic formula; it reveals so much about the nature of solutions!
Complex Zeros
What happens when the discriminant is negative, like our \(-4\)? This means we deal with complex numbers. Complex zeros arise when dealing with numbers that include the imaginary unit \(i\), defined as \(\sqrt{-1}\). So, when resolving \[x = \frac{-2 \pm \sqrt{-4}}{2}\],it begins by evaluating \(\sqrt{-4}\) as \(2i\).

Let's continue:
  • \(x = \frac{-2 \pm 2i}{2}\)
  • Breaking it into parts gives \(-1 \pm i\)
These results, \(-1 + i\) and \(-1 - i\), are our complex zeros. They often appear in conjugate pairs when working with polynomials with real coefficients. Mastering complex zeros unlocks understanding of more sophisticated polynomial solutions!

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Most popular questions from this chapter

Use the boundedness theorem to show that the real zeros of \(P(x)\) satisfy the given conditions. \(P(x)=x^{4}-x^{3}+3 x^{2}-8 x+8\); no real zero greater than 2

Use the boundedness theorem to show that the real zeros of \(P(x)\) satisfy the given conditions. \(P(x)=x^{5}+2 x^{3}-2 x^{2}+5 x+5\); no real zero less than \(-1\)

Divide. $$\frac{5 x^{4}-2 x^{2}+6}{x^{2}+2}$$

Factor \(P(x)\) into linear factors given that \(k\) is a zero of \(P\). $$P(x)=6 x^{3}+25 x^{2}+3 x-4 ; \quad k=-4$$

Use the given zero to completely factor \(P(x)\) into linear factors. $$\text { Zero: } 5 i ; \quad P(x)=x^{4}-x^{3}+23 x^{2}-25 x-50$$
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