Chapter 3: Problem 48
Use the concepts of this section. Show analytically that \(-1\) is a zero of multiplicity 3 of \(P(x)=x^{5}+9 x^{4}+33 x^{3}+55 x^{2}+42 x+12,\) and find all complex zeros. Then, write \(P(x)\) in factored form.
Short Answer
Expert verified
Zero -1 has multiplicity 3; other zeros are
\(-3 \pm i\sqrt{3}\).
Factored form: \((x + 1)^3(x + 3 - i\sqrt{3})(x + 3 + i\sqrt{3})\).
Step by step solution
01
Verify Negative One is a Zero
To check if \(x = -1\) is a zero of \(P(x)\), substitute \(-1\) into \(P(x)\).\[P(-1) = (-1)^5 + 9(-1)^4 + 33(-1)^3 + 55(-1)^2 + 42(-1) + 12\]Simplify the expression:\[P(-1) = -1 + 9 - 33 + 55 - 42 + 12\]Combine the terms:\[P(-1) = 0\]Since \(P(-1) = 0\), \(-1\) is a zero of \(P(x)\).
02
Apply Synthetic Division for Multiplicity
Apply synthetic division using \(-1\) and \(P(x)\) to find the multiplicity. Perform synthetic division three times to confirm multiplicity.First division:- Coefficients: \(1, 9, 33, 55, 42, 12\).- Resulting polynomial: \(x^4 + 8x^3 + 25x^2 + 30x + 12\).Repeat with new polynomial form:- Coefficients from new polynomial \(1, 8, 25, 30, 12\).- Resulting polynomial: \(x^3 + 7x^2 + 18x + 12\).Repeat one more time:- Coefficients: \(1, 7, 18, 12\).- Resulting polynomial: \(x^2 + 6x + 12\).Since division stops cleanly after three attempts as \(P(-1) = 0\), thus \(-1\) is a zero with multiplicity of 3.
03
Find Remaining Zeros
Now solve the remaining polynomial: \(x^2 + 6x + 12 = 0\).Use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1, b = 6, c = 12\).Calculate:\[x = \frac{-6 \pm \sqrt{6^2 - 4 \times 1 \times 12}}{2 \times 1} = \frac{-6 \pm \sqrt{36 - 48}}{2}\]\[x = \frac{-6 \pm \sqrt{-12}}{2}\]Simplify using imaginary numbers:\[x = \frac{-6 \pm 2i\sqrt{3}}{2}\]\[x = -3 \pm i\sqrt{3}\]The zeros are \(-1, -1, -1\) each with multiplicity of 3, and \(-3 + i\sqrt{3}, -3 - i\sqrt{3}\).
04
Write Factored Form of Polynomial
Given the zeros found, write \(P(x)\) in factored form:\[P(x) = (x + 1)^3(x + 3 - i\sqrt{3})(x + 3 + i\sqrt{3})\]This accounts for the zero \(-1\) with multiplicity 3 and the complex zeros \(-3 \pm i\sqrt{3}\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Multiplicity of Zeros
The multiplicity of a zero in a polynomial refers to how many times a particular zero appears. If \(x = a\) is a zero of a polynomial \(P(x)\), it means that when you substitute \(a\) into the polynomial, the result is zero, implying \(P(a) = 0\). This concept helps us understand the behavior of the polynomial at that zero.
For example, in the case of the polynomial \(P(x) = x^5 + 9x^4 + 33x^3 + 55x^2 + 42x + 12\), we verified that \(-1\) is a zero of multiplicity 3. This means that \(-1\) is a root three times over.
The multiplicity affects the shape of the graph at that point. If the multiplicity is odd, the graph crosses the x-axis at the zero. If it’s even, the graph merely touches the x-axis but doesn’t actually cross it.
For example, in the case of the polynomial \(P(x) = x^5 + 9x^4 + 33x^3 + 55x^2 + 42x + 12\), we verified that \(-1\) is a zero of multiplicity 3. This means that \(-1\) is a root three times over.
The multiplicity affects the shape of the graph at that point. If the multiplicity is odd, the graph crosses the x-axis at the zero. If it’s even, the graph merely touches the x-axis but doesn’t actually cross it.
Synthetic Division
Synthetic division is a shortcut method for dividing a polynomial by a linear divisor of the form \(x - c\). It simplifies the division process compared to long division, especially when you repeatedly check for zeros or factor a polynomial.
To perform synthetic division, follow these steps:
To perform synthetic division, follow these steps:
- Identify the root \(c\) of the divisor \(x - c\) (in our example, \(c = -1\)).
- Write down the coefficients of the polynomial \(1, 9, 33, 55, 42, 12\).
- Using the root, transform these coefficients step by step.
Quadratic Formula
The quadratic formula is a powerful tool used to find the zeros of quadratic polynomials. A quadratic equation in standard form is given by \(ax^2 + bx + c = 0\). The quadratic formula provides a way to calculate the zeros directly using the coefficients of the quadratic:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
The part under the square root, \(b^2 - 4ac\), is known as the discriminant. It determines the nature of the roots:
The part under the square root, \(b^2 - 4ac\), is known as the discriminant. It determines the nature of the roots:
- If the discriminant is positive, there are two distinct real roots.
- If it is zero, there is exactly one real root (a repeated root).
- If it is negative, there are two complex roots.
Complex Zeros
Complex zeros arise when the solutions to a polynomial equation include imaginary numbers. This occurs when the discriminant in the quadratic formula is negative, producing an expression under the square root that requires a solution in terms of complex numbers.
Complex numbers are expressed in the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit satisfying \(i^2 = -1\). In the polynomial from our exercise, after applying the quadratic formula, we found the zeros \(-3 + i\sqrt{3}\) and \(-3 - i\sqrt{3}\).
These roots come in conjugate pairs, which means if \(a + bi\) is a zero of a polynomial with real coefficients, then \(a - bi\) will also be a zero. This symmetry is crucial when factoring polynomials into linear factors, particularly for those of higher degrees.
Complex numbers are expressed in the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit satisfying \(i^2 = -1\). In the polynomial from our exercise, after applying the quadratic formula, we found the zeros \(-3 + i\sqrt{3}\) and \(-3 - i\sqrt{3}\).
These roots come in conjugate pairs, which means if \(a + bi\) is a zero of a polynomial with real coefficients, then \(a - bi\) will also be a zero. This symmetry is crucial when factoring polynomials into linear factors, particularly for those of higher degrees.