/*! 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 68 Find all of the real and imagina... [FREE SOLUTION] | 91Ó°ÊÓ

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Find all of the real and imaginary zeros for each polynomial function. $$U(x)=x^{4}-4 x^{3}+x^{2}+12 x-12$$

Short Answer

Expert verified
The zeros are 2, 2, i, -i.

Step by step solution

01

Identify the Polynomial and Its Degree

The given polynomial is \[ U(x) = x^{4} - 4x^{3} + x^{2} + 12x - 12. \] The polynomial is of degree 4.
02

Use the Rational Root Theorem

According to the Rational Root Theorem, the possible rational roots of the polynomial are the factors of the constant term (which is -12) divided by the factors of the leading coefficient (which is 1). Therefore, the possible rational roots are: \[ \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12. \]
03

Test the Possible Rational Roots

Use synthetic division or direct substitution to test each possible rational root. Let's begin with 1: - Substituting 1 into \( U(x) \): \[ U(1) = 1^{4} - 4(1)^{3} + 1^{2} + 12(1) - 12 = 1 - 4 + 1 + 12 - 12 = -2. \] Therefore, 1 is not a root.
04

Continue Testing the Possible Rational Roots

Next, test -1: - Substituting -1 into \( U(x) \): \[ U(-1) = (-1)^{4} - 4(-1)^{3} + (-1)^{2} + 12(-1) - 12 = 1 + 4 + 1 - 12 - 12 = -18. \] Therefore, -1 is not a root.
05

Test Other Possible Rational Roots

Continue testing until one is found two suitable factors that satisfy \( U(x) \): - Substituting 2 into \( U(x) \): \[ U(2) = 2^{4} - 4(2)^{3} + 2^{2} + 12(2) - 12 = 16 - 32 + 4 + 24 - 12 = 0. \] Therefore, 2 is a root. Substituting -2 into \( U(x) \): \[ U(-2) = (-2)^{4} - 4(-2)^{3} + (-2)^{2} + 12(-2) - 12 = 16 + 32 + 4 - 24 - 12 = 16. \] Therefore, -2 is not root.
06

Factor the Polynomial

Since 2 is a root, then \( x - 2 \) is a factor. Use synthetic division to divide the polynomial by \( x - 2 \). Perform the division: Set up the division as follows: \begin{array}{r|rrrrr} 2 & 1 & -4 & 1 & 12 & -12 \ & & 2 & -4 & -6 & 12 \ \ \text{} & 1 & -2 & -3 & 6 & 0 \end{array} The quotient is: \[ x^{3} - 2x^{2} - 3x + 6. \]
07

Find Zeros of Quotient Polynomial

The new polynomial is \( x^{3} - 2x^{2} - 3x + 6 \). Use the Rational Root Theorem and synthetic division to find its roots. Repeating steps 2 through 4, let's start with 1: - Substituting 1 into \( x^{3} - 2x^{2} - 3x + 6 \): \[ 1^{3} - 2(1)^{2} - 3(1) + 6 = 1 - 2 - 3 + 6 = 2. \]Not a root. Try -1: - Substituting -1 into \( x^{3} - 2x^{2} - 3x + 6 \): \[ (-1)^{3} - 2(-1)^{2} - 3(-1) + 6 = -1 - 2 + 3 + 6 = 6. \]Not a root. Try 2: - Substituting 2 into \( x^{3} - 2x^{2} - 3x + 6 \): \[ 2^{3} - 2(2)^{2} - 3(2) + 6 = 8 - 8 - 6 + 6 = 0. \]So, 2 is a root.
08

Further Factorization

\[ x^{3} - 2x^{2} - 3x + 6 \) can be factored as \( (x - 2)(x^{2} + 1) \]. Compute the roots of \(x^{2} + 1 = 0\):\[ x^{2} + 1 = 0, \] thus \[ x = \pm i \].

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Key Concepts

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

polynomial degree
Understanding the degree of a polynomial is crucial in finding its zeros. The degree is the highest power of the variable in the polynomial. In the given example, \[ U(x) = x^{4} - 4x^{3} + x^{2} + 12x - 12, \] the polynomial degree is 4 because the highest power of \(x\) is 4. The degree tells us the maximum number of roots (real and complex) that we can expect. So, for a fourth-degree polynomial, you can find up to 4 roots.
Rational Root Theorem
The Rational Root Theorem helps in identifying possible rational roots of a polynomial. According to this theorem, if \(p/q\) is a root of the polynomial, then:
  • \(p,\) is a factor of the constant term.
  • \(q,\) is a factor of the leading coefficient.
For the polynomial \[ U(x) = x^{4} - 4x^{3} + x^{2} + 12x - 12, \] the constant term is -12, and the leading coefficient is 1. Thus, the possible rational roots are: \[ \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12. \] By testing each of these, we determine if they are indeed roots of the polynomial.
synthetic division
Synthetic division is a method for dividing a polynomial by a binomial of the form \(x - c\). It simplifies polynomial division significantly compared to long division. Let's see it in action for the polynomial \[ U(x) = x^{4} - 4x^{3} + x^{2} + 12x - 12, \] where we found that 2 is a root. This shows that \(x - 2\) is a factor of the polynomial.
Setting up the synthetic division:

\begin{array}{r|rrrrr} 2 & 1 & -4 & 1 & 12 & -12 \ & & 2 & -4 & -6 & 12 \ \text{} & 1 & -2 & -3 & 6 & 0 \end{array}\end{array} The quotient from the division is another polynomial: \( x^{3} - 2x^{2} - 3x + 6\). We can now repeat the process with this new polynomial to find its roots, continuing until we fully factorize the original polynomial.
complex roots
Complex roots come in pairs known as conjugate pairs. These occur when solving quadratic factors that do not have real solutions. For example, if we factorize \[ x^{2} + 1, \] we set it to zero to find the roots: \[ x^{2} + 1 = 0 \] Solving this gives us: \[ x = \pm i, \] where \(i\) is the imaginary unit (\(i = \sqrt{-1}\)). In the given polynomial, after performing synthetic division, one of the factors was \[ x^{2} + 1, \] leading us to the complex roots \(i\) and \(-i\). Therefore, the polynomial \[ U(x) = x^{4} - 4x^{3} + x^{2} + 12x - 12, \] has 2 real roots and 2 complex roots.

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