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Find all complex zeros of each polynomial function. Give exact values. List multiple zeros as necessary. $$f(x)=x^{4}+2 x^{3}-3 x^{2}+24 x-180$$

Short Answer

Expert verified
The complex zeros are: \(6, -3, -\frac{5}{2} + \frac{\sqrt{15}}{2}i, -\frac{5}{2} - \frac{\sqrt{15}}{2}i\).

Step by step solution

01

Identify Possible Rational Roots

Use the Rational Root Theorem to list all possible rational roots. The possible rational roots are given by the factors of the constant term (-180) over the factors of the leading coefficient (1). Hence, possible rational roots are: \(\frac{\text{factors of } -180}{\text{factors of } 1}\). These values are: \(\text{±1, ±2, ±3, ±4, ±5, ±6, ±9, ±10, ±12, ±15, ±18, ±20, ±30, ±36, ±45, ±60, ±90, ±180}\).
02

Use Synthetic Division to Test Rational Roots

Test these possible rational roots to see which ones are actually roots of the polynomial. Start with +1 and -1, then move to +2 and -2, etc., until a root is found. Use synthetic division: - For x = 2: \[\begin{array}{r|rrrrr} 2 & 1 & 2 & -3 & 24 & -180 \ & & 2 & 8 & 10 & 68 \ \hline & 1 & 4 & 5 & 34 & -112 \end{array}\] Remainder is not zero.- For x = -3: \[\begin{array}{r|rrrrr} -3 & 1 & 2 & -3 & 24 & -180 \ & & -3 & 3 & 0 & -36 \ \hline & 1 & -1 & 0 & 24 & -144 \end{array}\] Remainder is not zero.- For x = 6:\[\begin{array}{r|rrrrr} 6 & 1 & 2 & -3 & 24 & -180 \ & & 6 & 48 & -240 \ \hline & 1 & 8 & 45 & 0 \end{array}\] Remainder is zero. So, x = 6 is a root and we can rewrite the polynomial as \( (x - 6)(x^3 + 8x^2 + 45x + 30) \).
03

Factor the Depressed Polynomial

Now, factor the cubic polynomial \(x^3 + 8x^2 + 45x + 30\). To find more roots, use the Rational Root Theorem again:- For x = -1:\[\begin{array}{r|rrrr}-1 & 1 & 8 & 45 & 30 \ & & -1 & -7 & -38 \ \hline & 1 & 7 & 38 & -8 \end{array}\]Remainder is not zero.- For x = -2:\[\begin{array}{r|rrrr}-2 & 1 & 8 & 45 & 30 \ & & -2 & -12 & -66 \ \hline & 1 & 6 & 33 & -36 \end{array}\]Remainder is not zero.- For x = -3:\[\begin{array}{r|rrrr}-3 & 1 & 8 & 45 & 30 \ & & -3 & -15 & -90 \ \hline & 1 & 5 & 30 & 0 \end{array}\]Remainder is zero. So, \(x = -3\) is a root and we can rewrite the cubic polynomial as \( (x + 3)(x^2 + 5x + 10) \).
04

Solve the Quadratic Equation

The quadratic polynomial can be solved using the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]For \(x^2 + 5x + 10\): \[a = 1, b = 5, c = 10 \]Thus, \[x = \frac{-5 \pm \sqrt{25 - 40}}{2} = \frac{-5 \pm \sqrt{-15}}{2} = \frac{-5 \pm i\sqrt{15}}{2} \]Hence the roots are \(x = -\frac{5}{2} + \frac{\sqrt{15}}{2}i\) and \(x = -\frac{5}{2} - \frac{\sqrt{15}}{2}i\).
05

Combine All Roots

Combine all roots found. The roots of the polynomial are: \[x = 6, x = -3, x = -\frac{5}{2} + \frac{\sqrt{15}}{2}i, x = -\frac{5}{2} - \frac{\sqrt{15}}{2}i \]

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Key Concepts

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

Rational Root Theorem
The Rational Root Theorem is a useful tool to identify possible rational roots of polynomial equations. It states that any potential rational root of a polynomial equation \[P(x) = a_n x^n + a_{n-1} x^{n-1} + \text{...} + a_1 x + a_0\] must be a factor of the constant term \(a_0\) divided by a factor of the leading coefficient \(a_n\).To illustrate, consider our polynomial \(f(x)=x^4+2x^3-3x^2+24x-180\). Here, the constant term is \(-180\) and the leading coefficient is \(1\).
  • The factors of \(-180\) are: ±1, ±2, ±3, ±4, ±5, ±6, ±9, ±10, ±12, ±15, ±18, ±20, ±30, ±36, ±45, ±60, ±90, ±180
  • The factors of \(1\) are: ±1
Hence, the potential rational roots can be listed as the factors of \(-180\) over the factors of \(1\), which result in the possible values: ±1, ±2, ±3, ±4, ±5, ±6, ±9, ±10, ±12, ±15, ±18, ±20, ±30, ±36, ±45, ±60, ±90, ±180.Identifying these potential roots is essential to finding actual rational roots.
Synthetic Division
After listing the possible rational roots, synthetic division is an efficient method to test which values are roots of the polynomial.Unlike long division, synthetic division uses only the coefficients to perform the division.For example, to test if 6 is a root of our polynomial \(f(x)\), we set up synthetic division as follows:\[\begin{array}{r|rrrrr}6 & 1 & 2 & -3 & 24 & -180 \ & & 6 & 48 & 270 & 112 \hline & 1 & 8& 45 & 0 & \end{array}\]Here:
  • We bring the first coefficient (1) down as-is.
  • Multiply it by 6 and write it under the next coefficient (2).
  • Add these values, continue multiplying the result by 6 and writing below the subsequent coefficients.
When the remainder is zero (as in this case), it confirms that 6 is a root.
Quadratic Formula
Once the polynomial is reduced, if a quadratic equation remains, the quadratic formula can be used to find its roots.The quadratic formula is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Let's apply it to our reduced quadratic polynomial \(x^2 + 5x + 10\):
  • Here, \(a = 1\)
  • \(b = 5\)
  • \(c = 10\)
Substituting these values into the formula gives:\[x = \frac{-5 \pm \sqrt{25 - 40}}{2}\]Simplifying further:\[x = \frac{-5 \pm \sqrt{-15}}{2}\]Since \(-15\) under the square root results in an imaginary number, we simplify it to get:\[x = \frac{-5 \pm i\sqrt{15}}{2}\]Thus, the roots are complex: \(x = -\frac{5}{2} + \frac{\sqrt{15}}{2}i\) and \(x = -\frac{5}{2} - \frac{\sqrt{15}}{2}i\).
Factoring Polynomials
Factoring polynomials involves breaking down a polynomial into simpler polynomials that multiply together to give the original polynomial.After identifying a rational root using synthetic division or the Rational Root Theorem, the polynomial is 'depressed' (reduced) by that factor, simplifying it further.When factoring \(f(x)=x^4+2x^3-3x^2+24x-180\), after finding the root \(x = 6\) and depressing the polynomial, we rewrite it as:\[(x - 6)(x^3 + 8x^2 + 45x + 30)\]Next, we repeat the process to factor the cubic polynomial \(x^3 + 8x^2 + 45x + 30\). By testing further rational roots and using synthetic division, we find \(x = -3\):\[(x - 6)(x + 3)(x^2 + 5x + 10)\]Factoring simplifies solving higher-degree polynomials and finding their roots, revealing all rational and complex zeros in a structured manner.

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Most popular questions from this chapter

Solve each problem. AIDS Deaths in the United States The table* lists the total (cumulative) number of known deaths caused by AIDS in the United States up to 2007 (a) Plot the data. Let \(x=0\) correspond to the year 1990 . (b) Would a linear or a quadratic function model the data better? Explain. (c) Find a quadratic function defined by \(g(x)=a x^{2}+b x+c\) that models the data. (d) Plot the data together with \(g\) on the same coordinate plane. How well does \(g\) model the number of AIDS cases? (e) Use \(g\) to estimate the total number of AIDS deaths in the year 2010 . (f) Consider the last two entries in the table for the years 2006 and 2007 . Is it safe to assume that the quadratic model given for \(g(x)\) will continue for years 2008 and beyond? $$\begin{array}{c|c||c|c} \hline \text { Year } & \text { AIDS Deaths } & \text { Year } & \text { AIDS Deaths } \\ \hline 1990 & 119,821 & 1999 & 419,234 \\ \hline 1991 & 154,567 & 2000 & 436,373 \\ \hline 1992 & 191,508 & 2001 & 454,099 \\ \hline 1993 & 220,592 & 2002 & 471,417 \\ \hline 1994 & 269,992 & 2003 & 489,437 \\ \hline 1995 & 320,692 & 2004 & 507,536 \\ \hline 1996 & 359,892 & 2005 & 524,547 \\ \hline 1997 & 381,738 & 2006 & 565,927 \\ \hline 1998 & 400,743 & 2007 & 583,298 \\ \hline \end{array}$$

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Work each problem.Suppose \(y\) is directly proportional to \(x,\) and \(x\) is replaced by \(\frac{1}{3} x .\) What happens to \(y ?\)

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