Polynomial Functions
Definition of Polynomial Function
A polynomial function is any function specified by a rule that can be written as $ f: x a_{n} x^{n} + a_{n-1} x^{n-1} + + a_{1} x + a_{0} $ , where $ a_{n} $ . n is the degree of the polynomial function. The domain of a polynomial function, unless otherwise specified, is R.
Special Polynomial Functions
Special polynomial function types have already been discussed:
| DEGREE | EQUATION | NAME | GRAPH | ||
| n = 0 | $ f(x) = a_{0} \(</td><td>Constant function</td><td>Horizontal straight line</td></tr><tr><td>n = 1</td><td>\) f(x) = a_{1}x + a_{0} $ | Linear function | Straight line with slope $ a_{1} \(</td></tr><tr><td>n = 2</td><td>\) f(x) = a_{2}x^{2} + a_{1}x + a_{0} $ | Quadratic function | Parabola |
I nteger Power Functions
If f has degree n and all coefficients except $ a_{n} $ are zero, then $ f(x) = ax^{n} $ , where $ a = a_{n} $ . Then if n = 1, the graph of the function is a straight line through the origin. If n = 2, the graph of the function is a parabola with vertex at the origin. If n is an odd integer, the function is an odd function. If n is an even integer, the function is an even function.
EXAMPLE 14.1 Draw graphs of (a) $ f(x) = x^{3} $ ; (b) $ f(x) = x^{5} $ ; (c) $ f(x) = x^{7} $
- Fig. 14-1; (b) Fig. 14-2; (c) Fig. 14-3.
EXAMPLE 14.2 Draw graphs of (a) $ f(x) = x^{4} $ ; (b) $ f(x) = x^{6} $ ; (c) $ f(x) = x^{8} $ .
- Fig. 14-4; (b) Fig. 14-5; (c) Fig. 14-6.
Zeros of Polynomials
If $ f(c) = 0 $ , c is called a zero of the polynomial $ f(x) $ .
Division of Polynomials
If a polynomial $ g(x) $ is a factor of another polynomial $ f(x) $ , then $ f(x) $ is said to be divisible by $ g(x) $ . Thus $ x^{3}-1 $ is divisible both by x-1 and by $ x^{2}+x+1 $ . If a polynomial is not divisible by another, it is possible to apply the technique of long division to find a quotient and remainder, as in the following examples:
EXAMPLE 14.3 Find the quotient and remainder for $ (2x{4}-x{2}-2)/(x^{2}+2x-1) $ .
Arrange the dividend and divisor in descending powers of the variable. Insert terms with zero coefficients and use the long division scheme.
\[ \begin{aligned}x^{2}+2x-1\sqrt{\begin{array}{l}2x^{2}-4x+9\\2x^{4}+0x^{3}-x^{2}+0x-2\\\hline2x^{4}+4x^{3}-2x^{2}\\-4x^{3}+x^{2}+0x\\\hline-4x^{3}-8x^{2}+4x\\9x^{2}-4x-2\\\hline9x^{2}+18x-9\\-22x+7\\\end{array}}\end{aligned} \]
Divide first term of dividend by first term of divisor
Multiply divisor by $ 2x^{2} $ ; subtract
Bring down next term; repeat division step
Multiply divisor by -4x; subtract
Bring down next term; repeat division step
Multiply divisor by 9; subtract
Remainder; degree is less than degree of divisor
The quotient is \(2x^{2}-4x+9\) and the remainder is \(-22x+7\). Thus:
\[ \frac{2x^{4}-x^{2}-2}{x^{2}+2x-1}=2x^{2}-4x+9+\frac{-22x+7}{x^{2}+2x-1} \]
Division Algorithm for Polynomials
If $ f(x) $ and $ g(x) $ are polynomials, with $ g(x) $ , then there exist unique polynomials $ q(x) $ and $ r(x) $ such that
\[ f(x)=g(x)q(x)+r(x)\quad and\quad\frac{f(x)}{g(x)}=q(x)+\frac{r(x)}{g(x)} \]
Either $ r(x) = 0 ( f(x) g(x) ) $ or the degree of $ r(x) $ is less than the degree of $ g(x) $ .
Therefore, if the degree of $ g(x) $ is 1, the degree of $ r(x) $ is 0, and the remainder is a constant polynomial r.
EXAMPLE 14.4 Find the quotient and remainder for $ (x{3}-5x{2}+7x-9)/(x-4) $ .
Use the long division scheme:
\[ \begin{aligned}x-4\sqrt{\begin{array}{c}x^{2}\quad-\quad x+\quad3\\x^{3}-5x^{2}\quad+\quad7x\quad-\quad9\end{array}}&\quad Divide first term of dividend by first term of divisor\\ \frac{x^{3}-4x^{2}}{\quad-x^{2}\quad+\quad7x}&\quad Multiply divisor by x^{2};subtract\\ \frac{-x^{2}+4x}{\quad3x\quad-\quad9}&\quad Bring down next term;repeat division step\\ \frac{3x-12}{\quad3}&\quad Multiply divisor by3;subtract\\ \quad&Remainder;degree is less than degree of divisor\end{aligned} \]
The quotient is $ x^{2} - x + 3 $ and the remainder is the constant 3. Thus
\[ \frac{x^{3}-5x^{2}+7x-9}{x-4}=x^{2}-x+3+\frac{3}{x-4} \]
Synthetic Division
Division of a polynomial $ f(x) $ by a polynomial of form x - c is accomplished efficiently by the synthetic division scheme. Arrange coefficients of the dividend $ f(x) $ in descending order in the first row of a three-row array.
\[ c\mid a_{n}a_{n-1}\cdots a_{1}a_{0} \]
The third row is formed by bringing down the first coefficient of $ f(x) $ , then successively multiplying each coefficient in the third row by c, placing the result in the second row, adding this to the corresponding coefficient in the first row, and placing the result in the next position in the third row.
\[ \begin{array}{c}c\mid a_{n}\quad a_{n-1}\quad\cdots\quad a_{1}\quad a_{0}\\\frac{c a_{n}\quad c b_{1}\quad\cdots\quad c b_{n-2}\quad c b_{n-1}}{a_{n}\quad b_{1}\quad\cdots\quad b_{n-1}\quad r}\end{array} \]
The last coefficient in the third row is the constant remainder; the other coefficients are the coefficients of the quotient, in descending order.
EXAMPLE 14.5 Use synthetic division to find the quotient and remainder in the previous example.
In this case, c = 4. Arrange the coefficients of $ x^{3} - 5x^{2} + 7x - 9 $ in the first row of a three-row array; proceed to bring down the first coefficient, 1, then multiply by 4, place the result in the second row, add to -5, place the result in the third row. Continue to the last coefficient of the array.
\[ \begin{aligned}&4\left|\begin{array}{cccc}{{{1}}}&{{{-5}}}&{{{7}}}&{{{-9}}} \\{{{\hline}}}&{{{4}}}&{{{-4}}}&{{{12}}} \\{{{1}}}&{{{-1}}}&{{{3}}}&{{{3}}} \\\end{array}\right.\\ \end{aligned} \]
As before, the quotient is $ x^{2}-x+3 $ and the remainder is 3.
Remainder Theorem
When the polynomial $ f(x) $ is divided by x - c, the remainder is $ f(c) $ .
EXAMPLE 14.6 Verify the remainder theorem for the polynomial $ f(x) = x^{3} - 5x^{2} + 7x - 9 $ divided by x - 4.
Calculate $ f(4) = 4^{3} - 5 ^{2} + 7 - 9 = 3 $ . The remainder in division has already been shown to be 3; thus, the conclusion of the theorem holds.
Factor Theorem
A polynomial $ f(x) $ has a factor of x - c if and only if $ f(c) = 0 $ . Thus, x - c is a factor of a polynomial if and only if c is a zero of the polynomial.
EXAMPLE 14.7 Use the factor theorem to verify that $ x + 2 $ is a factor of $ x^{5} + 32 $ .
Let $ f(x) = x^{5} + 32 $ ; then $ f(-2) = (-2)^{5} + 32 = 0 $ ; hence $ x - (-2) = x + 2 $ is a factor of $ f(x) $ .
Fundamental Theorem of Algebra
Every polynomial of positive degree with complex coefficients has at least one complex zero.
Corollaries of the Fundamental Theorem
- Every polynomial of positive degree n has a factorization of the form
\[ P(x)=a_{_{n}}\left(x-r_{_{1}}\right)\left(x-r_{_{2}}\right)\cdot\cdot\cdot\left(x-r_{_{n}}\right) \]
where the $ r_{i} $ are not necessarily distinct. If in the factorization $ x - r_{i} $ occurs m times, $ r_{i} $ is called a zero of multiplicity m. However, it is not necessarily possible to find the factorization using exact algebraic methods.
- A polynomial of degree n has at most n complex zeros. If a zero of multiplicity m is counted as m zeros, then a polynomial of degree n has exactly n zeros.
Further Theorems about Zeros
Further theorems about zeros of polynomials:
If $ P(x) $ is a polynomial with real coefficients, and if z is a complex zero of $ P(x) $ , then the complex conjugate $ {z} $ is also a zero of $ P(x) $ . That is, complex zeros of polynomials with real coefficients occur in complex conjugate pairs.
Any polynomial of degree n > 0 with real coefficients has a complete factorization using linear and quadratic factors, multiplied by the leading coefficient of the polynomial. However, it is not necessarily possible to find the factorization using exact algebraic methods.
If $ P(x) = a_{n}x^{n} + a_{n-1}x^{n-1} + + a_{1}x + a_{0} $ is a polynomial with integral coefficients and r = p/q is a rational zero of $ P(x) $ in lowest terms, then p must be a factor of the constant term $ a_{0} $ and q must be a factor of the leading coefficient $ a_{n} $ .
EXAMPLE 14.8 Find a polynomial of least degree with real coefficients and zeros 2 and 1 - 3i.
By the factor theorem, c is a zero of a polynomial only if x - c is a factor. By the theorem on zeros of polynomials with real coefficients, if 1 - 3i is a zero of this polynomial, then so is $ 1 + 3i $ . Hence the polynomial can be written as
\[ P(x)=a(x-2)[x-(1-3i)][x-(1+3i)] \]
Simplifying yields:
\[ \begin{aligned}P(x)&=a(x-2)[(x-1)+3i][(x-1)-3i]\\&=a(x-2)[(x-1)^{2}-(3i)^{2}]\\&=a(x-2)(x^{2}-2x+10)\\&=a(x^{3}-4x^{2}+14x-20)\\ \end{aligned} \]
EXAMPLE 14.9 List the possible rational zeros of $ 3x^{2} + 5x - 8 $ .
From the theorem on rational zeros of polynomials with integer coefficients, the possible rational zeros are:
\[ \frac{Factors of-8}{Factors of3}=\frac{\pm1,\pm2,\pm4,\pm8}{\pm1,\pm3}=\pm1,\pm2,\pm4,\pm8,\pm\frac{1}{3},\pm\frac{2}{3},\pm\frac{4}{3},\pm\frac{8}{3} \]
Note that the actual zeros are 1 and $ - $ .
Theorems Used In Locating Zeros
Theorems used in locating zeros of polynomials:
INTERMEDIATE VALUE THEOREM: Given a polynomial $ f(x) $ with a < b, if $ f(a) f(b) $ , then $ f(x) $ takes on every value c between a and b in the interval $ (a, b) $ .
COROLLARY: For a polynomial $ f(x) $ , if $ f(a) $ and $ f(b) $ have opposite signs, then $ f(x) $ has at least one zero between a and b.
DESCARTES’ RULE OF SIGNS: If $ f(x) $ is a polynomial with terms arranged in descending order, then the number of positive real zeros of $ f(x) $ is either equal to the number of sign changes between successive terms of $ f(x) $ or is less than this number by an even number. The number of negative real zeros of $ f(x) $ is found by applying this rule to $ f(-x) $ .
If the third line of a synthetic division of $ f(x) $ by x - r is all positive for some r > 0, then r is an upper bound for the zeros of $ f(x) $ ; that is, there are no zeros greater than r. If the terms in the third line of a synthetic division of $ f(x) $ by x - r alternate in sign for some r < 0, then r is a lower bound for the zeros of $ f(x) $ ; that is, there are no zeros less than r. (0 may be regarded as positive or negative for the purpose of this theorem.)
Solving Polynomial Equations
Solving polynomial equations and graphing polynomials:
The following statements are equivalent:
c is a zero of $ P(x) $ .
c is a solution of the equation $ P(x) = 0 $ .
x - c is a factor of $ P(x) $ .
For real c, the graph of $ y = P(x) $ has an x-intercept at c.
Graphing a Polynomial
To graph a polynomial function for which all factors can be found:
Write the polynomial in factored form.
Determine the sign behavior of the polynomial from the signs of the factors.
Enter the x-intercepts of the polynomial on the x-axis.
If desired, form a table of values.
Sketch the graph of the polynomial as a smooth curve.
EXAMPLE 14.10 Sketch a graph of $ y = 2x(x - 3)(x + 2) $
The polynomial is already in factored form. Use the methods of Chapter 6 to obtain the sign chart shown in Fig. 14-7.
The graph has x-intercepts -2,0,3 and is below the x-axis on the intervals $ (-∞, -2) $ and $ (0,3) $ and above the x-axis on the intervals $ (-2,0) $ and $ (3,∞) $ . Form a table of values as shown and sketch the graph as a smooth curve (Fig. 14-8).
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
| y | -36 | 0 | 8 | 0 | -12 | -16 | 0 | 48 |
SOLVED PROBLEMS
14.1. Prove the remainder theorem.
By the division algorithm, there exist polynomials $ q(x) $ and $ r(x) $ such that $ f(x) = q(x)(x - c) + r(x) $ . Since the degree of $ r(x) $ is less than the degree of x - c, that is, less than 1, the degree of $ r(x) $ must be zero. Thus $ r(x) $ is a constant; call it r. Thus, for all x,
\[ f(x)=q(x)(x-c)+r \]
In particular, let \(x = c\). Then \(f(c) = q(c)(c - c) + r\), that is, \(f(c) = r\). Thus,
\[ f(x)=q(x)(x-c)+f(c) \]
In other words, the remainder when $ f(x) $ is divided by c is $ f(c) $ .
14.2. Find the quotient and remainder when $ 2x^{3} + 3x^{2} - 13x + 5 $ is divided by 2x - 3. Use the long division scheme:
\[ 2x - 3 \sqrt{\begin{array}{c|c} x^{2} + 3x - 2 \\ 2x^{3} + 3x^{2} - 13x + 5 & Divide first term of dividend by first term of divisor \\ 2x^{3} - 3x^{2} & Multiply divisor by x^{2}; subtract \\ 6x^{2} - 13x & Bring down next term; repeat division step \\ 6x^{2} - 9x & Multiply divisor by 3; subtract \\ -4x + 5 & Bring down next term; repeat division step \\ -4x + 6 & Multiply divisor by -2; subtract \\ -1 & Remainder; degree is less than degree of divisor \end{array}} \]
The quotient is $ x^{2} + 3x - 2 $ and the remainder is -1.
14.3. Find the quotient and remainder when $ 3x{5}-7x{3}+5x^{2}+6x-6 $ is divided by $ x^{3}-x+2 $ .
Use the long division scheme:
\[ \begin{aligned}x^{3}-x+2\sqrt{\begin{array}{r}3x^{2}\quad-4\\3x^{5}\quad-7x^{3}+5x^{2}+6x-6\\\end{array}}\\\frac{3x^{5}\quad-3x^{3}+6x^{2}}{-4x^{3}-\quad x^{2}+6x-6}\\\frac{-4x^{3}\quad+4x-8}{-x^{2}+2x+2}\end{aligned} \]
Divide first term of dividend by first term of divisor
Multiply divisor by $ 3x^{2} $ ; subtract
Bring down next term; repeat division step
Multiply divisor by -4; subtract
Remainder; degree is less than degree of divisor
The quotient is $ 3x^{2}-4 $ and the remainder is $ -x^{2}+2x+2 $ .
14.4. Find the quotient and remainder when $ 2x^{3} + 5x^{2} - 10x + 9 $ is divided by $ x + 2 $ .
Use the synthetic division scheme. Note that in dividing by x - c, the coefficient c is placed in the upper-left corner and used to multiply the numbers generated in the third line. In division by $ x + 2 $ , that is, $ x - (-2) $ , use c = -2.
\[ \begin{aligned}\left.\begin{array}{r r r r r}{{{-2}}}&{{{2}}}&{{{5}}}&{{{-10}}}&{{{9}}} \\{{{2}}}&{{{-4}}}&{{{-2}}}&{{{24}}} \\{{{2}}}&{{{1}}}&{{{-12}}}&{{{33}}}\end{array}\right.\end{aligned} \]
The quotient is $ 2x^{2} + x - 12 $ and the remainder is 33.
14.5. Find the quotient and remainder when $ -3t^{5} + 10t^{4} + 15t^{2} + 18t - 6 $ is divided by t - 4.
Use the scheme for synthetic division by \(t-c\), with \(c=4\). Enter a zero for the missing coefficient of \(t^{3}\).
\[ \begin{aligned}&4\left|\begin{array}{ccccc}{{{-3}}}&{{{10}}}&{{{0}}}&{{{15}}}&{{{18}}}&{{{-6}}} \\{{{-12}}}&{{{-8}}}&{{{-32}}}&{{{-68}}}&{{{-200}}} \\{{{-3}}}&{{{-2}}}&{{{-8}}}&{{{-17}}}&{{{-50}}}&{{{-206}}} \\\end{array}\right.\\ \end{aligned} \]
The quotient is \(-3t^{4}-2t^{3}-8t^{2}-17t-50\) and the remainder is \(-206\).
14.6. Find the quotient and remainder when $ 2x{3}-5x{2}+6x-3 $ is divided by $ x- $ .
Use the scheme for synthetic division by x - c, with $ c = $ .
\[ \begin{aligned}\left.\frac{\frac{1}{2}\right|&\left.\begin{array}{cccc}{{{2}}}&{{{-5}}}&{{{6}}}&{{{-3}}} \\{{{1}}}&{{{-2}}}&{{{2}}} \\{{{2}}}&{{{-4}}}&{{{4}}}&{{{-1}}} \\\end{array}\right.\end{aligned} \]
The quotient is $ 2x^{2}-4x+4 $ and the remainder is -1.
14.7. Find the quotient and remainder when $ 3x^{4} + 8x^{3} - x^{2} + 7x + 2 $ is divided by $ x + $ .
Use the scheme for synthetic division by x - c, with $ c = - $ .
\[ \begin{array}{c|cccc}{{{-\frac{2}{3}}}}&{{{3}}}&{{{8}}}&{{{-1}}}&{{{7}}}&{{{2}}} \\{{{3}}}&{{{-2}}}&{{{-4}}}&{{{\frac{10}{3}}}}&{{{-\frac{62}{9}}}} \\{{{3}}}&{{{6}}}&{{{-5}}}&{{{\frac{31}{3}}}}&{{{-\frac{44}{9}}}} \\\end{array} \]
The quotient is $ 3x^{3} + 6x^{2} - 5x + $ and the remainder is $ - $ .
14.8. Prove the factor theorem.
By the remainder theorem, when \(f(x)\) is divided by \(x - c\), the remainder is \(f(c)\).
Assume c is a zero of \(f(x)\); then \(f(c)=0\). Therefore, \(f(x)=q(x)(x-c)+f(c)=q(x)(x-c)\), that is, \(x-c\) is a factor of \(f(x)\).
Conversely, assume x - c is a factor of $ f(x) $ ; then the remainder when $ f(x) $ is divided by x - c must be zero. By the remainder theorem, this remainder is $ f(c) $ ; hence, $ f(c) = 0 $ .
14.9. Show that \(x - a\) is a factor of \(x^{n} - a^{n}\) for all integers \(n\).
Let $ f(x) = x^{n} - a^{n} $ ; then $ f(a) = a^{n} - a^{n} = 0 $ . By the factor theorem, since a is a zero of $ f(x) $ , x - a is a factor.
14.10. Use the quadratic formula and the factor theorem to factor (a) $ x^{2} - 12x + 3 $ ; (b) $ x^{2} - 4x + 13 $ .
- The zeros of $ x^{2} - 12x + 3 $ , that is, the solutions of $ x^{2} - 12x + 3 = 0 $ , are found from the quadratic formula. Using a = 1, b = -12, c = 3 yields
\[ x=\frac{-(-12)\pm\sqrt{(-12)^{2}-4(1)(3)}}{2(1)}=\frac{12\pm\sqrt{132}}{2}=6\pm\sqrt{33} \]
Since the zeros are \(6 \pm \sqrt{33}\), the factors are \(x - (6 + \sqrt{33})\) and \(x - (6 - \sqrt{33})\). Thus
\[ x^{2}-12x+3=[x-(6+\sqrt{33})][x-(6-\sqrt{33})]or[(x-6)-\sqrt{33}][(x-6)+\sqrt{33}] \]
- Proceeding as in (a), use the quadratic formula with \(a = 1\), \(b = -4\), \(c = 13\) to obtain
\[ x=\frac{-(-4)\pm\sqrt{(-4)^{2}-4(1)(13)}}{2(1)}=\frac{4\pm\sqrt{-36}}{2}=2\pm3i \]
Since the zeros are $ 2 3i $ , the factors are $ x - (2 + 3i) $ and $ x - (2 - 3i) $ . Thus
\[ x^{2}-4x+13=[x-(2+3i)][x-(2-3i)]or[(x-2)-3i][(x-2)+3i] \]
14.11. Write the polynomial $ P(x) = x^{4} - 7x^{3} + 13x^{2} + 3x - 18 $ as a product of first-degree factors, given that 3 is a zero of multiplicity 2.
Since 3 is a zero of multiplicity 2, there exists a polynomial $ g(x) $ with $ P(x) = (x - 3)^{2} g(x) $ . To find $ g(x) $ , use the scheme for synthetic division by x - c, with c = 3, twice:
\[ \begin{aligned}3\left|\begin{array}{ccccc}{{{1}}}&{{{-7}}}&{{{13}}}&{{{3}}}&{{{-18}}} \\{{{3}}}&{{{-12}}}&{{{3}}}&{{{18}}} \\{{{3}}}&{{{1}}}&{{{-4}}}&{{{1}}}&{{{6}}}&{{{0}}} \\{{{\hline}}}&{{{3}}}&{{{-3}}}&{{{-6}}} \\{{{1}}}&{{{-1}}}&{{{-2}}}&{{{0}}} \\\end{array}\right.\end{aligned} \]
Thus
\[ \begin{aligned}P(x)&=(x-3)(x-3)(x^{2}-x-2)\\&=(x-3)(x-3)(x-2)(x+1)\end{aligned} \]
14.12. Write the polynomial $ P(x) = 2x^{3} + 2x^{2} - 40x - 100 $ as a product of first-degree factors, given that -3 - i is a zero. Find all zeros of $ P(x) $ .
Since $ P(x) $ has real coefficients and -3 - i is a zero, $ -3 + i $ is also a zero. Therefore, there exists a polynomial $ g(x) $ with $ P(x) = [x - (-3 - i)][x - (-3 + i)]g(x) $ . To find $ g(x) $ , use the scheme for synthetic division by x - c with, in turn, c = -3 - i and $ c = -3 + i $ .
\[ \begin{array}{ccc}{{{-3-i\left|\quad2\quad2}}}&{{{-40}}}&{{{-100}}} \\{{{-3+i\left|\quad\frac{-6-2i\quad10+10i\quad100}{2-4-2i-30+10i\quad0}\quad\frac{-6+2i\quad30-10i}{2-10\quad0}}} \\{{{\hline}}} \\\end{array} \]
Thus
\[ P(x)=[x-(-3-i)][x-(-3+i)](2x-10) \]
and the zeros of $ P(x) $ are $ -3 i $ and 5.
14.13. Find a polynomial $ P(x) $ of lowest degree, with real coefficients, such that 4 is a zero of multiplicity 3, -2 is a zero of multiplicity 2, 0 is a zero, and $ 5 + 2i $ is a zero.
Since $ P(x) $ has real coefficients and $ 5 + 2i $ is a zero, 5 - 2i is also a zero. Thus, write
\[ \begin{aligned}P(x)&=a(x-4)^{3}[x-(-2)]^{2}(x-0)[x-(5+2i)][x-(5-2i)]\\&=a(x-4)^{3}(x+2)^{2}x[(x-5)-2i][(x-5)+2i]\\&=a(x-4)^{3}(x+2)^{2}x(x^{2}-10x+29)\\ \end{aligned} \]
Here a can be any real number.
14.14. Find a polynomial $ P(x) $ of lowest degree, with integer coefficients, such that $ , $ , and $ - $ are zeros.
Write
\[ \begin{aligned}P(x)&=a\bigg(x-\frac{2}{3}\bigg)\bigg(x-\frac{3}{4}\bigg)\bigg[x-\bigg(-\frac{1}{2}\bigg)\bigg]\\&=a\bigg(\frac{3x-2}{3}\bigg)\bigg(\frac{4x-3}{4}\bigg)\bigg(\frac{2x+1}{2}\bigg)\\&=24b\bigg(\frac{3x-2}{3}\bigg)\bigg(\frac{4x-3}{4}\bigg)\bigg(\frac{2x+1}{2}\bigg)\\&=b(3x-2)(4x-3)(2x+1)\\ \end{aligned} \]
Here b can be any integer.
14.15. Show that $ f(x) = x^{3} - 5 $ has a zero between 1 and 2.
Since $ f(1) = 1^{3} - 5 = -4 $ and $ f(2) = 2^{3} - 5 = 3 $ , $ f(1) $ and $ f(2) $ have opposite signs. Hence the polynomial has at least one zero between 1 and 2.
14.16. Show that $ f(x)=2x{4}+3x{3}+x^{2}-2x-8 $ has a zero between -2 and -1.
Use the scheme for synthetic division with c = -2 and c = -1.
\[ \begin{aligned}\left.\begin{array}{r r r r r r}-2\left|\quad2\quad3\quad1\quad-2\quad-8\right.\right.\\\left.\quad-4\quad2\quad-6\quad16\right.\\\left.\quad2-1\quad3\quad-8\quad8\right.\end{array}\right.\end{aligned} \]
\[ \begin{aligned}&-1\left|\begin{array}{cccc}{{{2}}}&{{{3}}}&{{{1}}}&{{{-2}}}&{{{-8}}} \\{{{-2}}}&{{{-1}}}&{{{0}}}&{{{2}}} \\{{{2}}}&{{{1}}}&{{{0}}}&{{{-2}}}&{{{-6}}} \\\end{array}\right.\end{aligned} \]
Since $ f(-2) = 8 $ and $ f(-1) = -6 $ , $ f(-2) $ and $ f(-1) $ have opposite signs. Hence the polynomial has at least one zero between -2 and -1.
14.17. Use Descartes’ rule of signs to analyze the possible combinations of positive, negative, and imaginary zeros for $ f(x) = x^{3} - 3x^{2} + 2x + 8 $ .
The coefficients of $ f(x) $ exhibit two changes of sign. Thus there could be two or zero positive real zeros for f. To find the possible number of negative zeros, consider $ f(-x) $ .
\[ f(-x)=(-x)^{3}-3(-x)^{2}+2(-x)+8=-x^{3}-3x^{2}-2x+8 \]
The coefficients of $ f(-x) $ exhibit one change of sign. Thus there must be one negative real zero for f. Since there are either three or one real zeros, there can be either no or two imaginary zeros.
The table indicates the possible combinations of zeros:
| POSITIVE | NEGATIVE | IMAGINARY |
| 2 | 1 | 0 |
| 0 | 1 | 2 |
14.18. Use Descartes’ rule to signs of analyze the possible combinations of positive, negative, and imaginary zeros for $ f(x) = -2x^{6} + 3x^{5} - 3x^{3} + 5x^{2} - 6x + 9 $ .
The coefficients of $ f(x) $ exhibit five changes of sign. Thus there could be five or three positive real zeros for f or one positive real zero for f.
To find the possible number of negative zeros, consider $ f(-x) $ .
\[ \begin{aligned}f(-x)&=-2(-x)^{6}+3(-x)^{5}-3(-x)^{3}+5(-x)^{2}-6(-x)+9\\&=-2x^{6}-3x^{5}+3x^{3}+5x^{2}+6x+9\end{aligned} \]
The coefficients of $ f(-x) $ exhibit one change of sign. Thus there must be one negative real zero for f. Since there are either six, four, or two real zeros, there can be either no, two, or four imaginary zeros. The table indicates the possible combinations of zeros:
| POSITIVE | NEGATIVE | IMAGINARY |
| 5 | 1 | 0 |
| 3 | 1 | 2 |
| 1 | 1 | 4 |
14.19. Use Descartes’ rule of signs to show that $ f(x) = x^{3} + 7 $ has no positive real zeros and must have a real negative zero.
Since $ f(x) $ exhibits no changes of sign, there can be no positive real zeros. To find the possible number of negative zeros, consider $ f(-x) $ .
\[ f(-x)=(-x)^{3}+7=-x^{3}+7 \]
The coefficients of $ f(-x) $ exhibit one change of sign. Thus there must be one negative real zero for f.
14.20. Use Descartes’ rule of signs to show that $ f(x) = x^{4} + 2x^{2} + 1 $ has no real zeros.
Since $ f(x) $ exhibits no changes of sign, there can be no positive real zeros. To find the possible number of negative zeros, consider $ f(-x) $ .
\[ f(-x)=(-x)^{4}+2(-x)^{2}+1=x^{4}+2x^{2}+1 \]
Since $ f(-x) $ exhibits no changes of sign, there can be no negative real zeros. Since 0 is not a zero, there can be no real zeros of $ f(x) $ .
14.21. Find the smallest positive integer and the largest negative integer that are, respectively, upper and lower bounds for the zeros of $ f(x) = x^{3} + 2x^{2} - 3x - 5 $ .
Use the scheme for synthetic division by x - c, with c = successive positive integers (only the last line in the synthetic division is shown).
| 1 | 2 | -3 | -5 | |
| 1 | 1 | 3 | 0 | -5 |
| 2 | 1 | 4 | 5 | 5 |
Since the last line in synthetic division by x - 2 is all positive and the last line in synthetic division by x - 1 is not, 2 is the smallest positive integer that is an upper bound for the zeros of f.
Use the scheme for synthetic division by x - c, with c = successive negative integers (only the last line in the synthetic division is shown).
\[ \begin{array}{c|cccc}{{{1}}}&{{{2}}}&{{{-3}}}&{{{-5}}} \\{{{\hline-1}}}&{{{1}}}&{{{-4}}}&{{{-1}}} \\{{{-2}}}&{{{1}}}&{{{-3}}}&{{{1}}} \\{{{-3}}}&{{{1}}}&{{{-1}}}&{{{0}}}&{{{-5}}} \\\end{array} \]
Since the last line in synthetic division by $ x + 3 $ alternates in sign (recall that 0 can be regarded as positive or negative in this context) and the last line in synthetic division by $ x + 2 $ does not, -3 is the largest negative integer that is a lower bound for the zeros of f.
14.22. Use the corollary of the intermediate value theorem to locate, between successive integers, the zeros of f in the previous problem.
From the synthetic divisions carried out in the previous problem, since $ f(1) $ and $ f(2) $ have opposite signs, there is a zero of f between 1 and 2. Similarly, since $ f(-1) $ and $ f(-2) $ have opposite signs, there is a zero of f between -1 and -2. Finally, since $ f(-2) $ and $ f(-3) $ have opposite signs, there is a zero of f between -2 and -3.
14.23. Use the corollary of the intermediate value theorem to locate, between successive integers, the zeros of $ f(x)=x{4}-3x{3}-6x^{2}+33x-35 $ .
Use the scheme for synthetic division by x - c, with c = successive positive integers, then 0, then successive negative integers (only the last line in the synthetic divisions is shown).
\[ \begin{array}{c|cccc}{{{1}}}&{{{-3}}}&{{{-6}}}&{{{33}}}&{{{-35}}} \\{{{\hline1}}}&{{{1}}}&{{{-2}}}&{{{-8}}}&{{{25}}}&{{{-10}}} \\{{{2}}}&{{{1}}}&{{{-1}}}&{{{-8}}}&{{{17}}}&{{{-1}}} \\{{{3}}}&{{{1}}}&{{{0}}}&{{{-6}}}&{{{15}}}&{{{10}}} \\{{{4}}}&{{{1}}}&{{{1}}}&{{{-2}}}&{{{25}}}&{{{65}}} \\{{{5}}}&{{{1}}}&{{{2}}}&{{{4}}}&{{{53}}}&{{{230}}} \\{{{0}}}&{{{1}}}&{{{-3}}}&{{{-6}}}&{{{33}}}&{{{-35}}} \\{{{-1}}}&{{{1}}}&{{{-4}}}&{{{-2}}}&{{{35}}}&{{{-70}}} \\{{{-2}}}&{{{1}}}&{{{-5}}}&{{{4}}}&{{{25}}}&{{{-85}}} \\{{{-3}}}&{{{1}}}&{{{-6}}}&{{{12}}}&{{{-3}}}&{{{-26}}} \\{{{-4}}}&{{{1}}}&{{{-7}}}&{{{22}}}&{{{-55}}}&{{{185}}} \\\end{array} \]
Since $ f(2) $ and $ f(3) $ have opposite signs, there is a zero of f between 2 and 3. No other positive real zeros can be isolated from the data in the table (5 is an upper bound for the positive real zeros).
Since $ f(-3) $ and $ f(-4) $ have opposite signs, there is a zero of f between -3 and -4. No other negative real zeros can be isolated from the data in the table (-4 is a lower bound for the negative real zeros).
14.24. List the possible rational zeros of $ x{3}-5x{2}+7x-12 $
From the theorem on rational zeros of polynomials with integer coefficients, the possible rational zeros are:
\[ \frac{Factors of-12}{Factors of1}=\frac{\pm1,\pm2,\pm3,\pm4,\pm6,\pm12}{\pm1}=\pm1,\pm2,\pm3,\pm4,\pm6,\pm12 \]
14.25. List the possible rational zeros of $ 4x^{3} + 5x^{2} + 7x - 18 $
From the theorem on rational zeros of polynomials with integer coefficients, the possible rational zeros are:
\[ \frac{Factors of-18}{Factors of4}=\frac{\pm1,\pm2,\pm3,\pm6,\pm9,\pm18}{\pm1,\pm2,\pm4}=\pm1,\pm2,\pm3,\pm6\pm9,\pm18,\pm\frac{1}{2},\pm\frac{3}{2},\pm\frac{9}{2},\pm\frac{1}{4},\pm\frac{3}{4},\pm\frac{9}{4} \]
14.26. Find all zeros of $ f(x)=x{3}+3x{2}-10x-24 $
From Descartes’ rule of signs, the following combinations of positive, negative, and imaginary zeros are possible.
| POSITIVE | NEGATIVE | IMAGINARY |
| 1 | 2 | 0 |
| 1 | 0 | 2 |
From the theorem on rational zeros of polynomials with integer coefficients, the possible rational zeros are $ $ , $ $ , $ $ , $ $ $ $ , $ $ , $ $ , $ $ .
Use the scheme for synthetic division by x - c, with c = successive positive integers from this list (only the last line in the synthetic division is shown).
\[ \begin{array}{c|cccc}{{{1}}}&{{{3}}}&{{{-10}}}&{{{-24}}} \\{{{\hline1}}}&{{{1}}}&{{{4}}}&{{{-6}}}&{{{-30}}} \\{{{2}}}&{{{1}}}&{{{5}}}&{{{0}}}&{{{-24}}} \\{{{3}}}&{{{1}}}&{{{6}}}&{{{8}}}&{{{0}}} \\\end{array} \]
Thus, 3 is a zero and the polynomial can be factored as follows:
\[ \begin{aligned}f(x)&=(x-3)\left(x^{2}+6x+8\right)\\&=(x-3)\left(x+2\right)(x+4)\end{aligned} \]
Hence the zeros are 3, -2, and -4.
14.27. Find all zeros of $ f(x) = 3x^{4} + 16x^{3} + 20x^{2} - 9x - 18 $
From Descartes’ rule of signs, the following combinations of positive, negative, and imaginary zeros are possible.
| POSITIVE | NEGATIVE | IMAGINARY |
| 1 | 3 | 0 |
| 1 | 1 | 2 |
From the theorem on rational zeros of polynomials with integer coefficients, the possible rational zeros are
\[ \frac{Factors of-18}{Factor of3}=\frac{\pm1,\pm2,\pm3,\pm6,\pm9,\pm18}{\pm1,\pm3}=\pm1,\pm2,\pm3,\pm6,\pm9,\pm18,\pm\frac{1}{3},\pm\frac{2}{3} \]
Use the scheme for synthetic division by x - c, with c = successive positive integers from this list (only the last line in the synthetic division is shown).
| 3 | 16 | 20 | -9 | -18 | |
| 1 | 3 | 19 | 39 | 30 | 12 |
Since \(f(0)\) and \(f(1)\) have opposite signs, the positive zero is between 0 and 1.
Now use the scheme for synthetic division by x - c, with c = successive negative integers from the list (only the last line in the synthetic division is shown).
| 3 | 16 | 20 | -9 | -18 | |
| -1 | 3 | 13 | 7 | -16 | -2 |
| -2 | 3 | 10 | 0 | -9 | 0 |
Thus, -2 is a zero and the polynomial can be factored as follows:
\[ f(x)=(x+2)(3x^{3}+10x^{2}-9) \]
The possible rational zeros of the depressed polynomial $ 3x^{3} + 10x^{2} - 9 $ that have not been eliminated are -3, -9, and $ $ . Synthetic division by x - c, with c = -3, yields (only the last line in the synthetic division is shown):
| 3 | 10 | 0 | -9 | |
| -3 | 3 | 1 | -3 | 0 |
Thus, -3 is a zero and the polynomial can be factored as follows:
\[ f(x)=(x+2)(x+3)(3x^{2}+x-3) \]
The remaining zeros can be found by solving $ 3x^{2} + x - 3 = 0 $ by the quadratic formula to obtain $ $ in addition to -2 and -3.
14.28. Find all zeros of $ f(x)=4x{4}-4x{3}-7x^{2}-6x+18 $
From Descartes’ rule of signs, the following combinations of positive, negative, and imaginary zeros are possible.
| POSITIVE | NEGATIVE | IMAGINARY |
| 2 | 2 | 0 |
| 2 | 0 | 2 |
| 0 | 2 | 2 |
| 0 | 0 | 4 |
From the theorem on rational zeros of polynomials with integer coefficients, the possible rational zeros are $ $ , $ $ , $ $ , $ $ , $ $ , $ $ , $ $ , $ $ , $ $ , $ $ , $ $ , $ $ .
Use the scheme for synthetic division by x - c, with c = successive positive integers from this list (only the last line in the synthetic division is shown).
| 4 | -4 | -7 | -6 | 18 | |
| 1 | 4 | 0 | -7 | -13 | 5 |
| 2 | 4 | 4 | 1 | -4 | 10 |
| 3 | 4 | 8 | 17 | 45 | 153 |
3 is an upper bound for the positive zeros of f. Now use the scheme for synthetic division by x - c, with c = successive positive rational numbers from this list (only the last line in the synthetic division is shown).
| 4 | -4 | -7 | -6 | 18 | |
| $ \(</td><td>4</td><td>-2</td><td>-8</td><td>-10</td><td>13</td></tr><tr><td>\) $ | 4 | 2 | -4 | -12 | 0 |
Thus, $ $ is a zero and the polynomial can be factored as follows:
\[ f(x)=\left(x-\frac{3}{2}\right)(4x^{3}+2x^{2}-4x-12)=(2x-3)(2x^{3}+x^{2}-2x-6) \]
The only possible rational zero of the depressed polynomial $ 2x^{3} + x^{2} - 2x - 6 $ that has not been eliminated from consideration is $ $ . Synthetic division by x - c, with $ c = $ , yields (only the last line in the synthetic division is shown):
| 2 | 1 | -2 | -6 | |
| $ $ | 2 | 4 | 4 | 0 |
Thus, \(\frac{3}{2}\) is a double zero of the original polynomial, which can be factored as follows:
\[ f(x)=(2x-3)\bigg(x-\frac{3}{2}\bigg)(2x^{2}+4x+4)=(2x-3)^{2}(x^{2}+2x+2) \]
The remaining zeros can be found by solving $ x^{2} + 2x + 2 = 0 $ by the quadratic formula to obtain $ -1 i $ in addition to the double zero $ $ .
14.29. Sketch the graphs of the following polynomial functions:
$ f(x) = 2x^{3} - 9 $
$ f(x) = (x + 1)^{4} $
$ f(x) = -(x + 3)^{3} + 4 $
f(x) = x3 + 3x2 − 10x − 24
\[ \left(\mathrm{e}\right)f(x)=3x^{4}+16x^{3}+20x^{2}-9x-18 \]
f(x) = 4x4 − 4x3 − 7x2 − 6x + 18
The graph of $ f(x) = 2x^{3} - 9 $ is the same as the graph of $ f(x) = x^{3} $ stretched by a factor of 2 with respect to the y-axis and shifted down 9 units (see Fig. 14-9).
- The graph of $ f(x)=(x+1)^{4} $ is the same as the graph of $ f(x)=x^{4} $ shifted 1 unit to the left and compressed by a factor of 3 (see Fig. 14-10).
- The graph of $ f(x) = -(x + 3)^{3} + 4 $ is the same as the graph of $ f(x) = x^{3} $ shifted 3 units to the left, compressed by a factor of 2, reflected with respect to the y-axis, and shifted up 4 units (see Fig. 14-11).
- In Problem 14-26 it was shown that $ f(x) = x^{3} + 3x^{2} - 10x - 24 = (x - 3)(x + 2)(x + 4) $ . Use the methods of Chapter 6 to obtain the sign chart shown in Fig. 14-12.
The graph has x-intercepts -4, -2, 3 and is below the x-axis on the intervals $ (-∞, -4) $ and $ (-2, 3) $ and above the x-axis on the intervals $ (-4, -2) $ and $ (3, ∞) $ . Form a table of values and sketch the graph as a smooth curve.
See Fig. 14-13 and the accompanying table.
| x | -5 | -4 | -3 | -2 | -1 |
| y | -24 | 0 | 6 | 0 | -12 |
| x | 0 | 1 | 2 | 3 | 4 |
| y | -24 | -30 | -24 | 0 | 48 |
- In Problem 14-27 it was shown that $ f(x) = (x + 2)(x + 3)(3x^{2} + x - 3) $ . It was further shown that $ $ are zeros of the polynomial, hence; by the factor theorem, $ f(x) $ can be completely factored as
\[ f(x)=(x+2)(x+3)\bigg(x-\frac{-1+\sqrt{37}}{6}\bigg)\bigg(x-\frac{-1-\sqrt{37}}{6}\bigg)3 \]
For graphing purposes, the irrational zeros may be approximated as 0.85 and -1.2. A sign chart shows that the graph has x-intercepts -3, -2, -1.2, 0.85 and is below the x-axis on the intervals $ (-3, -2) $ and $ (-1.2, 0.85) $ and above the x-axis on the intervals $ (-, -3) $ , $ (-2, -1.2) $ , and $ (0.85, ) $ . Form a table of values and sketch the graph as a smooth curve.
See Fig. 14-14 and the accompanying table.
| x | -4 | -3 | -2 | -1 | 0 | 1 |
| y | 82 | 0 | 0 | -2 | -18 | 12 |
- In Problem 14-28 it was shown that $ f(x) = (2x - 3){2}(x{2} + 2x + 2) $ . Thus the graph of the polynomial has an x-intercept at $ x = $ and is above the x-axis for all other values of x. Form a table of values and sketch the graph as a smooth curve.
See Fig. 14-15 and the accompanying table.
| x | -1.5 | -1 | -0.5 | 0 |
| y | 45 | 25 | 20 | 18 |
| x | 0.5 | 1 | 1.5 | 2 |
| y | 13 | 5 | 0 | 10 |
SUPPLEMENTARY PROBLEMS
14.30. Find the quotient and remainder for the following:
$ (5x^{4} + x^{2} - 8x + 2)/(x^{2} - 3x + 1) $ (b) $ (x^{5} + x^{4} + 3x^{3} - x^{2} - x - 3)/(x^{2} + x + 1) $
$ (x{3}-3x{2}+8x-7)/(2x-5) $
$ (x{6}-x{4}-8x^{3}+x+2)/(x+3) $
Ans. (a) Quotient: 5x2 + 15x + 41, remainder: 100x − 39; (b) quotient: x3 + 2x − 3, remainder: 0;
- quotient: $ x^{2}-x+ $ , remainder: $ $ ; (d) quotient: $ x{5}-3x{4}+8x{3}-32x{2}+96x-287 $ , remainder: 863
14.31. Given $ f(x) = x^{4} + 2x^{3} + 6x^{2} + 8x + 8 $ , find (a) $ f(-3) $ ; (b) $ f(2i) $ ; (c) $ f(3 - i) $ ; (d) $ f(-1 + i) $ .
Ans. (a) 65; (b) 0; (c) 144 - 192i; (d) 0.
14.32. Find a polynomial $ P(x) $ of lowest degree, with integer coefficients, such that $ $ and -3 - 2i are zeros.
Ans. $ P(x) = a(5x^{3} + 27x^{2} + 47x - 39) $ , where a is any integer.
14.33. Show that $ x + a $ is a factor of $ x^{n} + a^{n} $ for all odd n.
14.34. Show that $ x + a $ is a factor of $ x^{n} - a^{n} $ for all even n.
14.35. Assuming the validity of the fundamental theorem of algebra, prove the first corollary stated above.
14.36. Prove: If $ P(x) $ is a polynomial with real coefficients, then if z is a complex zero of $ P(x) $ , then the complex conjugate $ {z} $ is also a zero of $ P(x) $ . Hint: Assume that z is a zero of $ P(x) = a_{n}x^{n} + a_{n-1}x^{n-1} + + a_{1}x + a_{0} $ , and use the facts that $ {a} = a $ if a is real and that $ z + w = {z} + {w} $ and $ zw = {z}{w} $ for all complex numbers.
14.37. Locate the zeros of $ f(x) = 6x^{3} + 32x^{2} + 41x + 12 $ between successive integers.
Ans. The zeros are in the intervals $ (-4,-3) $ , $ (-2,-1) $ , and $ (-1,0) $ .
14.38. Find all zeros exactly for the following polynomials:
\(2x^{3}-5x^{2}-2x+2\); (b) \(x^{4}+2x^{3}-2x^{2}-6x-3\); (c) \(x^{4}-x^{3}-3x^{2}+17x-30\);
\(x^{5} + 5x^{3} + 6x\); (e) \(3x^{5} - 2x^{4} - 9x^{3} + 6x^{2} - 12x + 8\)
Ans. (a) \(\left\{\frac{1}{2},1\pm\sqrt{3}\right\}\); (b) \(\{-1(\text{double}),\pm\sqrt{3}\}\); (c) \(\{2,-3,1\pm2i\}\);
- \(\{0,\pm i\sqrt{2},\pm i\sqrt{3}\}\); (e) \(\{\pm2,\frac{2}{3},\pm i\}\)
14.39. Solve the polynomial equations:
$ x^{3} - 19x - 30 = 0 $
$ 4x^{3} + 40x = 22x^{2} + 25 $
$ x^{5} - 5x^{4} - 4x^{3} + 36x^{2} + 27x - 135 = 0 $
$ -12x^{4} - 8x^{3} + 49x^{2} + 39x - 18 = 0 $
Ans. (a) $ {-3,-2,5} $ ; (b) $ {,} $ ; (c) $ {3,-2i} $ ; (d) $ {2,,-} $
14.40. Using the information in the previous problem, draw graphs of
$ f(x) = x^{3} - 19x - 30 $
$ f(x) = 4x^{3} - 22x^{2} + 40x - 25 $
$ f(x) = x^{5} - 5x^{4} - 4x^{3} + 36x^{2} + 27x - 135 $
$ f(x) = -12x^{4} - 8x^{3} + 49x^{2} + 39x - 18 $
Ans. (a) Fig. 14-16; (b) Fig. 14-17; (c) Fig. 14-18; (d) Fig. 14-19.
14.41. From a square piece of cardboard 20 inches on a side, an open box is to be made by removing squares of side x and turning up the sides. Find the possible values of x if the box is to have volume 576 cubic inches. (See Fig. 14-20.)
Ans. 4 inches, or $ 8 - $ inches.
14.42. A silo is to be built in the shape of a right circular cylinder with a hemispherical top (see Fig. 14-21). If the total height of the silo is 30 feet and the total volume is $ 1008$ cubic feet, find the radius of the cylinder.
