Precalculus Study Guide
Copyright Maplesoft, a division of Waterloo Maple Inc., 2024
Chapter 3: Graphing Polynomial Functions
Introduction
After straight lines, the simplest functions to study are the real polynomials, the general form of which is
fx=anxn+an−1xn−1+...+x2x2+a1x+a0
where the real numbers ak,k=0,1,...,n, are the coefficients, with an called the leading coefficient. Real polynomials are those whose coefficients are real numbers. The integer n is the degree of the polynomial.
If n=2, the polynomial function, which then has the form
fx=a2x2+a1x+a0
is called quadratic, whereas if n=3, the polynomial function, which then has the form
fx=a3x3+a2x2+a1x+a0
is called cubic.
If fr=0, then x=r is called a zero of the polynomial, and x−r is then a factor of the polynomial. For each zero rk, there is a corresponding factor x−rk, so that the factored form of the polynomial is
fx=x−r1x−r2⋯ x−rn
Hence, given the zeros of a polynomial, it is possible to reconstruct the polynomial itself.
(The fundamental theorem of algebra states that every polynomial of degree n has exactly n zeros, although they may not all be distinct - there can be repeated zeros.)
If the polynomial function is graphed, the real zeros are the x-intercepts.
Chapter Glossary
The following terms in Chapter 3 are linked to the Maple Math Dictionary.
coefficient
complete the square
complex conjugate
coordinate
cubic
factor
fundamental theorem of algebra
integer
intercept
leading coefficient
long division
monic polynomial
parabola
polynomial
quadratic formula
quartic
real number
synthetic division
tangent zero
Typical Problems
For the polynomial functions given in Problems 1 - 3, sketch a graph and determine both the y- and x-intercepts. Since the x-intercepts are the zeros of the polynomials, they can be found approximately from the graph, or exactly, by factoring the polynomial over the real numbers.
3.1. (a) fx=3x2−5x−2
(b) fx=2x2−4x+2
(c) fx=x2−2x+5
3.2. (a) fx=x+13
(b) fx=x3−5x2+4x+10
(c) fx=x3−3x+2
(d) fx=x3−2x2−x+2
3.3. (a) fx=x4−12x3+47x2−62x+26
(b) fx=x4−15x3+75x2−135x+74
3.4. For each of the following, find the real monic polynomial of least degree having the given numbers as its zeros. (A real polynomial is one whose coefficients are real. A monic polynomial has leading coefficient 1, that is, the coefficient of the term with the highest power is 1.)
(a) −2,3
(b) { −2 + i, 1 }
(c) { 1, 3, 1 + 2i }
(d) 3−4i,7
Maple Initializations
Before accessing the Maple version of the solutions to the typical problems stated above, initialize Maple by pressing the button provided on the right.
Solutions
Problem 3.1
3.1 - Mathematical Solution
3.1 (a) - Mathematical Solution
For the function fx=3x2−5x−2, the y-intercept is found to be f0=−2, and the x-intercepts are found from the quadratic formula to be
x1=−−5+−52−43−223 = 2
and
x2=−−5−−52−43−223 = −13
Moreover, fx factors to the product 3x+1x−2.
Since the function fx is a quadratic polynomial, its graph is that of a parabola, opening upward. The standard form of this parabola would be
Figure 3.1.1 Graph of fx=3x2−5x−2
y+4912=3x−562
obtained by completing the square in x. Thus, the coordinates of the vertex of the parabola are56,−4912.
From this information, Figure 3.1.1, a graph of the function, can be drawn.
3.1 (b) - Mathematical Solution
For the function fx=2x2−4x+2, the y-intercept is found to be f0=2, and the x-intercepts are found from the quadratic formula to be
x1=−−4+−42−42222 = 1
x2=−−4−−42−42222 = 1
Moreover, fx factors to the product 2x−12.
Figure 3.1.2 Graph of fx=2x2−4x+2
y=2x−12
obtained by completing the square in x. Thus, the coordinates of the vertex of the parabola are1,0.
From this information, Figure 3.1.2, a graph of the function, can be drawn.
3.1 (c) - Mathematical Solution
For the function fx=x2−2x+5, the y-intercept is found to be f0=5, and the x-intercepts are found from the quadratic formula to be
x1=−−2+−22−41521 = 1+2i
x2=−−2−−22−41521 = 1−2i
Moreover, fx does not factor over the real numbers.
Since the function fx is a quadratic polynomial, its graph is
Figure 3.1.3 Graph of fx=x2−2x+5
that of a parabola, opening upward. The standard form of this parabola would be
y−4=x−12
obtained by completing the square in x. Thus, the coordinates of the vertex of the parabola are1,4.
From this information, Figure 3.1.3, a graph of the function, can be drawn.
3.1 - Maplet Solution
3.1 (a) - Maplet Solution
A graph of the polynomial function fx=3x2−5x−2, along with its intercepts, can be obtained with Polynomial Tutor #1 .
Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 3.1.4.
To use this tutor, enter the coefficients of the polynomial in the appropriate boxes. Blanks will be interpreted as the number zero (0).
The Display Function button will display the polynomial entered, while the buttons labeled Graph, x-intercepts, and y-intercept will generate the corresponding results.
Figure 3.1.4 Thumbnail image of Polynomial Tutor #1
To launch Polynomial Tutor #1, click the following link: Polynomial Tutor #1
3.1 (b) - Maplet Solution
A graph of the polynomial function fx=2x2−4x+2, along with its intercepts, can be obtained with Polynomial Tutor #1 .
Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 3.1.5.
Figure 3.1.5 Thumbnail image of Polynomial Tutor #1
Since the x-intercepts are the zeros of the polynomial, if a zero is repeated, it will appear the appropriate number of times in the list of x-intercepts.
3.1 (c) - Maplet Solution
A graph of the polynomial function fx=x2−2x+5, along with its intercepts, can be obtained with Polynomial Tutor #1 .
Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 3.1.6.
Figure 3.1.6 Thumbnail image of Polynomial Tutor #1
If the polynomial has no real zeros, its graph will not cross the x-axis. Hence, there will be no x-intercepts returned.
3.1 - Interactive Solution
3.1 (a) - Interactive Solution
Enter the polynomial.
Context Panel: Assign to a Name≻fa
Obtain Figure 3.1.1
Type fa and press the Enter key.
Context Panel: Plots≻Plot Builder −1≤x≤3
Obtain x-intercept(s)
Type fa=0 and press the Enter key.
Context Panel: Solve≻Solve
Obtain the y-intercept
Context Panel: Evaluate at a Point≻x=0
3.1 (b) - Interactive Solution
Context Panel: Assign to a Name≻fb
Obtain Figure 3.1.2
Type fb and press the Enter key.
Type fb=0 and press the Enter key.
3.1 (c) - Interactive Solution
Context Panel: Assign to a Name≻fc
Obtain Figure 3.1.3
Type fc and press the Enter key.
Type fc=0 and press the Enter key.
3.1 - Programmatic Solution
3.1 (a) - Programmatic Solution
Ya≔3x2−5 x−2
Graph the polynomial as per Figure 3.1.1.
plotYa,x=−1..3
Obtain the x-intercepts.
solveYa=0,x
Obtain the y-intercept.
evalYa,x=0
3.1 (b) - Programmatic Solution
Yb≔2x2−4 x+2
plotYb,x=−1..3
solveYb=0,x
evalYb,x=0
3.1 (c) - Programmatic Solution
Yc≔x2−2 x+5
plotYc,x=−1..3
solveYc=0,x
evalYc,x=0
Problem 3.2
3.2 - Mathematical Solution
3.2 (a) - Mathematical Solution
The polynomial fx=x+13 has f0=1 for its y-intercept, and x=−1 for its x-intercept.
The graph is a translation to the left, by one unit, of the graph of x3. Hence, the graph of fx can be sketched, as shown in Figure 3.2.1.
Figure 3.2.1 Graph of fx=x+13
3.2 (b) - Mathematical Solution
The polynomial fx=x3−5x2+4x+10 has y-intercept f0=10, and x-intercept x=−1. This is found by searching for integer zeros, computing the values of fk,k=−3,−2,....
Although there are formulas expressing the zeros of a cubic in terms of its coefficients, they are very cumbersome to apply. That x=−1 is the only zero of fx is determined by dividing out the factor x+1, leaving the quadratic polynomial x2−6x+10 whose zeros are the complex conjugate pair 3 + i.
The division of fx by the factor x+1 can be done by synthetic division, the tableau for which appears in Table 3.2.1
Figure 3.2.2 Graph of fx=x3−5x2+4x+10
The final zero is the remainder, so the factor divides evenly. The coefficients of the depressed polynomial are the numbers to the left of the final zero in the bottom row of the tableau.
The zeros of the depressed polynomial are found with the quadratic formula, and they are
1 −5 4 10 [-1
−1 6 −10
________________
1 −6 10 0
Table 3.2.1 Synthetic division
x1=−−6+−62−41102 = 6+−42=6+2i2 = 3+i
x2=−−6−−62−41102 = 6−−42=6−2i2 = 3−i
Without a plotting device such as a calculator or computer, a graph can only be drawn from computed values such as those appearing in Table 3.2.2.
x
−3.0
−2.5
−2.0
−1.5
−1.0
−.5
0.
fx
−74.0
−46.9
−26.0
−10.6
6.62
10.0
0.5
1.0
1.5
2.0
2.5
3.0
10.9
8.12
6.0
4.38
4.0
Table 3.2.2 Values of the function fx=x3−5x2+4x+10
The graph of fx in Figure 3.2.2 was obtained with Maple. Drawing such a graph with just a pencil and paper is a daunting task.
3.2 (c) - Mathematical Solution
The polynomial fx=x3−3x+2 has y-intercept f0=2 and an x-intercept at x=−2. This is found by searching for zeros amongst the integers, and fortunately, this zero is an integer. Thus, x−−2=x+2 is a factor of the polynomial.
The companion factor is found by dividing fx by x+2. This division can be done by long division or by the synthetic division shown in the tableau in Table 3.2.3.
The final zero in the last row of the tableau is the remainder, so the factor divides evenly. The coefficients of the depressed polynomial are the numbers to the left of the final zero in the bottom row of the tableau.
Figure 3.2.3 Graph of fx=x3−3x+2
The zeros of the depressed polynomial are found with the quadratic formula or by noticing that x2−2x+1=x−12. Thus, the remaining zeros are 1 and again 1, so that x=1 is a double zero. There are two x-intercepts, namely, x=−2 and x=1, the latter being a double zero of the polynomial.
1 0 −3 2 [−2
−2 4 −2
_________________
1 −2 1 0
Table 3.2.3 Synthetic division
From this information, a sketch of fx can be drawn. The graph in Figure 3.2.3 was obtained in Maple.
3.2 (d) - Mathematical Solution
The polynomial fx=x3−2x2−x+2 has y-intercept f0=2 and an x-intercept at x=−1. This is found by searching for zeros amongst the integers, and fortunately, this zero is an integer. Thus, x−−1=x+1 is a factor of the polynomial.
The companion factor is found by dividing fx by x+1. This division can be done by long division or by the synthetic division shown in the tableau in Table 3.2.4.
Figure 3.2.4 Graph of fx=x3−2x2−x+2
The zeros of the depressed polynomial x2−3x+2 are found with the quadratic formula, and they are
1 −2 −1 2 [−1
−1 4 −2
1 −3 2 0
Table 3.2.4 Synthetic division
x1=−−3+−32−4122 = 3+12=3+12 = 2
x2=−−3−−32−4122 = 3−12=3−12 = 1
Thus, the remaining x-intercepts are x=1 and x=2, so a sketch of fx could now be drawn. The graph in Figure 3.2.4 was obtained in Maple.
3.2 - Maplet Solution
3.2 (a) - Maplet Solution
A graph of the polynomial function fx=x+13, along with its intercepts, can be obtained with Polynomial Tutor #1 .
Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 3.2.5.
Figure 3.2.5 Thumbnail image of Polynomial Tutor #1
3.2 (b) - Maplet Solution
A graph of the polynomial function fx=x3−5x2+4x+10, along with its intercepts, can be obtained with Polynomial Tutor #1 .
Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail in Figure 3.2.6.
Figure 3.2.6 Thumbnail image of Polynomial Tutor #1
3.2 (c) - Maplet Solution
A graph of the polynomial function fx=x3−3x+2, along with its intercepts, can be obtained with Polynomial Tutor #1 .
Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 3.2.7.
Figure 3.2.7 Thumbnail image of Polynomial Tutor #1
3.2 (d) - Maplet Solution
A graph of the polynomial function fx=x3−2x2−x+2, along with its intercepts, can be obtained with Polynomial Tutor #1 .
Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 3.2.8.
Figure 3.2.8 Thumbnail image of Polynomial Tutor #1
3.2 - Interactive Solution
3.2 (a) - Interactive Solution
Obtain Figure 3.2.1
Context Panel: Plots≻Plot Builder −3≤x≤1
3.2 (b) - Interactive Solution
Obtain Figure 3.2.2
Context Panel: Plots≻Plot Builder −3/2≤x≤4
Context Panel: Solve≻Numerically Solve
3.2 (c) - Interactive Solution
Obtain Figure 3.2.3
Context Panel: Plots≻Plot Builder −3≤x≤3
3.2 (d) - Interactive Solution
Context Panel: Assign to a Name≻fd
Obtain Figure 3.2.4
Type fd and press the Enter key.
Context Panel: Plots≻Plot Builder −2≤x≤3
Type fd=0 and press the Enter key.
3.2 - Programmatic Solution
3.2 (a) - Programmatic Solution
Ya≔x+13
Graph the polynomial as per Figure 3.2.1.
plotYa,x=−3..1,y=−4..4
3.2 (b) - Programmatic Solution
Yb≔x3−5x2+4x+10
Graph the polynomial as per Figure 3.2.2
plotYb,x=−2..5,y=−10..12
removehas,solveYb=0,x,I
3.2 (c) - Programmatic Solution
Yc≔x3−3 x+2
Graph the polynomial as per Figure 3.2.3
plotYc,x=−3..3,y=−10..10
3.2 (d) - Programmatic Solution
Yd≔x3−2x2−x+2
Graph the polynomial as per Figure 3.2.4
plotYd,x=−2..3
solveYd=0,x
evalYd,x=0
Problem 3.3
3.3 - Mathematical Solution
3.3 (a) - Mathematical Solution
The polynomial fx=x4−12x3+47x2−62x+26 has f0=26 for its y-intercept, and x=1 for one of its x-intercepts. This intercept is found by searching amongst the integers for zeros of the polynomial. Since x=1 is a zero, the factor x−1 divides the polynomial evenly. The companion factor is found by either long division or synthetic division. The tableau in Table 3.3.1 implements synthetic division by the factor x−1.
1 −12 47 −62 26 [ 1
1 −11 36 −26
___________________________
1 −11 36 −26 0
Table 3.3.1 Synthetic division
Figure 3.3.1 Graph of fx=x4−12x3+47x2−62x
The companion factor, the "depressed polynomial," is x3−11x2+36x−26, which also has x=1 as a zero. Again, this is found by searching through the integers. Once again, the factor x−1 can be divided out, this time from the depressed polynomial. The synthetic division tableau for this division appears in Table 3.3.2.
1 −11 36 −26 [ 1
1 −10 26
______________________
1 −10 −26 0
Table 3.3.2 Synthetic division
The new depressed polynomial is x2−10x−26, whose zeros can be found by the quadratic formula. These remaining zeros are
x1=−−10+−102−41262 = 10+−42=5+i
x2=−−10−−102−41262 = 10−−42=5−i
Knowledge of just the y-intercept and the double zero at x=1 does not yet lead to an unambiguous graph. Even making use of the property of quartics that their graphs will at most resemble the letter "W" still leaves a need for a list of values such as found in Table 3.3.3.
−0.5
0.0
3.5
4.5
5.0
5.5
148.0
70.3
26.0
5.3
3.3
16.3
20.0
20.3
18.0
15.3
16.0
25.3
50.0
Table 3.3.3 Values of the function fx=x4−12x3+47x2−62x
On the basis of all this information, the sketch of fx seen in Figure 3.3.1 could now be drawn. (However, Figure 3.3.1 was, in fact, generated by Maple.)
3.3 (b) - Mathematical Solution
The polynomial fx=x4−15x3+75x2−135x+74 has f0=74 for its y-intercept, and x=1 for one of its x-intercepts. This intercept is found by searching amongst the integers for zeros of the polynomial. Since x=1 is a zero, the factor x−1 divides the polynomial evenly. The companion factor is found by either long division or synthetic division. The tableau in Table 3.3.4 implements synthetic division by the factor x−1.
1 −15 75 −135 74 [ 1
1 −14 61 −74
1 −14 61 −74 0
Table 3.3.4 Synthetic division
Figure 3.3.2 Graph of fx=x4−15x3+75x2−135x+74
The companion factor, the "depressed polynomial," is x3−14x2+61x−74, which has x=2 as a zero. Again, this is found by searching through the integers. The factor x−2 can be divided out from the depressed polynomial. The synthetic division tableau for this division appears in Table 3.3.5.
1 −14 61 −74 [ 2
2 −24 74
1 −12 37 0
Table 3.3.5 Synthetic division
The new depressed polynomial is x2−12x+37, whose zeros can be found by the quadratic formula. These remaining zeros are
x1=−−12+−122−41372 = 12+−42=6+i
x2=−−12−−122−41372 = 12−−42=6−i
Knowledge of just the y-intercept and the zeros at x=1 and x=2 does not yet lead to an unambiguous graph. Even making use of the property of quartics that their graphs will at most resemble the letter "W" still leaves a need for a list of values such as the one given in Table 3.3.6.
300.0
−5.3
30.0
6.5
30.9
162.2
28.4
7.0
60.0
74.0
9.9
24.0
7.5
116.2
23.4
19.7
8.0
210.0
27.2
Table 3.3.6 Values of fx=x4−15x3+75x2−135x+74
On the basis of all this information, the sketch of fx seen in Figure 3.3.2 could now be drawn. (However, Figure 3.3.2 was, in fact, generated by Maple.)
3.3 - Maplet Solution
3.3 (a) - Maplet Solution
A graph of the polynomial function fx=x4−12x3+47x2−62x+26, along with its intercepts, can be obtained with Polynomial Tutor #1 .
Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 3.3.3.
Figure 3.3.3 Thumbnail image of Polynomial Tutor #1
3.3 (b) - Maplet Solution
A graph of the polynomial function fx=x4−15x3+75x2−135x+74, along with its intercepts, can be obtained with Polynomial Tutor #1 .
Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 3.3.4.
Figure 3.3.4 Thumbnail image of Polynomial Tutor #1
3.3 - Interactive Solution
3.3 (a) - Interactive Solution
Obtain Figure 3.3.1
Context Panel: Plots≻Plot Builder −1≤x≤6 Options≻Range from: 0≤y≤30
Context Panel: Factor
3.3 (b) - Interactive Solution
Context Panel: Plots≻Plot Builder −1≤x≤6 Options≻Range from: −5≤y≤140
3.3 - Programmatic Solution
3.3 (a) - Programmatic Solution
Ya≔x4−12x3+47x2−62 x+26
plotYa,x=−1..6,y=0..30
Factor the polynomial.
factorYa
3.3 (b) - Programmatic Solution
Yb≔x4−15x3+75x2−135 x+74
plotYb,x=−1..6,y=−5..140
factorYb
Problem 3.4
3.4 - Mathematical Solution
3.4 (a) - Mathematical Solution
If the zeros of a polynomial are −2 and 3, then the polynomial's factored form is
x−−2x−3 = x+2x−3
and its expanded form is
x2−x−6
3.4 (b) - Mathematical Solution
If the zeros of a polynomial are −2 + i and 1, then the polynomial's factored form is
x−−2+ix−−2−ix−1
= x+2−ix+2+ix−1
= x2+4x+5x−1
x3+3x2+x−5
3.4 (c) - Mathematical Solution
If the zeros of a polynomial are 1, 3, and 1 + 2i, then the polynomial's factored form is
x−1+2ix−1−2ix−1x−3
= x−1−2ix−1+2ix−1x−3
= x2−2x+5x−1x−3
x4−6x3+16x2−26x+15
3.4 (d) - Mathematical Solution
If the zeros of a polynomial with real coefficients are 3−4i and 7, then there must be at least one other zero, namely, 3+4i, the complex conjugate of the given zero 3−4i. The factored form of the desired polynomial is then
x−3+4ix−3−4ix−7
= x−3−4ix−3+4ix−7
= x2−6x+25x−7
x3−13x2+67x−175
3.4 - Maplet Solution
3.4 (a) - Maplet Solution
The real monic polynomial of least degree having −2,3 as its zeros can be found with Polynomial Tutor #2 .
Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 3.4.1.
To use this tutor, enter the zeros in the appropriate box, and press the button labeled Linear factored form. This gives the polynomial as a product of its linear factors, that is, as x−−2x−3=x+2x−3.
Pressing the button labeled Factored over reals displays the polynomial in terms of factors containing just real coefficients. In this problem where the zeros are all real, the linear factored form and this form are the same.
Figure 3.4.1 Thumbnail image of Polynomial Tutor #2
Pressing the button labeled Expanded form displays the polynomial as a sum of "coefficients times powers of x." Colloquially, this is the "multiplied out" form x2−x−6.
As in Polynomial Tutor #1, there are buttons for producing a graph of the required polynomial, and buttons for displaying the intercepts.
To launch Polynomial Tutor #2, click the following link: Polynomial Tutor #2
3.4 (b) - Maplet Solution
The real monic polynomial of least degree having −2 ±i,1 as its zeros can be found with Polynomial Tutor #2 .
Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 3.4.2.
To use this tutor, enter the zeros in the appropriate box, using I for the complex unit i=−1. Each complex zero must be entered separately, since there is no "+" operator in Maple.
Pressing the button labeled Linear factored form gives the polynomial as a product of its linear factors, that is, as
x−−2+ix−−2−ix−3
=x+2−ix+2+ix−3
Figure 3.4.2 Thumbnail image of Polynomial Tutor #2
Pressing the button labeled Factored over reals displays the polynomial as
x2+4x+5x−1
a product of factors containing just real coefficients.
Pressing the button labeled Expanded form displays the polynomial as a sum of "coefficients times powers of x." Colloquially, this is the "multiplied out" form x3+3x2+x−5.
3.4 (c) - Maplet Solution
The real monic polynomial of least degree having 1,3,1 ±2 i as its zeros can be found with Polynomial Tutor #2 .
Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 3.4.3.
x−1x−3x−1+2ix−1−2i
=x−1x−3x−1−2ix−1+2i
Figure 3.4.3 Thumbnail image of Polynomial Tutor #2
x−1x−3x2−2x+5
Pressing the button labeled Expanded form displays the polynomial as a sum of "coefficients times powers of x." Colloquially, this is the "multiplied out" form x4−6x3+16x2−26x+15.
3.4 (d) - Maplet Solution
The real monic polynomial of least degree having 3−4i,7 as its zeros will also have 3+4i as a zero. Remember, polynomials with real coefficients will have any of its complex zeros appearing as complex conjugate pairs. The required polynomial can be found with Polynomial Tutor #2 .
Clicking this link will launch the tutor with the solution embedded as shown in the thumbnail image in Figure 3.4.4.
Figure 3.4.4 Thumbnail image of Polynomial Tutor #2
x−3+4ix−3−4ix−7 =x−3−4ix−3+4ix−7
x2−6x+25x−7
Pressing the button labeled Expanded form displays the polynomial as a sum of "coefficients times powers of x." Colloquially, this is the "multiplied out" form x3−13x2+67x−175.
3.4 - Interactive Solution
3.4 (a) - Interactive Solution
Enter the first zero.
Context Panel: Assign to a Name≻a
Enter the second zero.
Context Panel: Assign to a Name≻b
Write the polynomial as a product of linear factors.
Context Panel: Expand
3.4 (b) - Interactive Solution
Enter the third zero.
Context Panel: Assign to a Name≻c
Context Panel: Expand Context Panel: Factor
3.4 (c) - Interactive Solution
Enter the fourth zero.
Context Panel: Assign to a Name≻
3.4 (d) - Interactive Solution
Enter the real zero.
Enter the complex zero.
Enter the third zero, the complex conjugate of b.
3.4 - Programmatic Solution
3.4 (a) - Programmatic Solution
Enter the given zeros, assigning them to the names zk,k=1,2.
z1≔−2; z2≔3
Write the desired polynomial as a product of linear factors.
f≔x−z1 x−z2
Expand the product of factors.
expandf
3.4 (b) - Programmatic Solution
z1≔−2+I; z2≔conjugatez1; z3≔1
f≔x−z1 x−z2 x−z3
F≔expandf
Factor the polynomial over the reals.
factorF
3.4 (c) - Programmatic Solution
z1≔1+2 I; z2≔z1&conjugate0;; z3≔1; z4≔3
f≔∏k=14x−zk
3.4 (d) - Programmatic Solution
z1≔7; z2≔3−4 I; z3≔z2&conjugate0;
Exercises - Chapter 3
For the quadratic polynomial functions given in Exercises 3.1 - 3.6, sketch a graph and determine both the y- and x-intercepts. Since the x-intercepts are the zeros of the polynomials, they can be found by the quadratic formula.
3.1. fx=x2−8x+12
3.2. fx=x2+2x−63
3.3. fx=x2−3x−10
3.4. fx=x2−2x−8
3.5. fx=2x2−x+5
3.6. fx=7x2+6x−8
For the cubic polynomial functions given in Exercises 3.7 - 3.12, sketch a graph and determine both the y- and x-intercepts. Since the x-intercepts are the zeros of the polynomials, they can be found approximately from the graph, or exactly, by factoring the polynomial over the real numbers.
3.7. fx=x3+11x2+20x−32
3.8. fx=x3−23x2+175x−441
3.9. fx=x3−14x2+39x+54
3.10. fx=x3+4x2−29x+24
3.11. fx=4x3−8x2+4x−8
3.12. fx=6x3+x2−4x+1
For the quartic polynomial functions given in Exercises 3.13 - 3.18, sketch a graph and determine both the y- and x-intercepts. Since the x-intercepts are the zeros of the polynomials, they can be found approximately from the graph, or exactly, by factoring the polynomial over the real numbers.
3.13. fx=x4−6x3−45x2+122x+360
3.14. fx=x4−6x3−45x2+274x−336
3.15. fx=x4−7x3−24x2+112x+128
3.16. fx=x4−13x2+52x−40
3.17. fx=x4−13x2+52x−40
3.18. fx=x4−9x3+21x2+29x−114
In Exercises 3.19 - 3.20, one complex zero of a quartic polynomial whose coefficients are real is given. Find the remaining three zeros. Hint: If using some form of technology, simply find all zeros as in Exercises 3.13 - 3.18. Otherwise, find the quadratic factor corresponding to the given complex zero, and long divide to find the other quadratic factor, which will then yield to the quadratic formula. (Complex arithmetic is treated at greater length in Chapter 8.)
3.19. fx=6x4−56x3+406x2+276x+148 with zero 5+7i
3.20. fx=x4−5x3+29x2−13x−116 with zero 2−5i
In Exercises 3.21 - 3.24, find the real monic polynomial of least degree having the given numbers as its zeros.
3.21. −3,2
3.22. −2,3,3
3.23. 2,2,3−5i
3.24. 1+i,2−3i
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