Algebra 2 by Richard Wright
Are you not my student and
has this helped you?
Objectives:
SDA NAD Content Standards (2018): AII.4.2, AII.4.3, AII.5.1, AII.5.2, AII.6.4
Lagrange points are places where the gravitational pull of two large masses such as the earth and sun balance the centripetal force required to keep a small object in orbit. These points are useful for spacecraft because they can remain in position with minimal fuel usage. The James Webb telescope is placed at Lagrange point L2 as illustrated in figure 1. The positions of Lagrange points can be calculated using polynomial equations.
Another use for the Remainder Theorem is to test whether a rational number is a zero for a given polynomial. But first there needs to be a group of rational numbers to test. There are an infinite number of possible zeros to choose from. It would be nice to have fewer numbers to choose from. The Rational Zero Theorem narrows down the number of possible rational zeros using the ratio of the factors of the constant term and factors of the leading coefficient of the polynomial.
Think about a quadratic function with two zeros, \(x = \frac{2}{3}\) and \(x = \frac{4}{5}\). The Factor Theorem indicates that if a zero is k, then a factor is (x − k). If k is a fraction, then the factor can be found by setting x = k.
$$ x = \frac{2}{3}; x = \frac{4}{5} $$
Subtract to make these equal zero.
$$ x - \frac{2}{3} = 0; x - \frac{4}{5} = 0 $$
Multiply by the denominators to get rid of the fraction.
3x – 2 = 0; 5x – 4 = 0
Multiply these together to make the quadratic function.
(3x – 2)(5x – 4)
Multiply these together.
15x2 – 22x + 8
Notice that the leading coefficient and constant term can be factored.
(3 · 5)x2 – 22x + (2 · 4)
The zeros are made of ratios of the factors of the constant term to the factors of the leading coefficient: \(\frac{2}{3}\) and \(\frac{4}{5}\). This is true of all polynomials and is called the Rational Zero Theorem.
If the polynomial f(x) = anxn + an − 1xn − 1 + ⋯ + a1x + a0 has integer coefficients, then every rational zero f(x) has the form of \(\frac{p}{q}\) where p is a factor of the constant term a0 and q is a factor of the leading coefficient an.
When the leading coefficient is 1, the possible rational zeros are the factors of the constant term.
To find all the rational zeros of a polynomial function,
List all possible rational zeros of f(x) = 2x4 + 5x3 − 17x2 − 4.
Solution
The only possible rational zeros of f(x) are the ratios of the factors of the last term, –4, and the factors of the leading coefficient, 2.
Find the p's which are factors of the constant term, −4.
p = ±1, ±2, ±4.
Find the q's which are factors of the leading coefficient, 2.
q = ±1, ±2.
The rational zeros are the ratios of p to q.
$$ \frac{p}{q} = ±\frac{1}{1}, ±\frac{1}{2}, ±\frac{2}{1}, ±\frac{2}{2}, ±\frac{4}{1}, ±\frac{4}{2} $$
Notice that \(\frac{1}{1} = 1\) and \(\frac{4}{2} = 2\), which have already been listed. So the list can be shortened.
$$ \frac{p}{q} = ±1, ±\frac{1}{2}, ±2, ±4 $$
Use the Rational Zero Theorem to find the rational zeros of f(x) = 4x3 + 8x2 + x − 3.
Solution
The Rational Zero Theorem say that the only possible rational zeros of f(x) are the ratios of the factors of the last term, −3, and the factors of the leading coefficient, 4.
Find the p's which are factors of the constant term, −3.
p = ±1, ±3.
Find the q's which are factors of the leading coefficient, 4.
q = ±1, ±2, ±4.
The rational zeros are the ratios of p to q.
$$ \frac{p}{q} = ±\frac{1}{1}, ±\frac{1}{2}, ±\frac{1}{4}, ±\frac{3}{1}, ±\frac{3}{2}, ±\frac{3}{4} $$
Simplifying and removing duplicates shortens the list to
$$ \frac{p}{q} = ±1, ±\frac{1}{2}, ±\frac{1}{4}, ±3, ±\frac{3}{2}, ±\frac{3}{4} $$
Choose one of the possible rational zeros to test. The graph in figure 2 indicates that −1 would be a good choice because that is a x-intercept and x-intercepts are zeros. Start with x = −1.
$$ \begin{array}{rrrrr} \underline{-1}| & 4 & 8 & 1 & -3 \\ & & -4 & -4 & 3 \\ \hline & 4 & 4 & -3 & |\underline{\phantom{0}0} \end{array} $$
The original function was degree 3. After dividing, the depressed polynomial is degree 2. The depressed polynomial is 4x2 + 4x − 3. Since this is quadratic factor or use the quadratic formula to find the last two zeros.
4x2 + 4x − 3
(2x − 1)(2x + 3)
Set each factor equal to zero.
2x − 1 = 0 or 2x + 3 = 0
2x = 1 or 2x = −3
$$ x = \frac{1}{2} \text{ or } x = -\frac{3}{2} $$
The zeros are −1, \(\mathbf{\frac{1}{2}}\), and \(\mathbf{-\frac{3}{2}}\).
Use the Rational Zero Theorem to find the rational zeros of f(x) = x3 + 4x2 − 11x − 30.
−5, −2, 3
The Rational Zero Theorem narrows the number of possible zeros of a polynomial function from an infinite number of choices to a few options. Then the possible rational zeros can be tested with synthetic division to see which are actually zeros.
To find all the zeros of polynomial functions,
Some zeros and factors may be repeated. If they are repeated, they have that multiplicity. For example, if 3 is a zero twice, 3 is a zero with multiplicity of 2.
Find the zeros of f(x) = x3 + 4x2 − 3x − 18.
Solution
Start by using the Rational Zero Theorem to find the list of possible rational zeros. The p's are factors of the constant term, −18.
p = ±1, ±2, ±3, ±6, ±9, ±18
The q's are factors of the leading coefficient.
q = ±1
The possible rational zeros are in the form \(\frac{p}{q}\). Duplicates such as \(\frac{1}{2}\) and \(\frac{2}{4}\) have been removed from the list below.
\(\frac{p}{q}\) = ±1, ±2, ±3, ±6, ±9, ±18
Choose one of the possible rational zeros to test. Since real zeros are also x-intercepts of the graph of the function, use a graphing calculator to find an x-intercept that is in the list of possible rational zeros. Remember some real zeros will be irrational and not on the list.
The graph shows an x-intercept at 2, so choose 2 as the first test point. Test it with synthetic division.
$$ \begin{array}{rrrrr} \underline{2}| & 1 & 4 & -3 & -18 \\ & & 2 & 12 & 18 \\ \hline & 1 & 6 & 9 & |\underline{\phantom{0}0} \end{array} $$
The remainder is zero, so that means 2 is a zero of the function. The depressed polynomial is one degree less than the original function, so the degree is now 2. The depressed polynomial is quadratic. Factor or use the quadratic formula to find the last two zeros. If the depressed polynomial was not quadratic, then it would be used with synthetic division and another possible zero.
x2 + 6x + 9 = 0
(x + 3)2 = 0
x + 3 = 0
x = −3
The factor (x + 3) is squared which means it is in the function twice. Thus, its zero, −3, has multiplicity two. Notice that the graph in figure 2 crosses the x-axis at 2 which indicates odd multiplicity (1 in this case) and touches without crossing at −3 which indicates even multiplicity (2 in this case).
The zeros of f(x) are 2 and −3 with multiplicity 2.
After knowing how to find zeros of polynomial functions, the question is how many zeros should there be? The Fundamental Theorem of Algebra states that a polynomial function has at least one complex zero. However, the process of finding zeros involves dividing by linear factors of the polynomial. Each time a factor is divided out, the degree of the depressed polynomial is reduced by one. Thus, we can divide a third degree polynomial by three factors. This indicates that a nth degree polynomial function has n factors and, thus, n zeros.
If f(x) is a polynomial of degree n > 0, then f(x) has at least one complex zero.
A polynomial has the same number of zeros as its degree.
Find the zeros of f(x) = x3 – 2x2 + 5x + 26.
Solution
This function is not factorable by grouping, so it will be solved by using the Rational Zero Theorem.
Start by using the Rational Zero Theorem to find the list of possible rational zeros. The p's are factors of the constant term, 26.
p = ±1, ±2, ±13, ±26
The q's are factors of the leading coefficient, 1.
q = ±1
The possible rational zeros are in the form \(\frac{p}{q}\).
$$ \frac{p}{q} = ±1, ±2, ±13, ±26 $$
Choose one of the possible rational zeros to test. Since real zeros are also x-intercepts of the graph of the function, use a graphing calculator to find an x-intercept that is in the list of possible rational zeros. Remember some real zeros will be irrational and not on the list.
The graph shows an x-intercept at −2, so choose −2 as the first test point. Test it with synthetic division.
$$ \begin{array}{rrrrr} \underline{-2}| & 1 & -2 & 5 & 26 \\ & & -2 & 8 & -26 \\ \hline & 1 & -4 & 13 & |\underline{\phantom{0}0} \end{array} $$
The remainder is zero, so that means −2 is a zero of the function. The depressed polynomial is one degree less than the original function, so the degree is now 2. The depressed polynomial is quadratic. Set the depressed polynomial equal to zero and solve to find the last two zeros. If the depressed polynomial was not quadratic, then it would be used with synthetic division and another possible zero.
x2 − 4x + 13 = 0
$$ x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a} $$
$$ x = \frac{4 ± \sqrt{(-4)^2 - 4(1)(13)}}{2(1)} $$
$$ x = \frac{4 ± \sqrt{-36}}{2} $$
$$ x = \frac{4 ± 6i}{2} $$
x = 2 ± 3i
The last two zeros are imaginary. Notice that they are the same value, just positive and negative. Also notice that the graph in figure 3 shows the curve approaching the x-axis, but curving away before it actually reaches the axis. This indicates two imaginary zeros.
The zeros of f(x) are −2 and 2 ± 3i.
Find the zeros of f(x) = x3 – 3x2 + 7x − 5.
The zeros are 1, 1 ± 2i.
Another way to use the Fundamental Theorem of Algebra is to find a polynomial function with given zeros. The Linear Factorization Theorem states that if a zero of a polynomial is k, the a factor of the polynomial is (x – k). The polynomial can be found by mutliplying all the factors together.
However, if one of the zeros is imaginary in the form a + bi, then a factor will be (x – (a + bi)). Multiplying out the factors will produce coefficients with i. To avoid that and have only real coefficients, then another zero will have to be a – bi and the corresponding factor (x – (a – bi)). Multiplying (x – (a + bi)) · (x – (a – bi)) results in only real coefficients. a + bi and a – bi are called complex conjugates which are complex numbers with the same real and imaginary parts, but the imaginary part has the opposite sign. The same is true of irrational numbers such as \(2 + \sqrt{3}\) and \(2 - \sqrt{3}\).
If the polynomial function f has real coefficients and a complex zero in the form a + bi, then the complex conjugate of the zero, a – bi, is also a zero.
If the polynomial function f has real coefficients and a irrational zero in the form \(a + \sqrt{b}\), then the irrational conjugate of the zero, \(a - \sqrt{b}\), is also a zero.
Company that makes concrete statues and monuments wants to make a simple rectangular prism monument. Because of pricing, they want to use 756 cubic feet of concrete. They want the length of the monument to be three inches longer than the width of the monument and the height of the monument to be two inches shorter than the width. What should the dimensions of the monument be?
Solution
Start by using the formula for the volume of a rectangular prism, V = ℓwh. In this case, ℓ = w + 3 and h = w – 2.
V = ℓwh
756 = (w + 3)w(w – 2)
756 = w3 + w2 – 6w
To solve the polynomial, set the equation equal to zero.
w3 + w2 – 6w – 756 = 0
This is not easily factored so use the Rational Zero Theorem to list the possible rational zeros. p is factors of the constant term, 756.
p = ±1, ±2, ±3, ±4, ±6, ±7, ±9, ±12, ±14, ±18, ±21, ±27, ±28, ±36, ±42, ±54, ±63, ±84, -±108, ±126, ±189, ±252, ±378, ±756
q is factors of the leading coefficient, 1. The list of \(\frac{p}{q}\) is then the same as p
\(\frac{p}{q}\) = ±1, ±2, ±3, ±4, ±6, ±7, ±9, ±12, ±14, ±18, ±21, ±27, ±28, ±36, ±42, ±54, ±63, ±84, -±108, ±126, ±189, ±252, ±378, ±756
Choose one of the possible rational zeros to test. Since real zeros are also x-intercepts of the graph of the function, use a graphing calculator to find an x-intercept that is in the list of possible rational zeros. Remember some real zeros will be irrational and not on the list.
The graph shows an x-intercept at 9, so choose 9 as the first test point. Test it with synthetic division.
$$ \begin{array}{rrrrr} \underline{9}| & 1 & 1 & -6 & -756 \\ & & 9 & 90 & 756 \\ \hline & 1 & 10 & 84 & |\underline{\phantom{0}0} \end{array} $$
The remainder is zero, so that means 9 is a zero of the function. The depressed polynomial is one degree less than the original function, so the degree is now 2. The depressed polynomial is quadratic. Set the depressed polynomial equal to zero and solve to find the last two zeros. If the depressed polynomial was not quadratic, then it would be used with synthetic division and another possible zero.
w2 + 10w + 84 = 0
This does not factor, so use the quadratic formula.
$$ w = \frac{-b ± \sqrt{b^2-4ac}}{2a} $$
$$ w = \frac{-10 ± \sqrt{10^2-4(1)(84)}}{2(1)} $$
$$ w = -5 ±\sqrt{59}i $$
The width of a monument must be real, so the only real solution is 9. The dimensions of the monument should be 12 ft. × 9 ft. × 7 ft.
A shipping container in the shape of a rectangular solid must have a volume of 12 cubic meters. The client tells the manufacturer that, because of the contents, the length of the container must be one meter longer than the width, and the height must be two meters less than twice the width. What should the dimensions of the container be?
3 m × 2 m × 2 m
List all the possible rational zeros.
Find all the zeros. (You may factor if they are easily factored.)
Mixed Review