As you study these exercises, move your cursor over the light bulbs and the highlighted words.
Study with an active cursor

The definition that pops up when you move the cursor over a highlighted word is what should pop into your mind when you read, hear, or speak that word.

The material that pops up when you move the cursor over a light bulb is a suggested strategy for solving the problem. You should always formulate a similar strategy when you attempt to solve a problem.

When you move the cursor over an explanation of a step in a solution, the property that pops up is the mathematical justification for taking that action. You must always be able to provide such a justification for every step you take in mathematics.

26) Analyze the function f whose rule is f(x) = -3x3 + 20x2 - 36x + 16 Solution: Clearly f is a polynomial function , so its domain is all real numbers. The graph of f is a smooth continuous graph which tries to cross the x-axis at three points. The leading term dominates when far from the origin.     This may be written as:             This was determined by the following consideration: Note that when x is far to the left, x is negative, its cube is negative and negative three times its cube is positive. When x far to the right, x is positive, its cube is positive, and negative three times its cube is negative. And may be visualized as: We now attempt to find the zeros of the function f. To find the zeros of f we must solve the equation resulting from f(x) = 0. So we must solve the equation -3x3 + 20x2 -36x + 16 = 0. Because we have no direct method for solving this cubic equation we look at the Possible Rational Zeros of the function f. If is a rational zero of f, then p is a divisor of the constant term 16 and q is a divisor of the leading coefficient -3. Therefore: and , so that If the function f has any rational zeros, they must be in the above list of 20 domain elements. In view of the definition of zeros of a function we now compute range elements corresponding to the above domain elements in the hope of finding a zero of the function. f(1) = -3 + 20 - 36 + 16 0. Therefore 1 is not a zero of f f(-1) = 3 + 20 + 36 + 160. Therefore -1 is not a zero of f f(2) = -24 + 80 - 72 + 16 = 0. Therefore 2 is a zero of f We now know the following equivalent statements 1)   2 is a real rational zero of f 2)   2 is a solution of the equation -3x3 + 20x2 - 36x + 16 = 0 3)   (x - 2) is a factor of -3x3 + 20x2 - 36x + 16 4)   (2, 0) is an x-intercept of the graph of f What we know about f may be shown on a coordinate system as illustrated below Aside: Did you notice that this method of hunting for zeros of the polynomial has given us a method for solving third degree polynomial equations. We could continue checking the list of domain elements with the hope of finding other zeros, but it is generally more efficient to make use of Item 3 and polynomial division to determine a factorization of the polynomial. The long division process at the right shows the dividend, divisor, quotient, and remainder as dicussed in the Division Algorithm for Polynomials from which we conclude: f(x) = -3x3 + 20x2 - 36x + 16 = (x - 2)(-3x2 + 14x - 8) Remember that we are looking for zeros of the function f so we must solve the equation resulting from 0 = f(x) = -3x3 + 20x2 - 36x + 16 = (x - 2)(-3x2 + 14x - 8). From The Zero Factor Property we obtain x - 2 = 0 or -3x2 + 14x - 8 = 0. We may solve the quadratic equation either by factoring or by using the Quadratic Formula . -3x2 + 14x - 8 = 0 (-1)(3x2 - 14x + 8 (-1)(x - 4)(3x - 2) = 0 -1 = 0 OR x - 4 = 0 OR 3x - 2 = 0 Solutions are therefore 4 and We now have the following:  f(x) = -3x3 + 20x2 - 36x + 16 = (x - 2)(-3x2 + 14x - 8) = (- 1)(x - 2)(x - 4)(x - ) = (- 1)(x - 2)(x - 4)(3)(x - ) = (- 3)(x - 2)(x - 4)(x - ) In summary this is what we have discovered Three real rational zeros of f 2, 4 and Three x-intercepts of the graph of f (2, 0), (4, 0), and (, 0) All solutions to the third degree equation    0 = -3x3 + 20x2 - 36x + 16 2, 4 and Factorization of the third degree polynomial into prime linear factors -3x3 + 20x2 - 36x + 16 = (- 3)(x - 2)(x - 4)(x - ) A computer generated graph of f is shown at the right. The y-intercept of 16 is not shown on the graph.

It is instructive to observe the graph of f and the graphs of its factors on the same coordinate system. Shown below
Notice the x-intercepts.

The graph of f, a linear factor of f, and a quadratic factor of f

The graph of f, and four linear factors of f

The graph at the right is a blow-up of the graph showing the quadratic and linear factors which we found earlier. Areas where the quadratic (green) and the linear (red) factors are both above or both below the x-axis are shown in brown. Observe that these regions correspond to regions where the cubic (blue) is above the x-axis. Areas where the quadratic (green) and the linear (red) factors are on opposite sides of the x-axis are shown in light green. Observe that these regions correspond to regions where the cubic (blue) is below the x-axis. These two observations are nothing more than observing that the product of two numbers with the same sign is positive and the product of two numbers with different signs is negative.

47) Use the fact that 5i is a zero of the polynomial function f whose rule is

             f(x) = 2x³ + 3x² + 50x + 75
       to find all the zeros of the function f.
Solution: The fact that 5i is a zero of f implies that its conjugate -5i is also a zero of f.  It then follows that both (x - 5i) and (x + 5i) are factors of f and therefore their product x² + 25 is a factor of 2x³ + 3x² + 50x + 75.
The quotient when 2x³ + 3x² + 50x + 75 is divided by x² + 25 is 2x + 3.
Therefore f(x) = 2x³ + 3x² + 50x + 75 = (x - 5i)(x + 5i)(2x + 3) and it follows that the zeros of f are 5i, -5i, and -3/2

48) Use the fact that 3i is a zero of the polynomial function f whose rule is
           f(x) = x³ + x² + 9x + 9
       to find all the zeros of the function f.
Solution: The fact that 3i is a zero of f implies that its conjugate -3i is also a zero of f.  It then follows that both (x - 3i) and (x + 3i) are factors of f and therefore their product x² + 9 is a factor of x³ + x² + 9x + 9.
The quotient when x³ + x² + 9x + 9 is divided by x² + 9 is x + 1.
Therefore f(x) = x³ + x² + 9x + 9 = (x - 3i)(x + 3i)(x + 1) and it follows that the zeros of f are 5i, -5i, and -1

50) Use the fact that 5 + 2i is a zero of the polynomial function f whose rule is
           f(x) = x³ - 7x² - x + 87
       to find all the zeros of the function f.
Solution: The fact that 5 + 2i is a zero of f implies that its conjugate 5 - 2i is also a zero of f.  It then follows that both (x - [5 + 2i]) and (x - [5 - 2i]) are factors of f and therefore their product x² -10x + 29 is a factor of x³ - 7x² - x + 87.
The quotient when x³ - 7x² - x + 87 is divided by x² -10x + 29 is x + 3.
Therefore f(x) = x³ - 7x² - x + 87 = (x² -10x + 29)(x + 3) = (x - [5 - 2i])(x - [5 + 2i])(x + 3)
and it follows that the zeros of f are 5 - 2i, 5 + 2i, and -3.

Parenthetically:   (x - [5 - 2i])(x - [5 + 2i])(x + 3) = x² - [5-2i]x - [5+2i]x + [5+2i][5-2i]
                          = x² - 5x + 2ix - 5x - 2ix + 25 -10i + 10i - 4i² = x² - 10x + 25 + 4
                          = x² - 10x + 29