College
Algebra Exercises
Section 3.2
28) Find the real zeros (and their multiplicity) of f(x) = 49 – x2
Solution: The zeros are found by solving the equation resulting from f(x) = 0. Therefore we must solve the equation 0 = f(x) = 49 – x2 = (7 – x)(7 + x).
The solutions are 7 and –7. Each is a real number. Each has multiplicity 1, an odd number. Therefore each zero represents an x-intercept. The graph of f will cross the x-axis at x = 7 and at x = -7.
30) Find the real zeros (and their multiplicity) of f(x) = x2 + 10x + 25.
Solution: The zeros are found by solving the equation resulting from f(x) = 0. Therefore we must solve the equation 0 = f(x) = x2 + 10x + 25= (x + 5)(x + 5) = (x + 5)2.
The solution is 5. It is a real number. It has multiplicity 2, an even number. Therefore the zero represents an x-intercept. The graph of f will touch but not cross the x-axis at x = 5.
34) Find the real zeros (and their multiplicity) of f(x) = 5x(x2 - 2x – 1).
Solution: The
zeros are found by solving the equation resulting from f(x) = 0. Therefore we must solve the equation 0 = f(x) = 5x(x2 - 2x – 1). The quadratic formula yields th solutions
of x2 - 2x – 1 = 0 as 
The solutions are 0 and 
. Each is a real
number. The multiplicity of each zero
is 1, an odd number. Therefore each of
the three zeros represent an x-intercepts.
The graph of f will cross the x-axis at x = 0, 
.
36) Find the real zeros (and their multiplicity)
of f(x) = x4 – x3-
20x2
Solution: The zeros are found by solving the equation resulting from f(x) = 0. Therefore we must solve the equation 0 = f(x) = x4 – x3- 20x2 = x2(x2 – x – 20) = x2(x - 5)(x + 4).
The solutions are 0, -4, and 5. Each is a real number. The multiplicity of 0 is 2, an even number. The multiplicity of - 4 is 1 an odd number. The multiplicity of 5 is 1 an odd number. Therefore all three zeros represent x-intercepts. The graph of f will touch but not cross the x-axis at x = 0 and it will cross the x-axis at x = -4 and at x = 5.
38) Find the real zeros (and their multiplicity)
of f(x) = x5 + x3-
9x
Solution: The
zeros are found by solving the equation resulting from f(x) = 0.
Therefore we must solve the equation 0 = f(x) = x5 + x3- 9x = x(x4 + x2 – 9)
= x(x2 + 3)(x2 - 3).
The solutions are 0, 
, 
, 
and 
. Notre that 
and 
are complex numbers and therefore do not show up on the graph
of f. Each of the other three zeros
are real numbers. In each case their multiplicity is 1, an odd
number. Therefore the three real zeros
represent x-intercepts. The graph of f
will cross the x-axis at x = 0, x = 
, and x = 
.
40) Find the real zeros (and their multiplicity)
of f(x) = 2x4 – 2x2 –
40
Solution: The zeros are found by solving the equation resulting from f(x) = 0. Therefore we must solve the equation 0 = f(x) = 2x4 – 2x2 – 40 = 2(x4 – x2 – 20) = 2(x2 – 5)(x2 + 4)
The solutions are 
, 
, 2i, and –2i. Note
2i and –2i are complex numbers and therefore do not show up on the graph of
f. The multiplicity of 
is 1, an odd
number. The multiplicity of 
is 1, an odd
number. Therefore 
and 
represent x-intercepts. The graph of f will cross the x-axis at x = 
and at x = 
.
42) Find the real zeros (and their multiplicity)
of f(x) = x3 – 4x2
– 25 x + 100
Solution: The zeros are found by solving the equation resulting from f(x) = 0. Therefore we must solve the equation 0 = f(x) = x3 – 4x2 – 25 x + 100 = x2(x – 4) - 25(x – 4) = (x – 4)(x2 - 25) = (x – 4)(x – 5)(x + 5). The solutions are 4, 5, and –5. Each of these zeros are real numbers. Each of these zeros has multiplicity 1, an odd number. Therefore4, 5, and –5 each represent an x-intercept. The graph of f will cross the x-axis at –5, 4, and 5.