Group 2, Fourth Period
Joey Hamilton and Matt Lee
Chapter 2 Review May 16, 2011
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5. Both the equation of y=g(x) and the graph is shown.
g(x) = x 3 - 7x 2 + 25x - 39
5a. Find all of the roots of g(x) (real, imaginary)
g(x) = x 3 - 7x 2 + 25x - 39
g(x)=(x-3)(x 2 -4x+13) (Factor out x-3, factor found using graph)
Set (x-3) equal to zero and use the quadratic formula to solve (x 2 -4x+13) .
x 3  0
x 3

x 2  4 x  13

Using Quadratic Formula:
4  16  52
2
4  36
2
4  6i
2
2  3i
x  3, 2  3i
5b. Find g(3i)
i  1
g (3i)  (3i) 3  7(3i) 2  25  3i  39
g (3i )  27i  63  75i  39
g (3i )  48i  24
g (3i )  24(2i  1)
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6. The leading coefficient of a cubic polynomial P is 4, and the coefficient of the linear term is 5
If P(0) = -14 and P(2) = 4, find P( x) .
P(x)  4x3  kx2  5x  b
P(0)  4  03  k  02  5  0  b  14
P(0)  0  b  14
b  14
P(x)  4x3  kx2  5x 14
P (2)  4  23  k  2 2  5  2  14  4
P(2)  32  4k 10 14  4
P(2)  28  4k  4
4k  24

k  6
Then fill “k” and “b” back into the original equation
P(x)  4x3  6x2  5x 14

7. If 2 is a zero of f (x)  x3  x2  kx  2 , solve for k
f (2)  2 3  2 2  k 2  2  0
f (2)  2  2k  0
2 k  2
k  1
8. *The perimeter of the bottom of a rectangular box is 15 inches. The perimeter of one of the
sides of the box is 48 inches. Find all 3 dimensions of the box that would maximize the volume
of the box?
2c  2b  15
2a  2b  48
15
b
2
a  24  b
c


bb


volume  a b c

15

volume  (24  b) b   b 
 2



Using your calculator, take this equation and input it as “y1” in Y=. Then use “ZOOM FIT” by
pressing “F2” and scroll to “ZOOM FIT”. Then find the maximum by pressing “F5” and
scrolling to “maximum” and have the “lower bound” to the left of the curve and “upper bound”
to the right of the curve. The coordinate will be (3.411, 287.167). 3.411 is our “b” value
therefore, b=3.411. Then fill “b” back into the perimeter equations.
a  24  3.411  20.589
c
15
 3.411  4.089
2
a  20.589in




b  3.411in
c  4.089in
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14. Graphed to the left is the function
y  f ( x) with the following
characteristics:
a. Vertical asymptote x  1
b. y - intercept (0, 36)
c. Small hole/discontinuity at x = 6.
Find the equation of one such y  f ( x)
__________________________
x  6 because of small hole/discontinuity at x=6. The graph is undefined at this point, but
continues past the point.
x  1 due to vertical asymptote at x=-1. The graph will grow infinitely close to -1, but never
reach -1.
According to the graph, x  3 is a double root.
From the above, we know that (x-6) and (x+1) have to be in the denominator of the function and
(x-6) and ( x  3) 2 have to be in the numerator of the function.
( x  6)( x  3) 2
( x  1)( x  6)
Set x equal to 0 to check for y intercept:
(0  6)(0  3) 2
9
(0  1)(0  6)
From the given, Y intercept is (0,36), so we need to add a coefficient (a) to the numerator of the
function. 9a=36, so a=4.
f ( x) 
4( x  6)( x  3) 2
( x  1)( x  6)
Chapter 2 Formulas and Definitions
A polynomial in x is an expression that can be written in the form:
an x n  an 1 x n 1  ...  a2 x 2  a1 x  a0 where n is a nonnegative integer. Every polynomial defines a
function. Any value of x that makes the function equal zero is a root of the equation or a zero of
the function. In other words, the x intercepts of the function are the zeroes.
Synthetic Substitution
Functions can be solved using synthetic substitution. First you take the leading coefficient. Then,
you multiply it by the value of x. Then, you add the next coefficient. Finally, repeat the last two
steps until you reach the final coefficient.
Ex.:
If P( x)  3x 4  7 x 3  5 x 2  9 x  10 , find P (2)
3
3
-7
-5
9
10
6
-2
-14
-10
-1
-7
-5
0
Synthetic Division
If you know a factor of a polynomial equation, then you can use synthetic division to get the
other factor.
Ex.:
2 x 3  6 x 2  3x  1
x  3 2 x 4  0 x 3  15 x 2  10 x  5
2x 4  6x3
6 x 3  15 x 2
6 x 3  18 x 2
3x 2  10 x
3x 2  9 x
 1x  5
 1x  3
2- Remainder
The Remainder Theorem
When a polynomial P (x ) is divided by x  a , the remainder is P (a ) .
The Factor Theorem
For a polynomial P (x ) , x  a is a factor if and only if P (a ) = 0.
Cubic Functions
Graph of f ( x)  ax 3  bx 2  cx  d
Quartic Functions
Graph of f ( x)  ax 4  bx 3  cx 2  dx  e
Useful Way of Sketching a Cubic Function
Find and plot the zeroes of the function. ( x  1)( x  1)( x  2)  0
Then, perform a sign analysis of f (x ) by testing one value of x from each of the intervals
determined by the zeroes.
x  2 f ( x)  (3  1)(3  1)(3  2)
1< x< 2 f ( x)  (1.5  1)(1.5  1)(1.5  2)
1<x<1 f ( x)  (0  1)(0  1)(0  2)
x  1 f ( x)  (2  1)( 2  1)( 2  2)
Finally, sketch the graph using the sign analysis results to find out where the graph is above or
below the x-axis.
Double and Triple Roots
If a polynomial P (x ) has a squared factor such as ( x  c) 2 , then x  c is a double root of P( x)  0
If a polynomial P (x ) has a cubed factor such as ( x  c) 3 , then x  c is a triple root of P( x)  0 .
Maximums and Minimums of Polynomial Functions
The maximum or minimum value of the quadratic function f ( x)  ax 2  bx  c occurs at x 
b
2a
The maximum of a cubic function is the highest point of its peak and the minimum is the lowest
point of its valley. To approximate this value, graph the function on a computer or graphing
calculator.
The Rational Root Theorem
Let P (x ) be a polynomial of degree n with integral coefficients and a nonzero constant term:
P( x)  a n x n  a n 1 x n 1  ...  a0 , where a0  0 .
p
where p and q are nonzero integers with no
q
common factor other than 1, then p must be a factor of a 0 , and q must be a factor of a n .
If one of the roots of the equation P( x)  0 is x 
Five Theorems about Polynomial Equations
Theorem 1 The Fundamental Theorem of Algebra
In the complex number system consisting of all real and imaginary numbers, if P (x ) is a
polynomial of degree n (n > 0) with complex coefficients, then the equation P( x)  0 has exactly
n roots (provided a double root is counted as 2 roots, a triple root is counted as 3 roots, and so
on).
Ex:
The function f ( x)  ax 3  bx 2  cx  d has 3 roots.
Theorem 2 Complex Conjugates Theorem
If P (x ) is a polynomial with real coefficients, and a  bi is an imaginary root of the equation
P ( x)  0 , then a  bi is also a root.
Theorem 3
Suppose P (x ) is a polynomial with rational coefficients, and a and b are rational numbers such
that
b is irrational. If a  b is a root of the equation P( x)  0 , then a  b is also a root.
Theorem 4
If P (x ) is a polynomial of odd degree with real coefficients, then the equation P( x)  0 has at
least one real root.
Theorem 5
For the equation a n x n  a n 1 x n 1  a n  2 x n  2  ...  a 0  0 , with a0  0 : the sum of the roots is

a n 1
a
a
, the product of the roots is 0 if n is even or  0 if n is odd.
an
an
an