PRECALCULUS A
TEST #2 – POLYNOMIALS AND RATIONAL FUNCTIONS, PRACTICE
SECTION 2.3 – Polynomial and Synthetic Division
1)
Divide using long division: ( 6 x 3 + 11x 2 - 4 x - 9 ) ¸ ( 3x - 2 )
2)
Divide using long division: ( x 3 - 13x - 12 ) ¸ ( x - 4 )
3)
Divide using synthetic division: ( 2 x 3 + 5 x 2 - 7 x - 12 ) ¸ ( x + 3)
4)
Divide using synthetic division: ( x 4 - 5 x 2 - 10 x - 12 ) ¸ ( x - 2 )
5)
Find the other solutions of the equation 4 x 3 + 12 x 2 - x - 3 = 0 if one of the solutions is ½.
SECTION 2.4 – Real Zeros of Polynomial Functions
6)
Use Descartes’s Rule of Signs to determine the possible number of positive and negative real
zeros of the function f ( x) = 3 x 5 - 2 x 4 - 3x 3 + x - 1.
7)
Use the Rational Zero Test to list
function f ( x ) = 4 x 4 + 16 x 3 - 47 x 2 + 9 x + 18.
8)
Find all the real zeros of the function f ( x ) = x 3 - 5 x 2 + 5 x - 1.
9)
Find all the real zeros of the function f ( x ) = 2 x 4 + 7 x3 - 9 x 2 + 34 x + 20.
all
possible
rational
zeros
of
SECTION 2.5 – Complex Numbers
-72.
11)
Simplify ( 2 + 3i ) - ( 6 - 8i ) + (1 + i ) .
13)
Simplify ( 3 - 5i ) .
10)
Simplify
12)
Simplify ( 2 + 3i )( 4 - 5i ) .
14)
Simplify
15)
Solve the equation 2 x 2 + 4 x + 15 = 0 using the Quadratic Formula.
2 + 4i
.
5 - 3i
2
the
SECTION 2.6 – The Fundamental Theorem of Algebra
16)
Find all the zeros of the function f ( x) = x 4 + 2 x 3 + 6 x 2 + 18 x - 27.
17)
Use the answers from Problem #16 to write the factorization of x 4 + 2 x3 + 6 x 2 + 18 x - 27.
18)
Find all the zeros of the function f ( x ) = 2 x 4 - x 3 - x 2 + 4 x + 2.
19)
Use the answers from Problem #18 to write the factorization of 2 x 4 - x 3 - x 2 + 4 x + 2.
20)
Find all the zeros of the function f ( x ) = 3 x 4 + 25 x 3 + 57 x 2 + x - 34 if one of the solutions
is -4 - i.
SECTION 2.7 – Rational Functions
21)
If f ( x) =
2
, provide as much information as possible about the graph of this function.
x+3
22)
If f ( x) =
2x + 4
, provide as much information as possible about the graph of this function.
3x - 1
23)
If f ( x) =
x3 + 1
, provide as much information as possible about the graph of this function.
x2 - 9
24)
If f ( x) =
x2 - 5x + 6
, provide as much information as possible about the graph of this
2 x2 + 5x - 3
function.
25)
Graph f ( x ) =
8 x - 16
on the coordinate plane.
2x + 6
SECTION 2.8 – Partial Fractions
26)
Write the partial fraction decomposition of
9x + 3
.
x2 + x
27)
Write the partial fraction decomposition of
x+7
.
x - x-6
28)
Write the partial fraction decomposition of
5x - 7
.
( x - 2)2
29)
Write the partial fraction decomposition of
4x - 1
.
(2 x + 1)2
2
**********************ANSWERS**********************
1)
+ 5x + 2 +
2x2
3x - 2 6 x3
+ 11x 2
-
4x
- ( 6 x3
- 4x2 )
15 x 2
-
4x
- (15 x
-
10 x )
2
-5
3x - 2
9
6x
-
9
- (6x
-
4)
-5
2)
+ 4x
x2
x-4
x
+ 0x
3
- ( x3
- 13 x
2
+
-
3
12
- 4x2 )
- 13x
4x2
- ( 4 x2
- 16 x )
3x
- ( 3x
- 12
- 12 )
0
-3
2
3)
2
2
1
4)
1
5)
-7 -12
® 2 x2 - x - 4
-6
3
12
-1
-4
0
0
-5 -10 -12
® x 3 + 2 x 2 - x - 12 +
2
4
-2 -24
2
-1
- 12 - 36
-36
x-2
1
4 12 -1 -3
® Since there are only three terms left, use the Quadratic Formula ®
2
2 7
3
4
6)
5
14
6
0
-14 ± 100
-14 + 10 -4
1
-14 - 10 -24
®
=
=- &
=
= -3
8
8
8
2
8
8
f ( x ) = 3 x5 - 2 x 4 - 3x 3 + x - 1 ® 3 sign changes ® 3 or 1 positive real zeros
f (- x ) = -3 x5 - 2 x 4 + 3 x3 - x - 1 ® 2 sign changes ® 2 or 0 positive real zeros
7)
Factors of p = ±1, ± 2, ± 3, ± 6, ± 9, ± 18 ® Factors of q = ±1, ± 2, ± 4
1
3
9
1
3
9
Possible rational zeros = ± 1, ± 2, ± 3, ± 6, ± 9, ± 18, ± , ± , ± , ± , ± , ±
2
2
2
4
4
4
8)
Graph f ( x ) = x3 - 5 x 2 + 5 x - 1. The graph appears to cross the x -axis at 1. Divide the original
polynomial by 1 using synthetic division.
1 1 -5
5 -1
1 -4
1 -4
1
1
0
The remaining polynomial only has three terms, so use the Quadratic Formula.
9)
4 ± 12
4±2 3
®
® 2 ± 3 ® Zeros are 1, 2 ± 3
2
2
Graph f ( x ) = 2 x 4 + 7 x 3 - 9 x 2 + 34 x + 20. The graph appears to cross the x-axis at - 5. Divide
the original polynomial by –5 using synthetic division.
-5
2
7
-10
2
-9
34
20
15 -30 -20
-3
6
4
0
The remaining polynomial has more than three terms, so you can’t use the Quadratic
Formula yet. The graph also appears to cross the x-axis at –½, so divide the new
polynomial by –½ using synthetic division.
-0.5 2 -3
-1
2 -4
6
4
2 -4
8
0
The remaining polynomial only has three terms, so use the Quadratic Formula.
4 ± -48
1
® Stop here, as -48 is imaginary ® Zeros are -5 and 2
2
10)
-72 = 36 × -1 × 2 = 6i 2
11)
( 2 + 3i ) - ( 6 - 8i ) + (1 + i ) = ( 2 + 3i ) + ( -6 + 8i ) + (1 + i ) = -3 + 12i
12)
( 2 + 3i )( 4 - 5i ) = 8 - 10i + 12i - 15i 2 = 8 + 2i + 15 = 23 + 2i
13)
( 3 - 5i )
14)
2 + 4i 5 + 3i 10 + 6i + 20i + 12i 2 10 + 26i - 12 -2 + 26i -1 + 13i
×
=
=
=
=
5 - 3i 5 + 3i
25 - 9i 2
25 + 9
34
17
2
= ( 3 - 5i )( 3 - 5i ) = 9 - 15i - 15i + 25i 2 = 9 - 30i - 25 = -16 - 30i
15)
16)
-4 ± 42 - 4 × 2 × 15 -4 ± 16 - 120 -4 ± -104 -4 ± 2i 26 -2 ± i 26
=
=
=
=
2×2
4
4
4
2
Graph f ( x ) = x 4 + 2 x 3 + 6 x 2 + 18 x - 27. The graph appears to cross the x-axis at 1. Divide the
original polynomial by 1 using synthetic division.
1
1
1
18 -27
2
6
1
3
9
3
9
27
27
0
The remaining polynomial has more than three terms, so you can’t use the Quadratic
Formula yet. The graph also appears to cross the x-axis at –3, so divide the new
polynomial by –3 using synthetic division.
-3
1
3
-3
1
0
9
27
0 -27
9
0
The remaining polynomial only has three terms, so use the Quadratic Formula.
0 ± -36
0 ± 6i
0 + 6i 6i
0 - 6i -6i
®
®
= = 3i &
=
= -3i ®
2
2
2
2
2
2
Zeros are 1, –3, 3i, and –3i
17)
Change zeros into factors
x = 1 ® x - 1 = 0 ® Factor = ( x - 1)
x = -3 ® x + 3 = 0 ® Factor = ( x + 3)
x = 3i ® x - 3i = 0 ® Factor = ( x - 3i )
x = -3i ® x + 3i = 0 ® Factor = ( x + 3i )
x 4 + 2 x 3 + 6 x 2 + 18 x - 27 = ( x - 1)( x + 3)( x - 3i )( x + 3i )
18)
Graph f ( x ) = 2 x 4 - x3 - x 2 + 4 x + 2. The graph appears to cross the x -axis at - 1. Divide the
original polynomial by –1 using synthetic division.
-1 2
-1 -1
-2
2 -3
4
2
3 -2 -2
2
2
0
The remaining polynomial has more than three terms, so you can’t use the Quadratic
Formula yet. The graph also appears to cross the x-axis at –½, so divide the new
polynomial by –½ using synthetic division.
-0.5 2 -3
-1
2 -4
2
2
2 -2
4
0
The remaining polynomial only has three terms, so use the Quadratic Formula.
4 ± -16
4 ± 4i
4 + 4i
4 - 4i
1
®
®
= 1+ i &
= 1 - i ® Zeros are - 1, - , 1 + i , 1 - i
4
4
4
4
2
19)
Change zeros into factors
x = -1 ® x + 1 = 0 ® Factor = ( x + 1)
1
x = - ® 2 x = -1 ® 2 x + 1 = 0 ® Factor = ( 2 x + 1)
2
x = 1 + i ® x - 1 - i = 0 ® Factor = ( x - 1 - i ) x = 1 - i ® x - 1 + i = 0 ® Factor = ( x - 1 + i )
2 x 4 - x 3 - x 2 + 4 x + 2 = ( x + 1)( 2 x + 1)( x - 1 + i )( x - 1 - i )
20)
If - 4 - i is a zero, then - 4 + i is also a zero. Convert these zeros to factors.
x = -4 - i ® x + 4 + i = 0 ® Factor = ( x + 4 + i )
x = -4 + i ® x + 4 - i = 0 ® Factor = ( x + 4 - i )
Multiply these two factors ® ( x + 4 + i )( x + 4 - i ) ®
x 2 + 4 x + ix + 4 x + 16 + 4i - ix - 4i - i 2 = x 2 + 8 x + 16 + 1 = x 2 + 8 x + 17
Divide the original polynomial by x 2 + 8 x + 17.
+
3x 2
x + 8 x + 17 3 x
2
4
- ( 3x 4
+ 25 x
+ 57 x
3
+ 24 x3
+
x3
-( x
3
+
2
-
x
x
2
-
34
51x 2 )
+ 6 x2
+
+ 8x
+ 17 x )
2
-2 x
- ( -2 x
x
2
-
16 x
-
34
2
-
16 x
-
34 )
0
The remaining polynomial has three terms, so use the Quadratic Formula.
-1 ± 25 -1 ± 5 -1 + 5 4 2
-1 - 5 -6
=
=
= = &
=
= -1
6
6
6
6 3
6
6
Zeros are - 4 - i , - 4 + i , - 1,
21)
2
3
2
2
æ 2ö
® y = ® y -intercept = ç 0, ÷
0+3
3
è 3ø
To find x -intercept, set numerator = 0 ® Numerator has no variable ® No x-intercept
To find vertical asymptote, set denominator = 0 ® x + 3 = 0 ® x = -3
Degree of numerator = 0, degree of denominator = 1 → If degree of denominator is greater than
degree of numerator, the horizontal asymptote is the x-axis or y = 0
To find y -intercept, set x = 0 ® y =
22)
2×0 + 4
4
®y=
® y = -4 ® y -intercept = ( 0, - 4 )
3× 0 -1
-1
To find x -intercept, set numerator = 0 ® 2 x + 4 = 0 ® 2 x = -4 ®
To find y -intercept, set x = 0 ® y =
x = -2 ® x-intercept = ( -2, 0 )
1
3
Degree of numerator = 1, degree of denominator = 1 → If degree of denominator is equal to
degree of denominator, horizontal asymptote = leading coefficient of numerator over leading
2
coefficient of denominator → y =
3
To find vertical asymptote, set denominator = 0 ® 3 x - 1 = 0 ® 3x = 1 ® x =
23)
03 + 1
1
1
1ö
æ
®y=
® y = - ® y -intercept = ç 0, - ÷
2
0 -9
-9
9
9ø
è
3
3
To find x -intercept, set numerator = 0 ® x + 1 = 0 ® x = -1 ® x = -1 ® x-intercept = ( -1, 0 )
To find y -intercept, set x = 0 ® y =
To find vertical asymptote, set denominator = 0 ® x 2 - 9 = 0 ® ( x + 3)( x - 3) = 0 ®
x = -3 & x = 3
Degree of numerator = 3, degree of denominator = 2 → If degree of numerator is greater than
degree of denominator, there is no horizontal asymptote
24)
To find y -intercept, set x = 0 ® y =
y -intercept = ( 0, - 2 )
02 - 5 × 0 + 6
6
®y=
® y = -2 ®
2
2×0 + 5×0 -3
-3
To find x -intercept, set numerator = 0 ® x 2 - 5 x + 6 ® 0 = ( x - 2 )( x - 3) ®
x = 2 & x = 3 ® x-intercepts = ( 2, 0 ) & ( 3, 0 )
To find vertical asymptote, set denominator = 0 ® 2 x 2 + 5 x - 3 = 0 ® ( 2 x - 1)( x + 3) = 0 ®
1
& x = -3
2
If degree of denominator is equal to degree of denominator, horizontal asymptote = leading
1
coefficient of numerator over leading coefficient of denominator → y =
2
x=
25)
10
8
6
Horizontal Asymptote: y = 4
4
Vertical Asymptote: x = -3
-10
2
-5
5
10
-2
-4
-6
-8
26)
9x + 3 9x + 3
9x + 3
A
B
=
®
= +
® Multiply each term by the LCD
2
x + x x ( x + 1)
x ( x + 1) x x + 1
9x + 3
A
B
× x ( x + 1) = × x ( x + 1) +
× x ( x + 1) ® 9x + 3 = A × ( x + 1) + B × x
x ( x + 1)
x
x +1
Set x = -1 ® 9 × ( -1) + 3 = A × ( -1 + 1) + B × ( -1) ® -6 = - B ® 6 = B
Set x = 0 ® 9 × 0 + 3 = A × ( 0 + 1) + B × 0 ® 3 = A
9x + 3 3
6
= +
2
x + x x x +1
27)
x+7
x+7
x+7
A
B
=
®
=
+
® Multiply each term by the LCD
x - x - 6 ( x + 2 )( x - 3)
( x + 2 )( x - 3) x + 2 x - 3
2
x+7
A
B
× ( x + 2 )( x - 3) =
× ( x + 2 )( x - 3) +
× ( x + 2 )( x - 3) ®
x+2
x -3
( x + 2 )( x - 3)
x + 7 = A × ( x - 3) + B × ( x + 2 )
Set x = 3 ® 3 + 7 = A × ( 3 - 3) + B × ( 3 + 2 ) ® 10 = 5B ® 2 = B
Set x = -2 ® -2 + 7 = A × ( -2 - 3) + B × ( -2 + 2 ) ® 5 = -5 A ® -1 = A
x+7
-1
2
=
+
x - x-6 x+ 2 x -3
2
28)
5x - 7
( x - 2)
=
2
A
B
+
® Multiply each term by the LCD
x - 2 ( x - 2 )2
5x - 7
( x - 2)
2
× ( x - 2) =
2
A
B
2
2
× ( x - 2) +
× x - 2 ) ® 5x - 7 = A × ( x - 2) + B
2 (
x-2
( x - 2)
Set x = 2 ® 5 × 2 - 7 = A × ( 2 - 2 ) + B ® 3 = B
Set x = 3 & B = 3 ® 5 × 3 - 7 = A × ( 3 - 2 ) + 3 ® 8 = A + 3 ® 5 = A
5x - 7
( x - 2)
29)
4x -1
( 2 x + 1)
2
=
=
2
5
3
+
x - 2 ( x - 2) 2
A
B
+
® Multiply each term by the LCD
2 x + 1 ( 2 x + 1)2
4x -1
( 2 x + 1)
× ( 2 x + 1) =
2
2
A
B
2
2
× ( 2 x + 1) +
× 2 x + 1) ® 4 x - 1 = A × ( 2 x + 1) + B
2 (
2x + 1
( 2 x + 1)
1
æ æ 1ö ö
æ 1ö
Set x = - ® 4 × ç - ÷ - 1 = A × ç 2 × ç - ÷ + 1÷ + B ® -3 = B
2
è 2ø
è è 2ø ø
Set x = 1 & B = -3 ® 4 × 1 - 1 = A × ( 2 × 1 + 1) - 3 ® 3 = 3 A - 3 ® 6 = 3 A ® 2 = A
4x -1
( 2 x + 1)
2
=
2
-3
+
2 x + 1 (2 x + 1) 2