by x

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1. Use synthetic substitution to evaluate
f (x) = x3 + x2 – 3x – 10 when x = 2.
2
ANSWER
–4
1
1
2
–3
6
-10
6
1
3
3
-4
Check HW 5.4
multiples of 3
EXAMPLE 1
Use polynomial long division
Divide f (x) = 3x4 – 5x3 + 4x – 6 by x2 – 3x + 5.
SOLUTION
Write polynomial division in the same format you
use when dividing numbers. Include a “0” as the
coefficient of x2 in the dividend. At each stage,
divide the term with the highest power in what is
left of the dividend by the first term of the divisor.
This gives the next term of the quotient.
EXAMPLE 1
Use polynomial long division
3x2 + 4x – 3
x2 – 3x + 5 )3x4 – 5x3 + 0x2 + 4x – 6
quotient
-(3x4 – 9x3 + 15x2)
4x3 – 15x2 + 4x
-(4x3 – 12x2 + 20x)
–3x2 – 16x – 6
-(–3x2 + 9x – 15)
–25x + 9
ANSWER
remainder
3x4 – 5x3 + 4x – 6 = 3x2 + 4x – 3 + –25x + 9
x2 – 3x + 5
x2 – 3x + 5
EXAMPLE 2
Use polynomial long division with a linear divisor
Divide f(x) = x3 + 5x2 – 7x + 2 by x – 2.
x2 + 7x + 7
x – 2 ) x3 + 5x2 – 7x + 2
quotient
-(x3 – 2x2)
7x2 – 7x
-(7x2 – 14x)
7x + 2
-(7x – 14)
16
ANSWER
remainder
x3 + 5x2 – 7x +2
= x2 + 7x + 7 + 16
x–2
x–2
GUIDED PRACTICE
for Examples 1 and 2
Divide using polynomial long division.
1.
(2x4 + x3 + x – 1)
(x2 + 2x – 1)
ANSWER (2x2 – 3x + 8) + –18x + 7
x2 + 2x – 1
2.
(x3 – x2 + 4x – 10)  (x + 2)
ANSWER (x2 – 3x + 10) +
–30
x+2
GUIDED PRACTICE
for Examples 1 and 2
Divide using polynomial long division.
2.
(x3 – x2 + 4x – 10)  (x + 2)
x 2  3 x  10
x  2 x3  x 2  4 x  10
 ( x3  2 x 2 )
 3x 22  4 x
 (3x  6x)
10 x  10
 (10x  20)
 30
ANSWER (x2 – 3x + 10) +
–30
x+2
EXAMPLE 3
Use synthetic division
Divide f (x)= 2x3 + x2 – 8x + 5 by x + 3 using synthetic
division.
SOLUTION
–3
2
1
-6
–8
15
5
-21
2
-5
7
-16
2x  5x  7  x163
2
EXAMPLE 4
Factor a polynomial
Factor f (x) = 3x3 – 4x2 – 28x – 16 completely given that
x + 2 is a factor.
SOLUTION
–2
3
-4
-6
-28
20
-16
16
3
-10
-8
0
3x  10x  8
2
EXAMPLE 4
Factor a polynomial
Use the result to write f (x) as a product of two
factors and then factor completely.
f (x) = 3x3 – 4x2 – 28x – 16
Write original polynomial.
= (x + 2)(3x2 – 10x – 8)
Write as a product of two
factors.
= (x + 2)(3x + 2)(x – 4)
Factor trinomial.
for Examples 3 and 4
GUIDED PRACTICE
Divide using synthetic division.
3. (x3 + 4x2 – x – 1)  (x + 3)
–3
1
4
-3
-1
-3
-1
12
1
1
-4
11
ANSWER
x2
11
+x–4+
x+3
for Examples 3 and 4
GUIDED PRACTICE
Factor the polynomial completely given that x – 4 is a
factor.
5.
f (x) = x3 – 6x2 + 5x + 12
( x  4)(x 2  2x  3)
4
1
-6
4
5
-8
12
-12
1
-2
-3
0
x2  2x  3
(x – 4)(x –3)(x + 1)
GUIDED PRACTICE
for Examples 5 and 6
Find the other zeros of f given that f (–2) = 0.
7.
f (x) = x3 + 2x2 – 9x – 18
( x  2)(x2  9)  0
–2
1
1
2
-9
-18
-2
0
18
0
-9
0
x2  9
( x  2)(x  3)(x  3)  0
Zeros are -2, 3, -3
EXAMPLE 5
Standardized Test Practice
SOLUTION
Because f (3) = 0, x – 3 is a factor of f (x). Use synthetic
division.
3
1
1
–2
–23
60
3
3
–60
1
–20
0
EXAMPLE 5
Standardized Test Practice
Use the result to write f (x) as a product of two
factors. Then factor completely.
f (x) = x3 – 2x2 – 23x + 60
= (x – 3)(x2 + x – 20)
= (x – 3)(x + 5)(x – 4)
The zeros are 3, –5, and 4.
ANSWER The correct answer is A.
Class/Homework Assignment
WS 5.5 (1-24 mult. of 3)
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