HW: Pg. 341-342 #13-61 eoo. Quiz 1 Pg. 344 #13-26.
-
Upload
linda-burke -
Category
Documents
-
view
216 -
download
1
Transcript of HW: Pg. 341-342 #13-61 eoo. Quiz 1 Pg. 344 #13-26.
HW: Pg. 341-342 #13-61 eoo
Quiz 1 Pg. 344 #13-26
Vocabulary• Polynomial Long Division:
– When you divide a polynomial ______ by a divisor _____, you get a quotient polynomial ______ and a remainder polynomial _____.
– This can be written as: • Remainder Theorem:
– If a polynomial _____ is divided by ______, then the remainder is __________.
• Synthetic Division:– Only use the ___________ of the polynomial and the
_____________ must be in the form _________.• Factor Theorem:
– A polynomial _____ has a factor ________ if and only if ___________.
EXAMPLE 1Use polynomial long division
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.
4 3 2Di ( ) 3 5vi x 3de b 4 56 yf x x x x x
EXAMPLE 1Use polynomial long division
Multiply divisor by 3x4/x2 = 3x23x4 – 9x3 + 15x2
4x3 – 15x2 + 4xSubtract.
Bring down next term.
Multiply divisor by 4x3/x2 = 4x4x3 – 12x2 + 20x
–3x2 – 16x – 6Subtract.
Bring down next term.
Multiply divisor by – 3x2/x2 = – 3–3x2 + 9x – 15
–25x + 9
remainder
3x2 + 4x – 3
x2 – 3x + 5 3x4 – 5x3 + 0x2 + 4x – 6) quotient
EXAMPLE 1Use polynomial long division
You can check the result of a division problem by multiplying the quotient by the divisor and adding
the remainder. The result should be the dividend.
(3x2 + 4x – 3)(x2 – 3x + 5) + (–25x + 9)
= 3x2(x2 – 3x + 5) + 4x(x2 – 3x + 5) – 3(x2 – 3x + 5) – 25x + 9
CHECK
= 3x4 – 9x3 + 15x2 + 4x3 – 12x2 + 20x – 3x2 + 9x – 15 – 25x + 9
= 3x4 – 5x3 + 4x – 6
3x4 – 5x3 + 4x – 6
x2 – 3x + 5
= 3x2 + 4x – 3 +–25x + 9
x2 – 3x + 5ANSWER
EXAMPLE 2Use polynomial long division with a linear divisor
x2
x – 2 x3 + 5x2 – 7x + 2)
quotient
x3 – 2x2 Multiply divisor by x3/x = x2.
7x2 – 7x Subtract.
Multiply divisor by 7x2/x = 7x.7x2 – 14x
7x + 2 Subtract.
Multiply divisor by 7x/x = 7.
16remainder
7x – 14
ANSWER
x3 + 5x2 – 7x +2
x – 2= x2 + 7x + 7 +
16
x – 2
3 2( ) 5Divide by x7 2 2f x x x x
+ 7x + 7
GUIDED PRACTICEfor Examples 1 and 2
Divide using polynomial long division.
(2x2 – 3x + 8) + –18x + 7
x2 + 2x – 1
ANSWER
(x2 – 3x + 10) + –30
x + 2ANSWER
4 3 21. 2 12 1 x xx x x
3 22. 24 x 10x x x
EXAMPLE 3Use synthetic division
–3 2 1 –8 5
–6 15 –21
2 –5 7 –16
2x3 + x2 – 8x + 5
x + 3
= 2x2 – 5x + 7 –16
x + 3
ANSWER
SOLUTION
3 2Divide by using sy( ) 2 8 5 nthetx 3 ic division.f x x x x
EXAMPLE 4Factor a polynomial
SOLUTION
Because x + 2 is a factor of f (x), you know that f (–2) = 0. Use synthetic division to find the other
factors.
–2 3 –4 –28 –16
–6 20 16
3 –10 –8 0
3 2Factor completely given that is a fa( ) 3 4 x cto28 6 r2 .1f x x x x
EXAMPLE 4Factor 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.
GUIDED PRACTICEfor Examples 3 and 4
Divide using synthetic division.
x2 + x – 4 +11
x + 3ANSWER
4x2 + 5x + 2 +9
x – 1
ANSWER
3 2 (x 33. ( 4 1) ) x x x
3 2 (x-14 (4 3 7. ) )x x x
3 2( ) 6 5 1 25. f x x x x
3 2( ) 22 406. f x x x x
(x – 4)(x –3)(x + 1)
(x – 4)(x –2)(x +5)
ANSWER
ANSWER
Factor the polynomial completely given that
x – 4 is a factor.
EXAMPLE 5Standardized Test Practice
SOLUTION
Because f (3) = 0, x – 3 is a factor of f (x). Use synthetic division.3 1 –2 –23 60
3 3 –60
1 1 –20 0
Use the result to write f (x) as a product of two factors. Then factor completely.
f (x) = x3 – 2x2 – 23x + 60
The zeros are 3, –5, and 4.
The correct answer is A. ANSWER
= (x – 3)(x + 5)(x – 4)
= (x – 3)(x2 + x – 20)
GUIDED PRACTICEfor Example 5
Find the other zeros of f given that f (–2) = 0.
3 and –3 1 and –7
ANSWER ANSWER
3 2( ) 2 9 1 87. f x x x x 3 2( ) 8 5 1 48. f x x x x
HOMEWORK:
Pg. 356 #15-35 eoo