holt algebra 2 6-3 dividing polynomials 6-3 dividing polynomials holt algebra 2 warm up warm up...
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Holt Algebra 2
6-3 Dividing Polynomials6-3 Dividing Polynomials
Holt Algebra 2
Warm UpWarm Up
Lesson PresentationLesson Presentation
Lesson QuizLesson Quiz
Holt Algebra 2
6-3 Dividing Polynomials
2. Use Pascal’s Triangle to expand the expression. (y – 5)4
Lesson Quiz
1. (2a3 – a + 3)(a2 + 3a – 5)
2a5 + 6a4 – 11a3 + 14a – 15
y4 – 20y3 + 150y2 – 500y + 625
Find each product.
3. Use Pascal’s Triangle to expand the expression. (3a – b)3
27a3 – 27a2b + 9ab2 – b3
Holt Algebra 2
6-3 Dividing Polynomials
Use long division and synthetic division to divide polynomials.
Objective
Holt Algebra 2
6-3 Dividing Polynomials
Polynomial long division is a method for dividing a polynomial by another polynomials of a lower degree. It is very similar to dividing numbers.
Holt Algebra 2
6-3 Dividing Polynomials
Divide using long division.
Example 1: Using Long Division to Divide a Polynomial
(–y2 + 2y3 + 25) ÷ (y – 3)
2y3 – y2 + 0y + 25
Step 1 Write the dividend in standard form, includingterms with a coefficient of 0.
Step 2 Write division in the same way you would when dividing numbers.
y – 3 2y3 – y2 + 0y + 25
Holt Algebra 2
6-3 Dividing Polynomials
Notice that y times 2y2 is 2y3. Write 2y2 above 2y3.
Step 3 Divide.
2y2
–(2y3 – 6y2) Multiply y – 3 by 2y2. Then subtract. Bring down the next term. Divide 5y2 by y.
5y2 + 0y
+ 5y
–(5y2 – 15y) Multiply y – 3 by 5y. Then subtract. Bring down the next term. Divide 15y by y.
15y + 25
–(15y – 45)
70 Find the remainder.
+ 15
Multiply y – 3 by 15. Then subtract.
Example 1 Continued
y – 3 2y3 – y2 + 0y + 25
Holt Algebra 2
6-3 Dividing Polynomials
Step 4 Write the final answer.
Example 1 Continued
–y2 + 2y3 + 25y – 3 = 2y2 + 5y + 15 +
70y – 3
Holt Algebra 2
6-3 Dividing Polynomials
Check It Out! Example 1a
Divide using long division. (15x2 + 8x – 12) ÷ (3x + 1)
15x2 + 8x – 12
Step 1 Write the dividend in standard form, includingterms with a coefficient of 0.
Step 2 Write division in the same way you would when dividing numbers.
3x + 1 15x2 + 8x – 12
Holt Algebra 2
6-3 Dividing Polynomials
Check It Out! Example 1a Continued
Notice that 3x times 5x is 15x2. Write 5x above 15x2.
Step 3 Divide.
5x
–(15x2 + 5x) Multiply 3x + 1 by 5x. Then subtract. Bring down the next term. Divide 3x by 3x.
3x – 12
+ 1
–(3x + 1)
–13Find the remainder.
Multiply 3x + 1 by 1. Then subtract.
3x + 1 15x2 + 8x – 12
Holt Algebra 2
6-3 Dividing Polynomials
Check It Out! Example 1a Continued
Step 4 Write the final answer.
15x2 + 8x – 123x + 1 = 5x + 1 –
133x + 1
Holt Algebra 2
6-3 Dividing Polynomials
Synthetic division is a shorthand method of dividing a polynomial by a linear binomial by using only the coefficients. For synthetic division to work, the polynomial must be written in standard form, using 0 and a coefficient for any missing terms, and the divisor must be in the form (x – a).
Holt Algebra 2
6-3 Dividing Polynomials
Holt Algebra 2
6-3 Dividing Polynomials
Divide using synthetic division.
(3x4 – x3 + 5x – 1) ÷ (x + 2)
Step 1 Find a.
Use 0 for the coefficient of x2.
For (x + 2), a = –2.a = –2
Example 2B: Using Synthetic Division to Divide by a Linear Binomial
3 – 1 0 5 –1 –2
Step 2 Write the coefficients and a in the synthetic division format.
Holt Algebra 2
6-3 Dividing Polynomials
Example 2B Continued
Draw a box around the remainder, 45.
3 –1 0 5 –1 –2
Step 3 Bring down the first coefficient. Then multiply and add for each column.
–6
3 45
Step 4 Write the quotient.
3x3 – 7x2 + 14x – 23 +45
x + 2Write the remainder over the divisor.
46–2814
–2314–7
Holt Algebra 2
6-3 Dividing Polynomials
Check It Out! Example 2a
Divide using synthetic division.
(6x2 – 5x – 6) ÷ (x + 3)
Step 1 Find a.
Write the coefficients of 6x2 – 5x – 6.
For (x + 3), a = –3.a = –3
–3 6 –5 –6
Step 2 Write the coefficients and a in the synthetic division format.
Holt Algebra 2
6-3 Dividing Polynomials
Check It Out! Example 2a Continued
Draw a box around the remainder, 63.
6 –5 –6 –3
Step 3 Bring down the first coefficient. Then multiply and add for each column.
–18
6 63
Step 4 Write the quotient.
6x – 23 +63
x + 3Write the remainder over the divisor.
–23
69
Holt Algebra 2
6-3 Dividing Polynomials
You can use synthetic division to evaluate polynomials. This process is called synthetic substitution. The process of synthetic substitution is exactly the same as the process of synthetic division, but the final answer is interpreted differently, as described by the Remainder Theorem.
Holt Algebra 2
6-3 Dividing Polynomials
Example 3A: Using Synthetic Substitution
Use synthetic substitution to evaluate the polynomial for the given value.
P(x) = 2x3 + 5x2 – x + 7 for x = 2.
Write the coefficients of the dividend. Use a = 2.
2 5 –1 7 2
4
2 41
P(2) = 41
Check Substitute 2 for x in P(x) = 2x3 + 5x2 – x + 7.P(2) = 2(2)3 + 5(2)2 – (2) + 7
P(2) = 41
3418
179
Holt Algebra 2
6-3 Dividing Polynomials
Check It Out! Example 3a
Use synthetic substitution to evaluate the polynomial for the given value.
P(x) = x3 + 3x2 + 4 for x = –3.
Write the coefficients of the dividend. Use 0 for the coefficient of x2 Use a = –3.
1 3 0 4 –3
–3
1 4
P(–3) = 4
Check Substitute –3 for x in P(x) = x3 + 3x2 + 4.P(–3) = (–3)3 + 3(–3)2 + 4
P(–3) = 4
00
00
Holt Algebra 2
6-3 Dividing Polynomials
4. Find an expression for the height of a parallelogram whose area is represented by 2x3 – x2 – 20x + 3 and whose base is represented by (x + 3).
Lesson Quiz
2. Divide by using synthetic division. (x3 – 3x + 5) ÷ (x + 2)
1. Divide by using long division. (8x3 + 6x2 + 7) ÷ (x + 2)
2x2 – 7x + 1
194; –43. Use synthetic substitution to evaluate P(x) = x3 + 3x2 – 6 for x = 5 and x = –1.
8x2 – 10x + 20 – 33
x + 2
x2 – 2x + 1 + 3
x + 2