algebra 2 unit 5.4

21
UNIT 5.4 DIVIDING UNIT 5.4 DIVIDING POLYNOMIALS POLYNOMIALS

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Page 1: Algebra 2 unit 5.4

UNIT 5.4 DIVIDING UNIT 5.4 DIVIDING POLYNOMIALSPOLYNOMIALS

Page 2: Algebra 2 unit 5.4

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

Page 3: Algebra 2 unit 5.4

Use long division and synthetic division to divide polynomials.

Objective

Page 4: Algebra 2 unit 5.4

Polynomial long division is a method for dividing a polynomial by another polynomials of a lower degree. It is very similar to dividing numbers.

Page 5: Algebra 2 unit 5.4

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

Page 6: Algebra 2 unit 5.4

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

Page 7: Algebra 2 unit 5.4

Step 4 Write the final answer.

Example 1 Continued

–y2 + 2y3 + 25y – 3 = 2y2 + 5y + 15 +

70y – 3

Page 8: Algebra 2 unit 5.4

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

Page 9: Algebra 2 unit 5.4

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

Page 10: Algebra 2 unit 5.4

Check It Out! Example 1a Continued

Step 4 Write the final answer.

15x2 + 8x – 123x + 1 = 5x + 1 –

133x + 1

Page 11: Algebra 2 unit 5.4

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).

Page 12: Algebra 2 unit 5.4
Page 13: Algebra 2 unit 5.4

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.

Page 14: Algebra 2 unit 5.4

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

Page 15: Algebra 2 unit 5.4

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.

Page 16: Algebra 2 unit 5.4

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

Page 17: Algebra 2 unit 5.4

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.

Page 18: Algebra 2 unit 5.4

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

Page 19: Algebra 2 unit 5.4

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

Page 20: Algebra 2 unit 5.4

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

Page 21: Algebra 2 unit 5.4

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