example 3 multiply polynomials vertically and horizontally a. multiply –2y 2 + 3y – 6 and y –...

EXAMPLE 3 Multiply polynomials vertically and horizontally

a. Multiply –2y2 + 3y – 6 and y – 2 in a vertical format.

b. Multiply x + 3 and 3x2 – 2x + 4 in a horizontal format.

SOLUTION

a. –2y2+ 3y – 6

y – 2

4y2 – 6y + 12 Multiply –2y2 + 3y – 6 by –2 .

–2y3 + 3y2 – 6y Multiply –2y2 + 3y – 6 by y

–2y3 + 7y2 –12y + 12 Combine like terms.

EXAMPLE 3 Multiply polynomials vertically and horizontally

b. (x + 3)(3x2 – 2x + 4) = (x + 3)3x2 – (x + 3)2x + (x + 3)4

= 3x3 + 9x2 – 2x2 – 6x + 4x + 12

= 3x3 + 7x2 – 2x + 12

EXAMPLE 4Multiply three binomials

Multiply x – 5, x + 1, and x + 3 in a horizontal format.

(x – 5)(x + 1)(x + 3) = (x2 – 4x – 5)(x + 3)

= (x2 – 4x – 5)x + (x2 – 4x – 5)3

= x3 – 4x2 – 5x + 3x2 – 12x – 15

= x3 – x2 – 17x – 15

EXAMPLE 5 Use special product patterns

a. (3t + 4)(3t – 4) Sum and difference

= 9t2 – 16

b. (8x – 3)2 Square of a binomial

= 64x2 – 48x + 9

c. (pq + 5)3 Cube of abinomial

= p3q3 + 15p2q2 + 75pq + 125

= (3t)2 – 42

= (8x)2 – 2(8x)(3) + 32

= (pq)3 + 3(pq)2(5) + 3(pq)(5)2 + 53

GUIDED PRACTICE for Examples 3, 4 and 5

Find the product.

3. (x + 2)(3x2 – x – 5)

3x3 + 5x2 – 7x – 10

4. (a – 5)(a + 2)(a + 6)

(a3 + 3a2 – 28a – 60)

5. (xy – 4)3

x3y3 – 12x2y2 + 48xy – 64ANSWER

ANSWER

ANSWER

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

Multiply divisor by 3x4/x2 = 3x2

3x4 – 9x3 + 15x2

4x3 – 15x2 + 4x Subtract. Bring down next term.

Multiply divisor by 4x3/x2 = 4x4x3 – 12x2 + 20x

–3x2 – 16x – 6 Subtract. Bring down next term.

Multiply divisor by – 3x2/x2 = – 3–3x2 + 9x – 15

–25x + 9 remainder

3x2 + 4x – 3x2 – 3x + 5 3x4 – 5x3 + 0x2 + 4x – 6)

quotient

EXAMPLE 1 Use 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 – 6x2 – 3x + 5

= 3x2 + 4x – 3 + –25x + 9x2 – 3x + 5

ANSWER

EXAMPLE 2 Use polynomial long division with a linear divisor

Divide f(x) = x3 + 5x2 – 7x + 2 by x – 2.

x2 + 7x + 7x – 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 = 7.16 remainder

7x – 14

ANSWER x3 + 5x2 – 7x +2x – 2

= x2 + 7x + 7 + 16x – 2

GUIDED PRACTICE for Examples 1 and 2

Divide using polynomial long division.

1. (2x4 + x3 + x – 1) (x2 + 2x – 1)

(2x2 – 3x + 8) + –18x + 7x2 + 2x – 1

ANSWER

(x2 – 3x + 10) + –30 x + 2

ANSWER

2. (x3 – x2 + 4x – 10) (x + 2)

EXAMPLE 3Use synthetic division

Divide f (x)= 2x3 + x2 – 8x + 5 by x + 3 using synthetic division.

Steps1.Make sure the coefficient of the divisor (x + 3 ) must be 1x . (It checks!)2.Take the opposite number of the constant in the divisor (x + 3 ) so we use r=-3.3.Find the coefficients and the constant of the dividend (2x3 + x2 – 8x + 5 ) and we use these numbers to set up our synthetic division.

–3 2 1 –8 5

EXAMPLE 3Use synthetic division

Divide f (x)= 2x3 + x2 – 8x + 5 by x + 3 using synthetic division.

–3 2 1 –8 5

–6 15 –21

2 –5 7 –16

4. Bring down the first number of the column (add it to zero)

5. The number we brought down we will multiply it by “r” the opposite of the divisor. And we place it under column 2.

6. Add the second column and multiply this number by “r” and we place this number under column 3. Add column 3 and repeat until we do not have any more columns.

EXAMPLE 3Use synthetic division

Divide f (x)= 2x3 + x2 – 8x + 5 by x + 3 using synthetic division.

–3 2 1 –8 5

–6 15 –21

2 –5 7 –16

2x3 + x2 – 8x + 5x + 3

= 2x2 – 5x + 7 –16

x + 3ANSWER

SOLUTION

7. Include the necessary variables to our answer. Start with the first variable (column 1) being one power less than our highest power of the divisor.

EXAMPLE 4 Factor a polynomial

Factor f (x) = 3x3 – 4x2 – 28x – 16 completely given that x + 2 is a factor.

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

Putting variables to our answer we get: 3x2 – 10x – 8

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.

GUIDED PRACTICE for Examples 3 and 4

Divide using synthetic division.

3. (x3 + 4x2 – x – 1) (x + 3)

x2 + x – 4 +11

x + 3ANSWER

4. (4x3 + x2 – 3x + 7) (x – 1)

4x2 + 5x + 2 +9

x – 1ANSWER

GUIDED PRACTICE for Examples 3 and 4

Factor the polynomial completely given that x – 4 is a factor.

5. f (x) = x3 – 6x2 + 5x + 12

(x – 4)(x –3)(x + 1)

6. f (x) = x3 – x2 – 22x + 40

(x – 4)(x –2)(x +5)ANSWER

ANSWER

EXAMPLE 3Synthetic division when the divisor’sCoefficient is not one.Divide f (x)= 6x3 +7 x2 + x + 1 by 2x + 3 using synthetic

division.

Steps1.What number can we divide the coefficient so it becomes a one (2x + 3 ) . If we divide it by 2 we get ( x + 3/2 ).2. Now we can solve it like before.3. Take the opposite number of the constant in the divisor (x + 3/2 ) so we use r= - 3/2 .4. Find the coefficients and the constant of the dividend (6x3 +7 x2 + x + 1 ) and we use these numbers to set up our synthetic division.

–3/ 2 6 7 1 1

EXAMPLE 3Use synthetic division

Divide f (x)= 6x3 +7 x2 + x + 1 by 2x + 3 using synthetic division.

–3/2 6 7 1 1

–9 3 –6

6 –2 4 –5

4. Bring down the first number of the column (add it to zero)

5. The number we brought down we will multiply it by “r” the opposite of the divisor. And we place under column 2.

6. Add the second column and multiply this number by “r” and we place this number under column 3. add column 3 and repeat until we do not have any more columns.

EXAMPLE 3 Use synthetic division

= 6x2 –2 x + 4 +- 5

x + 3/2

SOLUTION

7. Include the necessary variables to our answer. Start with the first variable (column 1) being one power less than our highest power of the divisor.

Divide f (x)= 6x3 +7 x2 + x + 1 by 2x + 3 using synthetic division.

–3/2 6 7 1 1

–9 3 –6

6 –2 4 –5

EXAMPLE 3Use synthetic division

6x2 – 2x + 4 +- 5

x + 3/2

8. Because at the beginning we divided our divisor by 2. Now we must also divide our answer (quotient) by 2.

Divide f (x)= 6x3 +7 x2 + x + 1 by 2x + 3 using synthetic division.

32

523

23

5426

2

1 22

xxx

xxxANSWER

EXAMPLE 4

SOLUTION

Use synthetic division

Divide f (x)= 4x3 +2 x2 -4 x + 3 by 2x + 3 using synthetic division.

122 2 xx

SOLUTION

Divide f (x)= 5x 4 – 3 x3 +10 x + 2 by 5x - 3

35

823

xx

EXAMPLE 5 The Remainder Theorem

51353732)3(

:do would what weis This

3 when 1572)(

:polynomial agiven are esay that w usLet

23

23

P

xxxxxP

We can also solve it using synthetic division.

3 2 – 7 5 –1

6 – 3 6

2 –1 2 5

Write a paragraph what the remainder theorem is.

EXAMPLE 5 The remainder and factor theorem

The factor theorem will show us if the divisor is a factor of the dividend.

If there is no remainder then the divisor is a factor of the dividend. The remainder equals zero.

If the remainder is not zero then the divisor is not a factor of the dividend. The remainder is not equal to zero