1. Multiply polynomials vertically and horizontally
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.

2. 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

3. EXAMPLE 4
Multiply 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

4. Use special product patterns
EXAMPLE 5
Use special product patterns
a. (3t + 4)(3t – 4)
= (3t)2 – 42
Sum and difference
= 9t2 – 16
b. (8x – 3)2
= (8x)2 – 2(8x)(3) + 32
Square of a binomial
= 64x2 – 48x + 9
c. (pq + 5)3
Cube of a binomial
= (pq)3 + 3(pq)2(5) + 3(pq)(5)2 + 53
= p3q3 + 15p2q2 + 75pq + 125

5. GUIDED PRACTICE
for Examples 3, 4 and 5
Find the product.
3. (x + 2)(3x2 – x – 5)
ANSWER
3x3 + 5x2 – 7x – 10
4. (a – 5)(a + 2)(a + 6)
(a3 + 3a2 – 28a – 60)
ANSWER
5. (xy – 4)3
ANSWER
x3y3 – 12x2y2 + 48xy – 64

6. 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.

7. ) EXAMPLE 1 Use polynomial long division 3x2 + 4x – 3 x2 – 3x + 5
quotient
x2 – 3x + 5
3x4 – 5x3 + 0x2 + 4x – 6 )
Multiply divisor by 3x4/x2 = 3x2
3x4 – 9x3 + 15x2
Subtract. Bring down next term.
4x3 – 15x x
4x3 – 12x2 + 20x
Multiply divisor by 4x3/x2 = 4x
Subtract. Bring down next term.
–3x2 – 16x – 6
–3x2 + 9x – 15
Multiply divisor by – 3x2/x2 = – 3
–25x + 9
remainder

8. EXAMPLE 1
Use polynomial long division
3x4 – 5x3 + 4x – 6
x2 – 3x + 5
= 3x2 + 4x – 3 +
–25x + 9
ANSWER
CHECK
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
= 3x4 – 9x3 + 15x2 + 4x3 – 12x2 + 20x – 3x2 + 9x – 15 – 25x + 9
= 3x4 – 5x3 + 4x – 6

9. ) EXAMPLE 2 Use polynomial long division with a linear divisor
Divide f(x) = x3 + 5x2 – 7x + 2 by x – 2.
x2 + 7x
quotient
x – 2
x3 + 5x2 – 7x )
x3 – 2x2
Multiply divisor by x3/x = x2.
7x2 – 7x
Subtract.
7x2 – 14x
Multiply divisor by 7x2/x = 7x.
7x + 2
Subtract.
7x – 14
Multiply divisor by 7x/x = 7. 16
remainder
ANSWER
x3 + 5x2 – 7x +2
x – 2
= x2 + 7x + 7 + 16

10. GUIDED PRACTICE
for Examples 1 and 2
Divide using polynomial long division.
(2x4 + x3 + x – 1) (x2 + 2x – 1)
(2x2 – 3x + 8) +
–18x + 7
x2 + 2x – 1
ANSWER
(x3 – x2 + 4x – 10)  (x + 2)
(x2 – 3x + 10) +
–30
x + 2
ANSWER

11. EXAMPLE 3
Use synthetic division
Divide f (x)= 2x3 + x2 – 8x + 5 by x + 3 using synthetic
division.
Steps
Make sure the coefficient of the divisor (x + 3 ) must be 1x . (It checks!)
Take the opposite number of the constant in the divisor (x + 3 ) so we use r=-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.
– –

12. EXAMPLE 3
Use synthetic division
Divide f (x)= 2x3 + x2 – 8x + 5 by x + 3 using synthetic
division.
Bring down the first number of the column (add it to zero)
The number we brought down we will multiply it by “r” the opposite of the divisor. And we place it under column 2.
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.
– –
– –21
2 – –16

13. EXAMPLE 3
Use synthetic division
Divide f (x)= 2x3 + x2 – 8x + 5 by x + 3 using synthetic
division.
SOLUTION
– –
– –21
2 – –16
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.
2x3 + x2 – 8x + 5
x + 3
= 2x2 – 5x + 7 – 16
ANSWER

14. 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.
– –4 – –16 –
3 – –
Putting variables to our answer we get: 3x2 – 10x – 8

15. Use the result to write f (x) as a product of two
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.

16. GUIDED PRACTICE
for Examples 3 and 4
Divide using synthetic division.
3. (x3 + 4x2 – x – 1)  (x + 3)
x2 + x – 4 + 11
x + 3
ANSWER
4. (4x3 + x2 – 3x + 7)  (x – 1)
4x2 + 5x + 2 + 9
x – 1
ANSWER

17. GUIDED PRACTICE
for Examples 3 and 4
Factor the polynomial completely given that x – 4 is a factor.
f (x) = x3 – 6x2 + 5x + 12
ANSWER
(x – 4)(x –3)(x + 1)
f (x) = x3 – x2 – 22x + 40
ANSWER
(x – 4)(x –2)(x +5)

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

19. EXAMPLE 3
Use synthetic division
Divide f (x)= 6x3 +7 x2 + x + 1 by 2x + 3 using synthetic division.
Bring down the first number of the column (add it to zero)
The number we brought down we will multiply it by “r” the opposite of the divisor. And we place under column 2.
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.
–3/
– –6
– –5

20. EXAMPLE 3
Use synthetic division
Divide f (x)= 6x3 +7 x2 + x + 1 by 2x + 3 using synthetic division.
SOLUTION
–3/
– –6
– –5
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.
= 6x2 –2 x + 4 +
- 5
x + 3/2

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

22. EXAMPLE 4
Use synthetic division
Divide f (x)= 4x3 +2 x2 -4 x + 3 by 2x + 3 using synthetic division.
SOLUTION
Divide f (x)= 5x 4 – 3 x3 +10 x + 2 by 5x - 3
SOLUTION

23. Write a paragraph what the remainder theorem is.
EXAMPLE 5
The Remainder Theorem
We can also solve it using synthetic division.
– –1
6 – –
Write a paragraph what the remainder theorem is.

24. 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