Warm up!!

•Pg. 130 #’s 16, 22, 36, 42, 52, 96

1) Divide 2) Factor 3) Find Zeros

Example:Divide 6x3 – 19x2 + 16x – 4 by x – 2

Answer: 6x2 – 7x + 2Which means that…

6x3 – 19x2 + 16x – 4 = (x – 2)(6x2 – 7x + 2)6x3 – 19x2 + 16x – 4 = (x – 2)(2x – 1)(3x – 2)

Zeros: x = 2 x = ½ x = ⅔

Answer: 5x + 3

Can check zeros on

calculator

Rules of Long Division

If the divisor goes into the equation evenly then it is a factor and a zero.

If the quotient has a remainder then it is not a factor.The proper form of writing the remainder is adding it

to the quotient and over the divisor.

Also written as:

F(x) = q(x) + r(x) d(x) d(x)

Improper Proper

Example:Divide x3-1 by x-1

– Write in descending powers– Insert zeros where there are missing power

Because it equals 0, x3 -1 is divisible by x-1

x2 + x + 1x – 1 √(x3 + 0x2 + 0x – 1)

x3 – x2

x2 + 0x x2 – x

x – 1 x – 1

0

Using Division Algorithm you can write it like this:

x3 – 1 = x2 + x + 1 x – 1

• Synthetic division is a shorthand, or shortcut, method of polynomial division in the special case of dividing by a linear factor. – Used when finding zeroes of polynomials.

Example: Divide x2 + 5x + 6 by x – 1

*Rational roots test determines that +/- 1, 2, 3, and 6 are possible zeros.

• Synthetic provides the same quotient but in a quicker fashion.

Let’s divide:x2 + 5x + 6 by x – 1

• If r = 0, (x – k) is a factor.• If r = 0, (k, 0) is an x intercept of f. *k is a zero• The remainder r gives the value of f at x = k, if

r = f(k) (Remainder Theorem)

To evaluate polynomial function f(x) when x=k, divide f(x) by x-k

– Remainder will equal f(k)

Example 1:Evaluate f(x) = 3x3 + 8x2 + 5x – 7 for x = -2

So f(-2) = -9, r = -9

Example 2:Evaluate f(x) = x3 – x2 – 14x + 11 for x = 4 So f(4) = 3, r = 3

Repeated Division Multi-step synthetic division…

Find all the zeros of the following polynomial function given (x – 2) and (x + 3) are factors…

f (x) = 2x4 + 7x3 – 4x2 – 27x – 18.

Solution:Using synthetic division with the factor (x – 2), you obtain the following.

0 remainder, so f (2) = 0 and (x – 2) is a factor.

Repeated Division (Continued)Take the result of this division and perform synthetic division again using the

factor (x + 3).

Because the resulting quadratic expression factors as2x2 + 5x + 3 = (2x + 3)(x + 1)

the complete factorization of f (x) is f (x) = (x – 2)(x + 3)(2x + 3)(x + 1).

0 remainder, so f (-3) = 0 and (x +3 ) is a factor.

Try #60 on page 141

Example 1:Show that (x – 2) and (x + 3) are factors and find all the zeros off(x) = 2x4 + 7x3 – 4x2 – 27x – 18

x = 2, -3, -3/2, -1

Example 2:Show that (x + 2) and (x – 1) are factors and find all the zeros off(x) = 2x3 + x2 – 5x + 2

x = ½, -2, 1

• Graph should confirm zeros

Page 140 #’s 5 – 35 (5’s), 47, 51, 55, 59,

61, 69, 73