long division just like in the 3 rd grade 6x 3 + 19x 2 + 16x – 4 divided by x - 2 x - 2 6x 3 + 19x...

Long Division

• Just like in the 3rd grade

6x3 + 19x2 + 16x – 4 divided by x - 2

x - 2 6x3 + 19x2 + 16x - 4

U-Try

(4x3 -7x2 – 11x + 5) divided by (4x + 5)

Synthetic division

to divide ax3 + bx2 + cx +d by (x – k)

k a b c d

a

ka

b-ka remainder

Coefficients of Quotient

k(b-ka)

Lets try one x4 – 10x2 – 2x + 4 divided by ( x + 3)

( x + 3) k = -3

-3

x4 + 0x3 – 10x2 – 2x + 4

1

1 0 –10 – 2 4

-3

-3

9

-1

3

1

-3

1

-3

x4 + 0x3 – 10x2 – 2x + 4

1

1 0 –10 – 2 4

-3

-3

9

-1

3

1

-3

1remainder

__x3 + __x2 + __x + __+

__

(x + 3)

You try

• 3x3 -17x2 + 15x -25 divided by (x - 5)

Pretty Cool Remainder Theorem

• If a polynomial is divided by (x - k) then the remainder will be f(k)

Or the PCRT

Let’s try one

• Find the remainder of the problem

• 9x3 – 16x – 18x2 + 32 divided by (x – 2)

f(x) = 9x3 – 16x – 18x2 + 32

f(2) = 9(2)3 – 16(2) – 18(2)2 + 32

f(2) = 9(8) – 16(2) – 18(4) + 32

Is it a root?

• If you try synthetic division and there is no remainder, that means k is a solution to f(x) = 0 or … f(k) = 0

A great way to test for roots of higher degree polynomials

The most confusing instructions for any homework problem I’ve ever

seen!

What they want

Take f(x) and divide it by (x - k)

Then write (x - k) (quotient) + remainder

(x-k)

Page 233 problems 39 -46

Divide then divide again to factor 3rd degree polynomials given 2

factors

Lets try a few problems

• Page 235 problems

• 7 - 15

• 21 - 27

• 51 - 65