2.3Polynomial Division

andSynthetic Division

Ex. Long Division

4161962 23 −+−− xxxx

What times xequals 6x3?6x2

6x3 - 12x2

Change the signsand add.- +

- 7x2 + 16x - 4

- 7x

- 7x2 + 14x+-

2x - 4

+ 2

2x - 4

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

Dividend Divisor Quotient Remainder

6x3 – 19x2 + 16x – 4 = (x – 2) (6x2 – 7x + 2) + 0

Divide x3 – 1 by x - 1

1001 23 −++− xxxxx2

x3 - x2

x2 + 0x - 1

+ x

x2 - xx - 1

+ 1

x - 1

- +

- +

Synthetic Division

Use synthetic division to divide x4 – 10x2 – 2x + 4 by x + 3. First, write the coef’s.of the dividend.

Put zeros in for missing terms.

1 0 -10 -2 4

-3

Bring down the1, mult. then add diagonally.

1 -3 -1 1 1remainder

quotient

x3 - 3x2 - x + 13

1

++x

Remainder Theorem: If a polynomial f(x) is divided by x – k, then the remainder isr = f(k)

Use the remainder theorem to find f(-2) if f(x) = 3x3 + 8x2 + 5x - 7.

3 8 5 -7

-2 3 2 1 -9

f(-2) = -9

This means that (-2, -9) is a point on the graph of f.

Factor Theorem: A polynomial f(x) has a factor(x – k) if and only if f(k) = 0.

Show that (x – 2) and (x + 3) are factors of f(x) = 2x4 + 7x3 – 4x2 – 27x - 18

2 7 -4 -27 -18

2 2 11 18 9 0 f(2) = 0

-3 2 5 3 0

(x – 2)(x + 3)(2x2 + 5x + 3)

f(-3) = 0