Polynomial Factors

Polynomial Factor Theorem

Polynomial Factors 27/23/2013

The Remainder Theorem

Polynomial f(x) divided by x – k

yields a remainder of f(k)

Question:

Polynomial Factor Theorem

= (x – k)Q(x)

… leading to …

What if the remainder is 0 ?

Then we know that f(x) = (x – k)Q(x) + r(x)

and thus (x – k) is a factor of f(x)

Polynomial Factors 37/23/2013

The Factor Theorem

Polynomial f(x) has factor x – k

if and only if

f(k) = 0 Factor Theorem Corollary

If (x – k) is a factor of f(x) = anxn + an–1xn–1 + • • • + a1x + a0

then k is a factor of a0

Polynomial Factor Theorem

Question: What about the converse ?

Polynomial Factors 47/23/2013

Example: f(x) = x4 + x3 – 19x2 + 11x + 30

6 42 138 894

Polynomial Factor Theorem

1 1 –19 11 30 2

1

2

3

6

–13

–26

–15

–30

0

(x – 2) is a factor of f(x)

Try factor k = 6

... and 2 is a factor of 30

1 1 –19 11 30

1 7 23 149 924

(x – 6) NOT a factor of f(x)

... BUT 6 IS a factor of 30

Try factor k = 2

Question:

What if k = ±1, –2, ±3, ±5, –6, ±10, ±15 ?

What about other factors of 30 ?

≠ 0

6

Polynomial Factors 57/23/2013

Factor Theorem Example Given the graph of

polynomial f(x)

Estimate the degree of f(x) Even or odd degree ?

Polynomial Factor Theorem

x

y

● ● ●(–7, 0) (2, 0) (11, 0)

Odd degree ! WHY ?

Polynomial Factors 67/23/2013

Factor Theorem Example Given the graph of polynomial f(x) Find the factors of f(x)

Since f(–7) = f(2) = f(11) = 0

… then factors are

(x + 7), (x – 2), (x – 11)

Note:

(x + 7)(x – 2)(x – 11) = x3 – 6x2 – 69x + 154

Polynomial Factor Theorem

x

y

● ● ●(–7, 0) (2, 0) (11, 0)

Polynomial Factors 77/23/2013

Factor Theorem Example The graph of polynomial f(x)

Question:

Is f(x) = (x + 7)(x – 2)(x – 11) ?

Polynomial Factor Theorem

x

y

● ● ●(–7, 0) (2, 0) (11, 0)

Not necessarily !f(x) = (x + 7)Q1(x) = (x + 7)(x – 2)Q2(x)

= (x + 7)(x – 2)(x – 1)Q3(x)

Polynomial Factors 87/23/2013

Factor Theorem Example The graph of polynomial f(x)

Question:

Polynomial Factor Theorem

x

y

● ● ●(–7, 0) (2, 0) (11, 0)

f(x) = (x + 7)(x – 2)(x – 1)Q3(x)

What is Q3(x) ?

Is Q3(x) constant ?

Does Q3(x) have factors x – k ?

Polynomial Factors 97/23/2013

Factor Theorem Example Given the graph of polynomial f(x)

estimate the degree of f(x) Even or odd degree ?

Find the factors of f(x)

Polynomial Factor Theorem

x

y

(–3, 0) (5, 0) ● ●

Note: (x + 3)(x – 5) = x2 – 2x2 – 15

f(–3) = f(5) = 0so (x + 3) and (x – 5) are factors

WHY !

Polynomial Factors 107/23/2013

Factor Theorem Example Given the graph of f(x)

Polynomial Factor Theorem

x

y

(–3, 0) (5, 0) ● ●

Is f(x) equal to x2 – 2x – 15 ? Probably not !

Question:

(x + 3)(x – 5) = x2 – 2x2 – 15

What is the graph of x2 – 2x – 15 ?Note: f(x) = (x + 3)Q1(x) = (x + 3)(x – 5)Q2(x)

Which ones and how many ?

Does Q2(x) have factors (x – k) ?

Polynomial Factors 117/23/2013

These things are related

If f(k) = 0, then point (k, 0) on the graph of f is an x-intercept

The number k is a zero for f(x), i.e. f(k) = 0

(x – k) is a factor of f(x)

The number k is a factor of the constant term of f(x)

Intercepts, Zeros and Factors

WHY ?

Polynomial Factors 127/23/2013

Example: Completely Factor

f(x) = 2x4 + 14x3 + 18x2 – 54x – 108

Now use synthetic division to check out zeros

Completely Factored Polynomials

= 2(x4 + 7x3 + 9x2 – 27x – 54)

Polynomial Factors 137/23/2013

Completely Factored Polynomials

f(x) = 2(x4 + 7x3 + 9x2 – 27x – 54)

1

–3

4

–12 9

–18

54

0

1 7 9 –27 –54 –3

–3

x – k = x – (–3)

(x + 3) is a factor

1–3

1 –3 18

0

1 4 –3 –18

–6 (x + 3) is a second factor

Q1 –3

Polynomial Factors 147/23/2013

Completely Factored Polynomials

f(x) = 2(x4 + 7x3 + 9x2 – 27x – 54)1 7 9 –27 –54

0 1 4 –18 –3 (x + 3) is a factor

1 1 0 –6 (x + 3) is a second factor

1–3 –2

6 0

1 1 –6

(x + 3) is a third factor

(x – 2) is a fourth factor

2(x + 3)3(x – 2)

Note: x = –3 is a repeated zero of multiplicity 3

Q2 –3

Polynomial Factors 157/23/2013

Completely Factored Polynomials

f(x) = 2(x4 + 7x3 + 9x2 – 27x – 54)

1

–3

4

–12 9

–18

54

0

1 7 9 –27 –54 –3

–3

x – k = x – (–3)

(x + 3) is a factor

1–3

1 –3 18

0

1 4 –3 –18

–6 (x + 3) is a second factor

1–3 –2

6 0

1 1 –6

(x + 3) is a third factor

(x – 2) is a fourth factor

2(x + 3)3(x – 2)

Note:x = –3 is a repeated zero of multiplicity 3

Q1

Q2

–3

–3

Polynomial Factors 167/23/2013

Complete Factoring with Multiple Zeros General Form

f(x) = anxn + an–1 xn–1 + ... + a1x + a0

= an(x – kn)(x – kn–1 ) ... (x – k1)

Some of the zeros ki may be repeated

Polynomial Functions

Polynomial Factors 177/23/2013

Complete Factoring with Multiple Zeros

f(x) = an(x – kn)(x – kn–1 ) ... (x – k1)

Degree of f(x) = n

Number of zeros is m, m ≤ n

Real zeros occur at x-intercepts

Counting multiplicities, total number of zeros is n

Polynomial Functions

Polynomial Factors 187/23/2013

Complete Factoring with Multiple Zeros Example

f(x) = 2(x + 3)3(x – 2)

deg f(x) = 4

One zero at 2, one at –3 of multiplicity 3

Total zeros, counting multiplicities,

1 + 3 = 4 = deg f(x)

Polynomial Functions

Polynomial Factors 197/23/2013

Complete Factoring Examples 1. f(x) = 5x3 – 10x2 – 15x

= 5x(x2 – 2x – 3)

= 5x(x – 3)(x + 1)

= 5(x + 1)(x – 0)(x – 3)

Polynomial Functions

Zeros of f(x) are –1 , 0 , 3

Polynomial Factors 207/23/2013

Complete Factoring Examples 2. Given:

f(x) is a quadratic polynomial lead coefficient is 7 f(–3) = 0 and f(2) = 0

Write f(x) in completely factored form

Polynomial Functions

Polynomial Factors 217/23/2013

Complete Factoring Examples Write f(x) in completely factored form

Note that –3 and 2 are zeros of f(x) From the Factor Theorem

x –(–3) and x – 2 are factors of f(x) Thus

f(x) = 7(x + 3)(x – 2)

Polynomial Functions

Polynomial Factors 227/23/2013

Even/Odd Multiplicity Examples

Polynomial Functions

x

y(x)

x

y(x)

x

y(x)

x

y(x)

x

y(x)

●y = (x – 3)2

●y = x + 3

●y = (x – 3)3

●y = (x – 3)4

●y = (x – 3)5

●●

y = (x + 3)3(x – 3)

x

y(x)

y = (x + 2)3(x – 3)2

● ●

Polynomial Factors 237/23/2013

Think about it !