Solving Polynomial Equations by FactoringSolving Polynomial Equations by FactoringFactoring by grouping

Ex. 1. Solve:3 25 4 20 0x x x

3 2( 5 ) (4 20) 0x x x

2 5 4 5 0x x x 2 4 5 0x x

2 2 5 0x x x

2, 2, 5x x x

Variable substitutionAnother method of solving higher-degree polynomial equations involves recognizing polynomials that have quadratic form.

Ex. 2. Solve 4 22 3 0x x 2Let y x 22 22 3 0x x

22 3 0y y

2 3 1 0y y 3

2y 2 3

2x

3 6

22x

or 1y 2 1x

x i

The Rational Root Theorem

Let P(x) be a polynomial of degree n with integral coefficients and a nonzero constant term:

0

11 0, 0...n n

n n where aP x a x a x a

If one of the roots of the equation P(x) = 0 is where

P and q are nonzero integers with no common factor other than 1, then p must be a factor of and q must be a factor of

px

q

0ana

Ex. 3.

a. According to the rational root theorem, what are the possible rational roots of 4 3 23 13 15 4 0P x x x x

px

q is a possible rational root if p is a factor of –4 and q is a

factor of 3.

1, 2, 4

1, 3

p

q

1 2 41, 2, 4, , ,

3 3 3or

b. Determine whether any of the possible rational roots really are roots. Then find all other roots, real or imaginary.

Try synthetic substitution

Or substitute in to see if any P(x) = 0

4 3 21 3 1 13 1 15 1 4 27P

4 3 21 3 1 13 1 15 1 4 1P

4 3 22 3 2 13 2 15 2 4 208P

4 3 22 3 2 13 2 15 2 4 0P

Use synthetic division 2 3 13 15 0 4

3

6

7

14

1

22

4

0

3 22 3 7 2x x x x

Find the other roots of 3 23 7 2P x x x x

Check x = -2 again because it may be a double root

2 3 7 1 2

36

1

2

1

2

0

1 1 4 3 1 1 13

6 6x

23 1 0x x

1 132( ),

6So x a double root and x