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Chapter 4 Quadratic and Polynomial Equations
4.10
MATHPOWERTM 11, WESTERN EDITION 4.10.1
Solving Polynomial Equations
Solve by factoring:
x3 - x2 - 4x + 4 = 0
x2(x - 1) - 4(x - 1) = 0
(x - 1)(x2 - 4) = 0
(x - 1)(x - 2)(x + 2) = 0
Therefore the roots are:
x = 1, 2, -2
Solve: x3 - 7x + 6 = 0
Potential zeros: ±1, ±2, ±3, ±6
P(1) = (1)3 - 7(1) + 6
=0
Therefore, x - 1 is a factor.
-1
1
1
0
-7
6
-1
-1
6
1
-6
0
Factor x2 + x - 6:
(x + 3)(x - 2)
Therefore, the equation in factored form
is (x - 1)(x + 3)(x - 2) = 0.
The roots are x = 1, -3, 2.
4.10.2
Solving Polynomial Equations
Irrational Zeros
Solve x3 - 4x + 3 = 0.
Potential zeros: ±1, ±3
P(1) = (1)3 - 4(1) + 3
=0
Therefore, x - 1 is a factor.
-1
3
3
b  b2  4ac
x
2a
0
1  (1) 2  4(1)(3)
x
2(1)
1
1
0
-1
1
-4
-1
-3
Note: x2 + x - 3
does not factor.
Use the quadratic
formula.
1  13
x
2
Therefore, the zeros are:
1, -1± √13
2
4.10.3