Solving Polynomial Equations

a

We use the Zero Product Property. This means that one of the factors must be equal to $0 .$ It gives us three sub-equations.

(x + 3) (x - 6) (x + 1) = 0

ZeroProdPropUse the Zero Product Property

x + 3 = 0 x - 6 = 0 x + 1 = 0 (I) (II) (III)

(I):

LHS - 3 = RHS - 3

x = - 3 x - 6 = 0 x + 1 = 0

(II):

LHS + 6 = RHS + 6

x = - 3 x = 6 x + 1 = 0

(III):

LHS - 1 = RHS - 1

x_{1} = - 3 x_{2} = 6 x_{3} = - 1

b

We use the zero product property here as well. There are four factors. In this case, it means that there are four solutions.

(4 - x) (x + 8) (2 + x) (x - 14) = 0

ZeroProdPropUse the Zero Product Property

4 - x = 0 x + 8 = 0 2 + x = 0 x - 14 = 0 (I) (II) (III) (IV)

(I):

LHS - 4 = RHS - 4

- x = - 4 x + 8 = 0 2 + x = 0 x - 14 = 0

(I):

Change signs

x = 4 x + 8 = 0 2 + x = 0 x - 14 = 0

(II):

LHS - 8 = RHS - 8

x = 4 x = - 8 2 + x = 0 x - 14 = 0

(III):

LHS - 2 = RHS - 2

x = 4 x = - 8 x = - 2 x - 14 = 0

(IV):

LHS + 14 = RHS + 14

x_{1} = 4 x_{2} = - 8 x_{3} = - 2 x_{4} = 14

c

Nothing new. We proceed in the same manner.

x (6 + x) (x - 9) (x + 17) = 0

ZeroProdPropUse the Zero Product Property

x = 0 6 + x = 0 x - 9 = 0 x + 17 = 0 (I) (II) (III) (IV)

(II):

LHS - 6 = RHS - 6

x = 0 x = - 6 x - 9 = 0 x + 17 = 0

(III):

LHS + 9 = RHS + 9

x = 0 x = - 6 x = 9 x + 17 = 0

(IV):

LHS - 17 = RHS - 17

x_{1} = 0 x_{2} = - 6 x_{3} = 9 x_{4} = - 17

Solving Polynomial Equations 1.2 - Solution