Solving Polynomial Equations

a

We have a product that is equal to $0 .$ This means that any of the factors must be equal to $0$ and that we can use the Zero Product Property to solve the equation.

x (x + 4) (x - 94) = 0

ZeroProdPropUse the Zero Product Property

x = 0 x + 4 = 0 x - 94 = 0 (I) (II) (III)

SubEqn

(II):

LHS - 4 = RHS - 4

x = 0 x = - 4 x - 94 = 0

AddEqn

(III):

LHS + 94 = RHS + 94

x_{1} = 0 x_{2} = - 4 x_{3} = 94

b

We use the Zero Product Property again.

(x + 73) (x - 54) (x + 7) = 0

ZeroProdPropUse the Zero Product Property

x + 73 = 0 x - 54 = 0 x + 7 = 0 (I) (II) (III)

SubEqn

(I):

LHS - 73 = RHS - 73

x = - 73 x - 54 = 0 x + 7 = 0

AddEqn

(II):

LHS + 54 = RHS + 54

x = - 73 x = 54 x + 7 = 0

SubEqn

(III):

LHS - 7 = RHS - 7

x_{1} = - 73 x_{2} = 54 x_{3} = - 7

c

Nothing new. We proceed in the same manner.

(x - 11) (x + 5) (x - 1056) (x + 68) = 0

ZeroProdPropUse the Zero Product Property

x - 11 = 0 x + 5 = 0 x - 1056 = 0 x + 68 = 0 (I) (II) (III) (IV)

AddEqn

(I):

LHS + 11 = RHS + 11

x = 11 x + 5 = 0 x - 1056 = 0 x + 68 = 0

SubEqn

(II):

LHS - 5 = RHS - 5

x = 11 x = - 5 x - 1056 = 0 x + 68 = 0

AddEqn

(III):

LHS + 1056 = RHS + 1056

x = 11 x = - 5 x = 1056 x + 68 = 0

SubEqn

(IV):

LHS - 68 = RHS - 68

x_{1} = 11 x_{2} = - 5 x_{3} = 1056 x_{4} = - 68

Solving Polynomial Equations 1.12 - Solution