Solving Polynomial Equations

a

We have a product that is equal to $0 .$ It means that we can use the Zero Product Property.

(x - \frac{2}{3}) (4.67 + x) (x - 231) = 0

ZeroProdPropUse the Zero Product Property

x - 2 / 3 = 0 4.67 + x = 0 x - 231 = 0 (I) (II) (III)

(I):

LHS + 2 / 3 = RHS + 2 / 3

x = 2 / 3 4.67 + x = 0 x - 231 = 0

(II):

LHS - 4.67 = RHS - 4.67

x = 2 / 3 x = - 4.67 x - 231 = 0

(III):

LHS + 231 = RHS + 231

x_{1} = 2 / 3 x_{2} = - 4.67 x_{3} = 231

b

We proceed in the same manner and use the zero product property.

(\frac{4}{7} + x) (x - \frac{5}{9}) (x - \frac{6}{1 3}) = 0

ZeroProdPropUse the Zero Product Property

4 / 7 + x = 0 x - 5 / 9 = 0 x - 6 / 13 = 0 (I) (II) (III)

(I):

LHS - 4 / 7 = RHS - 4 / 7

x = - 4 / 7 x - 5 / 9 = 0 x - 6 / 13 = 0

(II):

LHS + 5 / 9 = RHS + 5 / 9

x = - 4 / 7 x = 5 / 9 x - 6 / 13 = 0

(III):

LHS + 6 / 13 = RHS + 6 / 13

x_{1} = - 4 / 7 x_{2} = 5 / 9 x_{3} = 6 / 13

c

We proceed in the same manner.

(x - \frac{1 9}{3}) (x + \frac{2 3}{4 9}) (x - \frac{7}{3}) = 0

ZeroProdPropUse the Zero Product Property

x - 19 / 3 = 0 x + 23 / 49 = 0 x - 7 / 3 = 0 (I) (II) (III)

(I):

LHS + 19 / 3 = RHS + 19 / 3

x = 19 / 3 x + 23 / 49 = 0 x - 7 / 3 = 0

(II):

LHS - 23 / 49 = RHS - 23 / 49

x = 19 / 3 x = - 23 / 49 x - 7 / 3 = 0

(III):

LHS + 7 / 3 = RHS + 7 / 3

x_{1} = 19 / 3 x_{2} = - 23 / 49 x_{3} = 7 / 3

Solving Polynomial Equations 1.15 - Solution