Solving Polynomial Equations

To solve the given equation, we will start by rewriting it in standard form.

x^{3} + 1 = x^{2} + x

Put in Standard Form

LHS - x^{2} = RHS - x^{2}

x^{3} + 1 - x^{2} = x

LHS - x = RHS - x

x^{3} + 1 - x^{2} - x = 0

RearrangeEqnRearrange equation

x^{3} - x^{2} - x + 1 = 0

Now we will factor out a common factor from each group and apply the Zero Product Property. Let's do it!

x^{3} - x^{2} - x + 1 = 0

Factor

FactorOutFactor out

x^{2}

x^{2} (x - 1) - x + 1 = 0

FactorOutFactor out

- 1

x^{2} (x - 1) - (x - 1) = 0

FactorOutFactor out

(x - 1)

(x - 1) (x^{2} - 1) = 0

WritePowWrite as a power

(x - 1) (x^{2} - 1^{2}) = 0

(a^{2} - b^{2}) = (a + b) (a - b)

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

CommutativePropMultCommutative Property of Multiplication

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

(x - 1)^{2} (x + 1) = 0

Solve using the Zero Product Property

ZeroProdPropUse the Zero Product Property

(x - 1)^{2} = 0 x + 1 = 0 (I) (II)

(I):

LHS = RHS

x - 1 = 0 x + 1 = 0

(I):

LHS + 1 = RHS + 1

x = 1 x + 1 = 0

(II):

LHS - 1 = RHS - 1

x = 1 x = - 1

There are two solutions for the given equation,

x = 1

and

x = - 1 .

Solving Polynomial Equations 1.17 - Solution