Solving Polynomial Equations

To solve the given equation by factoring, we will start by writing all the terms on the left-hand side. Then, we will factor out the GCF.

2 x^{4} = 4 x^{3} - 2 x^{2}

LHS - 4 x^{3} = RHS - 4 x^{3}

2 x^{4} - 4 x^{3} = - 2 x^{2}

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

2 x^{4} - 4 x^{3} + 2 x^{2} = 0

FactorOutFactor out

2 x^{2}

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

We have rewritten the left-hand side as a product of two factors. Now, we will apply the Zero Product Property to solve the equation.

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

ZeroProdPropUse the Zero Product Property

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

(I):

LHS / 2 = RHS / 2

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

(I):

LHS = RHS

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

From Equation (I), we found that one solution is

x = 0 .

To find other solutions, we will solve Equation (II). Note that this is a quadratic equation. Thus, we will use the Quadratic Formula.

a x^{2} + b x + c = 0 \Leftrightarrow x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

To do so, we first need to identify

a,

b,

and

c .

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

We see that

a = 1,

b = - 2,

and

c = 1 .

Let's substitute these values into the formula and solve for

x .

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

SubstituteValuesSubstitute values

x = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( 1 ) ( 1 )}{2 ( 1 )}

Solve for

x

- (- a) = a

x = \frac{2 \pm ( - 2 ) ^{2} - 4 ( 1 ) ( 1 )}{2 ( 1 )}

CalcPowCalculate power

x = \frac{2 \pm 4 - 4 ( 1 ) ( 1 )}{2 ( 1 )}

x = \frac{2 \pm 4 - 4}{2}

SubTermSubtract term

x = \frac{2 \pm 0}{2}

CalcRootCalculate root

x = \frac{2 \pm 0}{2}

AddSubTermsAdd and subtract terms

x = \frac{2}{2}

\frac{a}{a} = 1

x = 1

This solution to the quadratic equation is also solution for the given equation.

\underline{Solutions} x = 0, x = 1

Solving Polynomial Equations 1.1 - Solution