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Pearson Algebra 2 Common Core, 2011 View details

Chapter Review

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Exercise 26 Page 349

How can factoring out the GCF help you apply the Zero Product Property?

0, - 1 ± sqrt(37)/2

Practice makes perfect

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.

3x^3+3x^2=27x

SubEqn

LHS-27x=RHS-27x

3x^3+3x^2-27x=0

FactorOut

Factor out 3x

3x( x^2+x-9 ) =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.

3x( x^2+x-9 ) =0

ZeroProdProp

Use the Zero Product Property

lc3x=0 & (I) x^2+x-9=0 & (II)

DivEqn

(I): .LHS /3.=.RHS /3.

lx=0 x^2+x-9=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. ax^2+bx+c=0 ⇔ x=- b±sqrt(b^2-4ac)/2a To do so, we first need to identify a, b, and c. x^2+x-9=0 ⇔ 1x^2+ 1x+( - 9)=0 We see that a= 1, b= 1, and c= - 9. Let's substitute these values into the formula and solve for x.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 1±sqrt(1^2-4( 1)( - 9))/2( 1)

▼

Solve for x

CalcPow

Calculate power

x=- 1±sqrt(1-4(1)(- 9))/2(1)

MultNegPos

(- a)b = - ab

x=- 1±sqrt(1-4(- 9))/2(1)