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Pearson Algebra 2 Common Core, 2011

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Pearson Algebra 2 Common Core, 2011 View details

Cumulative Standards Review

Finish the assignment

Exercise 26 Page 494

Recall the formula a^2± 2ab+b^2=(a± b)^2.

- 4

Practice makes perfect

We want to solve the given equation by factoring. We will have to factor a perfect square trinomial. a^2± 2ab+b^2 ⇔ (a± b)^2

Factoring

Before we begin, let's rewrite the equation with all the non-zero terms on the left-hand side.

- 16 = x^2+8x

LHS+16=RHS+16

0=x^2+8x+16

Rearrange equation

x^2+8x+16=0

Since all the terms are on the left-hand side, we are ready to start factoring.

x^2+8x+16=0

FacPosPerfectSquare

a^2+2ab+b^2=(a+b)^2

(x+4)^2=0

SplitIntoFactors

Split into factors

(x+4)(x+4)=0

Solving

Now let's apply the Zero Product Property to solve.

(x+4)(x+4)=0

Use the Zero Product Property

lcx+4=0 & (I) x+4=0 & (II)

(I), (II): LHS-4=RHS-4

lx_1=- 4 x_2=- 4

The only solution to this equation is x=- 4.

Checking Our Answer

Checking the answer

To check our answer, we will substitute - 4 for x in the given equation.

x^2+8x+16=0

x= - 4

( - 4)^2+8( - 4)+16? =0

▼

Simplify left-hand side

NegBaseToPosBase

(- a)^2=a^2

4^2+8(- 4)+16? =0

Calculate power

16+8(- 4)+16? =0

a(- b)=- a * b

16-32+16? =0

Add and subtract terms

0=0 ✓

Since substituting and simplifying resulted in a true statement, we know that x=- 4 is a solution of the equation.

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