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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

Exercise 5 Page 355

Make sure you rewrite the equation leaving all the terms on one side, and that you factor out the GCF if it exists.

C

Practice makes perfect

We want to solve the given equation by factoring.

Factoring

Since we are finding the zeros, we can set the function equal to zero.

x^2+2x-15=0

Write as a sum

x^2+5x-3x-15=0

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Factor out x & (-3)

Factor out x

x(x+5)-3x-15=0

Factor out (- 3)

x(x+5)-3(x+5)=0

Factor out (x-3)

(x-3)(x+5)=0

Solving

To solve this equation, we will apply the Zero Product Property.

(x-3)(x+5)=0

Use the Zero Product Property

lcx-3=0 & (I) x+5=0 & (II)

(I): LHS+3=RHS+3

lx= 3 x+5=0

(II): LHS-5=RHS-5

lx_1=3 x_2=- 5

These zeros can be written as the points (3,0) and (-5,0). The point (-5,0) corresponds to option C.

Checking Our Answer

Checking our answer

We can substitute our solution back into the given equation and simplify to check if our answers are correct. We will start with x=-5.

x^2+2x-15=0

x= - 5

( -5)^2+2( -5)-15? =0

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Simplify

Calculate power

25+2(-5)-15? =

a(- b)=- a * b

25-10-15? =0

Subtract term

0=0 ✓

Again, we created a true statement. x=- 5 is indeed a solution of the equation.