body{margin:0;} noscript{z-index: 10000; position: fixed; display: block; height: 100%; width: 100%; background: #fff;} noscript .fa{font-size: 40px; display:block; color: #daa520;} noscript div{margin-top: 6%; padding-left: 20px; padding-right: 20px; text-align: center;} ! You must have JavaScript enabled to use this site.

McGraw Hill Glencoe Algebra 2, 2012

MH

McGraw Hill Glencoe Algebra 2, 2012 View details

Preparing for Standardized Tests

Finish the assignment

Continue to next subchapter

Exercise 1 Page 297

We want to solve the given equation by factoring.

Factoring

Let's start by writing all the terms on one side of the equals sign. We will also factor out a GCF, if we find one.

2x^2 + 10x - 48 = 0

.LHS /2.=.RHS /2.

x^2 + 5x - 24 = 0

Rewrite 5x as 8x - 3x

x^2 + 8x - 3x - 24 = 0

▼

Factor out x & -3

Factor out x

x(x+8)-3x - 24=0

Factor out -3

x(x+8)-3(x+8)=0

Factor out (x+8)

(x-3)(x+8)=0

Solving

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

(x-3)(x+8)=0

Use the Zero Product Property

lcx-3=0 & (I) x+8=0 & (II)

(I): LHS+3=RHS+3

lx=3 x+8=0

(II): LHS-8=RHS-8

lx_1=3 x_2=-8

Checking Our Answer

Checking our answer

We can substitute our solutions back into the given equation and simplify to check if our answers are correct. We will start with x=3.

2x^2+10x - 48 = 0

x= 3

2( 3)^2+10( 3)- 48 ? =0

▼

Simplify

Calculate power

18+10(3) - 48 ? = 0

Multiply

18 + 30 - 48? =0

Add terms

48 - 48 ? =0

Subtract term

0 = 0 ✓

Substituting and simplifying created a true statement, so we know that x=3 is a solution of the equation. Let's move on to x=- 8.

2x^2 + 10x - 48 = 0

x= - 8

2( -8)^2+10( -8) - 48? = 0

▼

Simplify

NegBaseToPosBase

(- a)^2=a^2

2( 8)^2+10( -8) - 48? = 0

Calculate power

128+10( -8) - 48? = 0

a(- b)=- a * b

128 - 80 - 48 ? = 0

Subtract terms

0 = 0 ✓

Again, we created a true statement. x=- 8 is indeed a solution of the equation.

Subchapter links

Exercises p.297