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McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
Study Guide and Review
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Exercise 44 Page 82

Use the Zero Product Property.

0, 25/3

Practice makes perfect

To solve the given equation, let's factor and solve using the Zero Product Property. At the end, we will check that our answer is correct.

Factoring

To begin, let's rewrite the equation with all the non-zero terms on the left-hand side. After that, we will proceed to factoring.
3x^2=5x
SubEqn

LHS-5x=RHS-5x

3x^2-5x=0
FactorOut

Factor out x

x(3x-5)=0

Solving

Now we can apply the Zero Product Property to solve.
x(3x-5)=0
ZeroProdProp

Use the Zero Product Property

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

(II): LHS+5=RHS+5

lx=0 3x=5
DivEqn

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

lx_1=0 x_2=5/3
The solutions for our equation are 0 and 53.

Checking

To check our answers, we will substitute them for x in the given equation. Let's start with x=0.
3x^2=5x
Substitute

x= 0

3( 0)^2? =5( 0)
CalcPow

Calculate power

3(0)? =5(0)
ZeroPropMult

Zero Property of Multiplication

0=0 âś“
Since substituting and solving resulted in a true statement, we know that x=0 is a solution of the equation. Let's now check x= 53.
3x^2=5x
Substitute

x= 5/3

3( 5/3)^2? =5( 5/3)
PowQuot

(a/b)^m=a^m/b^m

3(25/9)? =5(5/3)
MoveLeftFacToNum

a*b/c= a* b/c

3* 25/9? =5* 5/3
SplitIntoFactors

Split into factors

3* 25/3* 3? =5* 5/3
CancelCommonFac

Cancel out common factors

25/3? =5* 5/3
Multiply

Multiply

25/3=25/3 âś“
This is also a true statement, so we know that x= 253 is a solution of the equation.
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11
Adding and Subtracting Polynomials 12 (Page 81)
Adding and Subtracting Polynomials 13 (Page 81)
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Multiplying a Polynomial by a Monomial 19 (Page 81)
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arrow_right 4 Special Products p.82
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arrow_right 5 Using the Distributive Property p.82
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