Exercise 35 Page 174

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

x=- 2/5 and x= 6

Practice makes perfect

We want to solve the given equation by factoring.

Factoring

Let's start by factoring out a GCF, if we find one.

15x^2-84x-36=0

FactorOut

Factor out 3

3(5x^2-28x-12)=0

WriteSum

Write as a sum

3(5x^2-30x+2x-12)=0

▼

Factor out x & 4

FactorOut

Factor out 5x

3 ( 5x(x-6)+2x-12 )=0

FactorOut

Factor out 2

3 ( 5x(x-6)+2(x-6) )=0

FactorOut

Factor out (x-6)

3(5x+2)(x-6)=0

Solving

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

3(5x+2)(x-6)=0

DivEqn

.LHS /3.=.RHS /3.

(5x+2)(x-6)=0

ZeroProdProp

Use the Zero Product Property

lc5x+2=0 & (I) x-6=0 & (II)

SubEqn

(I): LHS-2=RHS-2

l5x=- 2 x-6=0

DivEqn

(I): .LHS /5.=.RHS /5.

lx=- 25 x-6=0

AddEqn

(II): LHS+6=RHS+6

lx_1=- 25 x_2= 6

Subchapter links

Exercises p.174-177

100

101

102

103

Exercise 35 Page 174

Hints

Check the answer

Practice exercises

Asses your skill

Factoring

Solving