g(x)=x+6×2+2x-35

Set x+6×2+2x-35 equal to 0.

x+6×2+2x-35=0

Factor x2+2x-35 using the AC method.

Consider the form x2+bx+c. Find a pair of integers whose product is c and whose sum is b. In this case, whose product is -35 and whose sum is 2.

-5,7

Write the factored form using these integers.

x+6(x-5)(x+7)=0

x+6(x-5)(x+7)=0

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

(x-5)(x+7),1

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

The number 1 is not a prime number because it only has one positive factor, which is itself.

Not prime

The LCM of 1,1 is the result of multiplying all prime factors the greatest number of times they occur in either number.

1

The factor for x-5 is x-5 itself.

(x-5)=x-5

(x-5) occurs 1 time.

The factor for x+7 is x+7 itself.

(x+7)=x+7

(x+7) occurs 1 time.

The LCM of x-5,x+7 is the result of multiplying all factors the greatest number of times they occur in either term.

(x-5)(x+7)

(x-5)(x+7)

Multiply each term by (x-5)(x+7) and simplify.

Multiply each term in x+6(x-5)(x+7)=0 by (x-5)(x+7) in order to remove all the denominators from the equation.

x+6(x-5)(x+7)⋅((x-5)(x+7))=0⋅((x-5)(x+7))

Cancel the common factor of (x-5)(x+7).

Cancel the common factor.

x+6(x-5)(x+7)⋅((x-5)(x+7))=0⋅((x-5)(x+7))

Rewrite the expression.

x+6=0⋅((x-5)(x+7))

x+6=0⋅((x-5)(x+7))

Simplify 0⋅((x-5)(x+7)).

Expand (x-5)(x+7) using the FOIL Method.

Apply the distributive property.

x+6=0⋅(x(x+7)-5(x+7))

Apply the distributive property.

x+6=0⋅(x⋅x+x⋅7-5(x+7))

Apply the distributive property.

x+6=0⋅(x⋅x+x⋅7-5x-5⋅7)

x+6=0⋅(x⋅x+x⋅7-5x-5⋅7)

Simplify and combine like terms.

Simplify each term.

Multiply x by x.

x+6=0⋅(x2+x⋅7-5x-5⋅7)

Move 7 to the left of x.

x+6=0⋅(x2+7⋅x-5x-5⋅7)

Multiply -5 by 7.

x+6=0⋅(x2+7x-5x-35)

x+6=0⋅(x2+7x-5x-35)

Subtract 5x from 7x.

x+6=0⋅(x2+2x-35)

x+6=0⋅(x2+2x-35)

Multiply 0 by x2+2x-35.

x+6=0

x+6=0

x+6=0

Subtract 6 from both sides of the equation.

x=-6

x=-6

Solve for g g(x)=(x+6)/(x^2+2x-35)