(3n+1)2+6(3n+1)-7=0

Let u=3n+1. Substitute u for all occurrences of 3n+1.

u2+6u-7

Factor u2+6u-7 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 -7 and whose sum is 6.

-1,7

Write the factored form using these integers.

(u-1)(u+7)

(u-1)(u+7)

Replace all occurrences of u with 3n+1.

(3n+1-1)(3n+1+7)

Simplify.

Combine the opposite terms in 3n+1-1.

Subtract 1 from 1.

(3n+0)(3n+1+7)

Add 3n and 0.

3n(3n+1+7)

3n(3n+1+7)

Add 1 and 7.

3n(3n+8)

3n(3n+8)

Replace the left side with the factored expression.

3n(3n+8)=0

3n(3n+8)=0

Divide each term in 3n(3n+8)=0 by 3.

3n(3n+8)3=03

Cancel the common factor.

3n(3n+8)3=03

Divide n(3n+8) by 1.

n(3n+8)=03

n(3n+8)=03

Divide 0 by 3.

n(3n+8)=0

If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.

n=0

3n+8=0

Set the first factor equal to 0.

n=0

Set the next factor equal to 0.

3n+8=0

Subtract 8 from both sides of the equation.

3n=-8

Divide each term by 3 and simplify.

Divide each term in 3n=-8 by 3.

3n3=-83

Cancel the common factor of 3.

Cancel the common factor.

3n3=-83

Divide n by 1.

n=-83

n=-83

Move the negative in front of the fraction.

n=-83

n=-83

n=-83

The final solution is all the values that make 3n(3n+8)3=03 true.

n=0,-83

The result can be shown in multiple forms.

Exact Form:

n=0,-83

Decimal Form:

n=0,-2.6‾

Mixed Number Form:

n=0,-223

Solve for n (3n+1)^2+6(3n+1)-7=0