Solve (3x^2+2x+1) (x+3)

x = - 3

Step-by-step explanation:

Solve for x over the real numbers:

(x + 3) (3 x^2 + 2 x + 1) = 0

Split into two equations:

x + 3 = 0 or 3 x^2 + 2 x + 1 = 0

Subtract 3 from both sides:

x = - 3 or 3 x^2 + 2 x + 1 = 0

Divide both sides by 3:

x = - 3 or x^2 + (2 x) / 3 + 1/3 = 0

Subtract 1/3 from both sides:

x = - 3 or x^2 + (2 x) / 3 = - 1/3

Add 1/9 to both sides:

x = - 3 or x^2 + (2 x) / 3 + 1/9 = - 2/9

Write the left hand side as a square:

x = - 3 or (x + 1/3) ^2 = - 2/9

(x + 1/3) ^2 = - 2/9 has no solution since for all x on the real line, (x + 1/3) ^2 >=0 and - 2/9<0:

Answer: x = - 3

(3x^2+2x+1) (x+3)

= (3x^2) (x) (+3) (3) + (2x) (x) + (2x) + 3

3x^3+9x^2+2x^2+6x+x+3

3x^3+11x^2+7x+3