92.5k views
5 votes
Solve (3x^2+2x+1)(x+3)

2 Answers

4 votes

(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

User David Tinker
by
7.0k points
6 votes

Answer:

x = -3

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

User Senthilnathan
by
7.3k points