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Ostrovityanka [42]
4 years ago
15

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

Mathematics
2 answers:
denis23 [38]4 years ago
7 0

(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

maksim [4K]4 years ago
6 0

Answer:

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

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