If (x + 3) is a factor of the polynomial \(x^{3} + 3x^{2} + nx - 12\), where n is a constant, find the value of n.
Answer Details
If (x + 3) is a factor of the polynomial, then we can write the polynomial as:
\[(x+3)(ax^2+bx+c)\]
Expanding the above equation, we get:
\[ax^3 + (b+3a)x^2 + (c+3b)x + 3c = x^3 + 3x^2 + nx - 12\]
Equating the coefficients of corresponding powers of x, we get:
\[a = 1, b + 3a = 3, c + 3b = n, 3c = -12\]
From the last equation, we have:
\[c = -4\]
Using this value of c in the third equation, we have:
\[n = c + 3b = -4 + 3b\]
From the second equation, we have:
\[b + 3a = 3\]
\[b + 3(1) = 3\]
\[b = 0\]
Hence, we have:
\[n = -4 + 3b = -4\]
Therefore, the value of n is -4.