To find the remainder when the polynomial x^3 + 4x^2 + 12x + 9 is divided by x^2, we can use polynomial long division or synthetic division. However, in this case, we can also simply calculate the remainder by substituting the divisors’ roots.
When we divide by x^2, the remainder will be a polynomial of lesser degree than the divisor. Therefore, the remainder will take the form Ax + B, where A and B are constants we need to determine.
To find these coefficients, we can evaluate the original polynomial at the roots of x^2 = 0. The roots of this equation are x = 0 (with a multiplicity of 2). So, we will evaluate the original polynomial at x = 0:
P(x) = x^3 + 4x^2 + 12x + 9
Substituting x = 0:
P(0) = 0^3 + 4(0^2) + 12(0) + 9 = 9
This means that the value of the polynomial at x = 0 is 9, which tells us that when we divide the polynomial by x^2, the constant term B in the remainder is 9.
Next, to find the coefficient A, we can take the derivative of the polynomial:
P'(x) = 3x^2 + 8x + 12
Evaluating the derivative at x = 0:
P'(0) = 3(0^2) + 8(0) + 12 = 12
The value of the derivative tells us the coefficient A, which is 12. Therefore, the remainder when the polynomial is divided by x^2 is:
R(x) = 12x + 9
So, the final answer is:
The remainder when x^3 + 4x^2 + 12x + 9 is divided by x^2 is 12x + 9.