what is the remainder when $(3x^{3}-2x^{2}+4x - 3)$ is divided by $(x^{2}+3x + 3)$?\n30\n$3x - 11$\n$28x…

Explanation:

Step1: Set up polynomial long - division

Let (3x^{3}-2x^{2}+4x - 3=(Ax + B)(x^{2}+3x + 3)+Cx+D). First, multiply ((Ax + B)(x^{2}+3x + 3)=Ax^{3}+3Ax^{2}+3Ax + Bx^{2}+3Bx + 3B=Ax^{3}+(3A + B)x^{2}+(3A + 3B)x+3B).

Step2: Equate the coefficients of like - terms

For the (x^{3}) term: (A = 3). Substitute (A = 3) into the coefficient of the (x^{2}) term: (3A + B=-2). Since (A = 3), then (3\times3 + B=-2), so (9 + B=-2), and (B=-11).

Step3: Find the remainder

The remainder is (Cx + D), where (C = 3A+3B) and (D = 3B). Substitute (A = 3) and (B=-11) into (C): (C=3\times3+3\times(-11)=9 - 33=-24) (This is wrong, let's use polynomial long - division directly).

Using polynomial long - division: [ \begin{align*} \frac{3x^{3}-2x^{2}+4x - 3}{x^{2}+3x + 3}&=3x-11+\frac{3x-11)(x^{2}+3x + 3)+(28x + 30)}{x^{2}+3x + 3}\ 3x^{3}-2x^{2}+4x-3&=(3x - 11)(x^{2}+3x + 3)+28x+30 \end{align*} ]

Answer:

(28x + 30)