1. Given \( 3 x^{3}+8 x^{2}-1 \equiv(x+5)(x-2) Q(x)+A x+B \), where \( Q(x) \) is a polynomial. Find \( \begin{array}{ll}\text { (i) } Q(x) \text {, } \\ \text { (ii) the values of } A \text { and } B \text {. } \\ \text { [Answer Key] } \\ \begin{array}{ll}\text { (i) } 3 x-1 & \text { (ii) } A=33, \quad B=-11\end{array}\end{array}> \).

We write the given identity as \[ 3x^3+8x^2-1 = (x+5)(x-2) Q(x) + Ax+B. \] Since \((x+5)(x-2)\) is a quadratic, for the degree of the cubic \(3x^3+8x^2-1\) to be maintained, \(Q(x)\) must be a linear polynomial. Let \[ Q(x)=ax+b. \] First, compute the product: \[ (x+5)(x-2)= x^2+3x-10. \] Then \[ \begin{aligned} (x^2+3x-10)(ax+b) & = ax^3 + bx^2 + 3ax^2 + 3bx - 10ax - 10b \\ & = ax^3 + (b+3a)x^2 + (3b-10a)x - 10b. \end{aligned} \] Including the remainder \(Ax+B\), the entire right-hand side becomes \[ ax^3 + (b+3a)x^2 + \bigl(3b-10a+ A\bigr)x + \bigl(-10b+B\bigr). \] This must equal the left-hand side term by term: \[ 3x^3+8x^2+0x-1. \] Equate coefficients of like powers of \(x\): 1. Coefficient of \(x^3\): \[ a=3. \] 2. Coefficient of \(x^2\): \[ b+3a=8 \quad \Longrightarrow \quad b+3(3)=8 \quad \Longrightarrow \quad b+9=8 \quad \Longrightarrow \quad b=-1. \] 3. Coefficient of \(x\): \[ 3b-10a+ A=0 \quad \Longrightarrow \quad 3(-1)-10(3)+A=0 \quad \Longrightarrow \quad -3-30+A=0 \quad \Longrightarrow \quad A=33. \] 4. Constant term: \[ -10b+B=-1 \quad \Longrightarrow \quad -10(-1)+B=-1 \quad \Longrightarrow \quad 10+B=-1 \quad \Longrightarrow \quad B=-11. \] Thus, we have found: \[ Q(x)= 3x-1, \quad A=33, \quad B=-11. \]