When the polynomial \( 6 x^{4}+p x^{2}+q \) is divided by \( x^{2}-5 x+6 \), the remainder is \( 400 x-693 \). Find the value of \( p \) and \( q \). [Answer Key] \( p=2, q=3 \)

We are given that \[ 6 x^4 + p x^2 + q \] divided by \[ x^2 - 5x + 6 \] leaves a remainder of \[ 400x - 693. \] Since the divisor is quadratic, the remainder must be of degree less than 2 (i.e. linear). We can express the division as \[ 6 x^4 + p x^2 + q = (x^2-5x+6) \cdot Q(x) + (400x-693), \] where \(Q(x)\) is the quotient polynomial. Because \(x^2-5x+6\) factors as \[ (x-2)(x-3)=0, \] its roots are \(x=2\) and \(x=3\). Substituting these roots into the equation will eliminate the \(Q(x)\) term. **Step 1. Substitute \(x=2\)** \[ 6(2^4) + p(2^2) + q = 400(2) - 693. \] Compute each term: \[ 6(16) = 96,\quad p(4)=4p,\quad 400(2)-693=800-693=107. \] Thus, we have: \[ 96 + 4p + q = 107. \] Subtracting 96 on both sides: \[ 4p + q = 11. \quad \text{(Equation 1)} \] **Step 2. Substitute \(x=3\)** \[ 6(3^4) + p(3^2) + q = 400(3) - 693. \] Compute each term: \[ 6(81) = 486,\quad p(9)=9p,\quad 400(3)-693=1200-693=507. \] Thus, we have: \[ 486 + 9p + q = 507. \] Subtracting 486 on both sides: \[ 9p + q = 21. \quad \text{(Equation 2)} \] **Step 3. Solve the system of equations** Subtract Equation 1 from Equation 2: \[ (9p+q)-(4p+q)=21-11, \] \[ 5p=10, \] \[ p=2. \] Substitute \(p=2\) into Equation 1: \[ 4(2)+q=11, \] \[ 8+q=11, \] \[ q=3. \] Thus, the values are \[ p=2,\quad q=3. \]