Question

The polynomials 5x³ − 3x² + kx − 25 and x³ + 15x + 2k + 18  leave the same remainder when divided by x − 3. Find the value of k.

Answer 1

Let p(x) = 5x³ − 3x² + kx − 25 and q(x) = x³ + 15x + 2k + 18

Step 1: Use the Remainder Theorem

For divisor x − 3,

x = 3

So we evaluate:

p(3) = q(3)

Step 2: Compute P(3)

p(x) = 5x³ − 3x² + kx − 25

p(3) = 5(3)³ − 3(3)² + k(3) − 25

= 5(27) − 3(9) + k(3) − 25

= 135 − 27 + 3k − 25

= 83 + 3k

Step 3: Compute q(3)

q(x) = x³ + 15x + 2k + 18

q(3) = 3³ + 15(3) + 2k + 18

= 27 + 45 + 2k + 18

= 90 + 2k

Step 4: Set the remainders equal

83 + 3k = 90 + 2k

Step 5: Solve for k

83 + 3k = 90 + 2k

3k − 2k = 90 − 83

k = 7