Question

When the polynomials x³ − px² + x + 6 and 2x³ − x² − (p + 3) x − 6 are divided by x − 3, they leave the same remainder, the value of p is ______.

Options

• 1

• −1

• 0

• 3

Answer 1

When the polynomials x³ − px² + x + 6 and 2x³ − x² − (p + 3)x − 6 are divided by x − 3, they leave the same remainder, the value of p is 1.

Explanation:

Let p(x) = x³ − px² + x + 6 and q(x) = 2x³ − x² − (p + 3)x − 6

Step 1: Use the Remainder Theorem

x − 3 = 0

x = 3

So we evaluate:

p(3) = q(3)

Step 2: Compute P(3)

p(x) = x³ − px² + x + 6

p(3) = 3³ − p(3)² + (3) + 6

= 27 − 9p + 3 + 6

= 36 − 9p

Step 3: Compute q(3)

q(x) = 2x³ − x² − (p + 3)x − 6

q(3) = 2(3)³ − 3² − (p + 3)(3) − 6

= 2(27) − 9 − (3p+ 9) − 6

= 54 − 9 − 3p − 9 − 6

= 30 − 3p

Step 4: Set the remainders equal

36 − 9p = 30 − 3p

Step 5: Solve for p

36 − 9p = 30 − 3p

36 − 30 = 9p − 3p

6 = 6p

p = `6/6`

p = 1