The remainder when 6p3 - p2 - 47p + 30 is divided by p - 3 is

Analyze the question: We need the remainder when a cubic polynomial is divided by (p − 3). The Remainder Theorem is the quickest way to do this.

Concept: If a polynomial f(p) is divided by (p − a), the remainder is f(a). Here, a = 3.

Solution:

f(p) = 6p³ − p² − 47p + 30

f(3) = 6(3)³ − (3)² − 47(3) + 30

• 6(27) = 162
• 3² = 9
• 47 × 3 = 141

f(3) = 162 − 9 − 141 + 30

f(3) = 153 − 141 + 30

f(3) = 12 + 30 = 42

Check with synthetic division (coefficients 6, −1, −47, 30, with a = 3):

• Bring down 6
• 6 × 3 = 18, and −1 + 18 = 17
• 17 × 3 = 51, and −47 + 51 = 4
• 4 × 3 = 12, and 30 + 12 = 42

Answer: B. 42

Exam tip: For any "remainder when divided by (p − a)" question, just substitute p = a into the polynomial. If the divisor is (p + a), substitute p = −a. This saves you from long division in JAMB, where time is tight.