Factorize 2x\(^2\) - 21x + 45

Assessment: WAEC SSCE - General Mathematics - 1993 (Objective) Subject: General Mathematics

Question 1 Report

Factorize 2x\(^2\) - 21x + 45

Answer Details
To factorize 2x\(^2\) - 21x + 45, we need to find two numbers that multiply to give 2\(\times\)45 = 90 and add to give -21. Let's factorize 90 to find such numbers: 90 = 1\(\times\)90 = 2\(\times\)45 = 3\(\times\)30 = 5\(\times\)18 = 6\(\times\)15 Out of these, the pair that adds up to -21 is 6 and 15. So, we can rewrite the expression as: 2x\(^2\) - 21x + 45 = 2x\(^2\) - 6x - 15x + 45 Now, we can group the first two terms and the last two terms separately and factorize them using the distributive law. That gives us: 2x\(^2\) - 6x - 15x + 45 = 2x(x - 3) - 15(x - 3) Notice that we have a common factor of (x - 3) in both terms. We can factor it out to get the final answer: 2x\(^2\) - 21x + 45 = (x - 3)(2x - 15) Therefore, the correct answer is (x - 3)(2x - 15).

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