What is the complete factorised form of (12a^2-27b^2)?
Answer and explanation
Correct answer: (3(2a-3b)(2a+3b))
First take out the common numerical factor 3 from both terms. This gives 12a²−27b²=3(4a²−9b²). The expression inside the bracket is a difference of two squares: 4a²=(2a)² and 9b²=(3b)². Using u²−v²=(u−v)(u+v), it becomes 3(2a−3b)(2a+3b). Thus option A is the complete factorised form.
The word complete is important. Option C stops after removing only the common factor 3, because 4a²−9b² can still be factorised. Expanding option A gives 3[(2a)²−(3b)²]=3(4a²−9b²)=12a²−27b². The other choices do not expand to the original expression, so the common-factor step and the difference-of-squares identity must both be used.
Frequently asked questions
What is the correct answer to this question?
(3(2a-3b)(2a+3b))
Why is this the correct answer?
First take out the common numerical factor 3 from both terms. This gives 12a²−27b²=3(4a²−9b²). The expression inside the bracket is a difference of two squares: 4a²=(2a)² and 9b²=(3b)². Using u²−v²=(u−v)(u+v), it becomes 3(2a−3b)(2a+3b). Thus option A is the complete factorised form.
The word complete is important. Option C stops after removing only the common factor 3, because 4a²−9b² can still be factorised. Expanding option A gives 3[(2a)²−(3b)²]=3(4a²−9b²)=12a²−27b². The other choices do not expand to the original expression, so the common-factor step and the difference-of-squares identity must both be used.
Which subject and chapter does this question cover?
This is a Class 9 Mathematics question. Chapter: Exploring Algebraic Identities. Topic: Factorisation.
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