To add to what notnek said, alternatively you can do this question without any work whatsoever by simply adding 0 creatively. More specifically,
(x−1)3x+4=(x−1)3(x−1)+5=(x−1)21+(x−1)35
I wish this technique was taught in schools more and part of A Level textbooks. It probably requires a textbook exercise for a student to become confident with it.
I wish this technique was taught in schools more and part of A Level textbooks. It probably requires a textbook exercise for a student to become confident with it.
Yes it definitely cuts workings short in most PF questions, and isn’t that difficult to get used to with little practice. Do you teach it to your students regardless of not being part of spec/textbooks?
To add to what notnek said, alternatively you can do this question without any work whatsoever by simply adding 0 creatively. More specifically,
(x−1)3x+4=(x−1)3(x−1)+5=(x−1)21+(x−1)35
thanks by the way i was wondering if you were given ∫ 2x +2 dx with limits, can you write it as ∫ 2x + ∫ 2 dx with limits on each integral sign? or are you only able to write it like this if it didnt have any limits
thanks by the way i was wondering if you were given ∫ 2x +2 dx with limits, can you write it as ∫ 2x + ∫ 2 dx with limits on each integral sign? or are you only able to write it like this if it didnt have any limits
The splitting is a propety of integrals, so yes you can as long as the limits are the same. You can also do it on indefinite integrals too.
Yes it definitely cuts workings short in most PF questions, and isn’t that difficult to get used to with little practice. Do you teach it to your students regardless of not being part of spec/textbooks?
That depends on time For further maths students I would always teach it.
thanks do you mind telling me your thought process of getting x(2x+1) = 2x(x+1) - x
Well we have a x+1 in the denominator so we would like to get this somehow on the numerator in order for them to cancel. So, you can notice that 2x+1 is almost 2(x+1). We can get it by saying that 2x+1 = 2x+2-1 hence we get x(2x+1) = x(2x+2-1) = x(2x+2)-x = 2x(x+1) - x
Well we have a x+1 in the denominator so we would like to get this somehow on the numerator in order for them to cancel. So, you can notice that 2x+1 is almost 2(x+1). We can get it by saying that 2x+1 = 2x+2-1 hence we get x(2x+1) = x(2x+2-1) = x(2x+2)-x = 2x(x+1) - x