The numerator is a difference of two squares (something I recommend knowing by heart) whereby a2−b2=(a+b)(a−b) hence 4x2−1=(2x+1)(2x−1).
To factorise the denominator 6x2+7x−5, do you have any methods you went over that you can use?
If not, then a nice way is to think of two numbers which multiply to make 6×−5=−30 and add to make 7. What are they?? Then split the middle 7 as a sum of these two numbers with coefficient of x. Then factorise.
The numerator is a difference of two squares (something I recommend knowing by heart) whereby a2−b2=(a+b)(a−b) hence 4x2−1=(2x+1)(2x−1).
To factorise the denominator 6x2+7x−5, do you have any methods you went over that you can use?
If not, then a nice way is to think of two numbers which multiply to make 6×−5=−30 and add to make 7. What are they?? Then split the middle 7 as a sum of these two numbers with coefficient of x. Then factorise.
so i got -3 and 10 as the two numbers so it was 6x^2 -3x +10x -5 and i factorised it to get (2x-1)(3x+5) so i canceled (2x-1) from the numerator and denominator to get (2x+1)/ (3x+5) what do i do next?