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I dont get this question

9. Show that 1/(〖6x〗^2+7x-5)÷1/(〖4x〗^2-1) simplifies to (ax+b)/(cx+d), where a, b¸ c and d are integers.
when you divide a fraction by another fraction, flip the second fraction and then multiply the 2 fractions together. then factorise and simplify.
Original post by confusedchildren
9. Show that 1/(〖6x〗^2+7x-5)÷1/(〖4x〗^2-1) simplifies to (ax+b)/(cx+d), where a, b¸ c and d are integers.


Rewrite the expression as a single fraction first.
i did the first part and got 4x^2 - 1 / 6x^2 + 7x -5 but idl how to factorise that
Original post by RDKGames
Rewrite the expression as a single fraction first.


i got 4x^2 - 1 / 6x^2 + 7x -5
Original post by confusedchildren
i got 4x^2 - 1 / 6x^2 + 7x -5


Great, now factorise the numerator and denominator.
Original post by confusedchildren
i did the first part and got 4x^2 - 1 / 6x^2 + 7x -5 but idl how to factorise that


i'd suggest you brush up on factorising quadratics then, there are plenty of videos on youtube etc
Original post by RDKGames
Great, now factorise the numerator and denominator.


im stuck i dont know how to do it :frown:
Original post by confusedchildren
im stuck i dont know how to do it :frown:


The numerator is a difference of two squares (something I recommend knowing by heart) whereby a2−b2=(a+b)(a−b)a^2 - b^2 = (a+b)(a-b) hence 4x2−1=(2x+1)(2x−1)4x^2 -1 = (2x+1)(2x-1).

To factorise the denominator 6x2+7x−56x^2+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=−306\times -5 = -30 and add to make 77. What are they?? Then split the middle 7 as a sum of these two numbers with coefficient of xx. Then factorise.
(edited 5 years ago)
Original post by RDKGames
The numerator is a difference of two squares (something I recommend knowing by heart) whereby a2−b2=(a+b)(a−b)a^2 - b^2 = (a+b)(a-b) hence 4x2−1=(2x+1)(2x−1)4x^2 -1 = (2x+1)(2x-1).

To factorise the denominator 6x2+7x−56x^2+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=−306\times -5 = -30 and add to make 77. What are they?? Then split the middle 7 as a sum of these two numbers with coefficient of xx. 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?
Original post by confusedchildren
what do i do next?


Celebrate.
Original post by RDKGames
Celebrate.

Cant celebrate if ive still got 3 more questions left ugh XD but thanks again

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