Hey, I did a past paper yesterday and i got this question:
Simplify fully x(sqrd) 8x +15 / 2x(sqrd)  7x  15
I don't get it will someone please give me an answer to this and some explanation PLEASE?
Help needed urgently with math question!!
Announcements  Posted on  

Fancy winning £500? Join the Summer Bucket List Challenge and beat the holiday boredom!  06072015  
Waiting on IB results? Our IB results hub explains everything you need to know  01062015 


(Original post by Qu33n)
Hey, I did a past paper yesterday and i got this question:
Simplify fully x(sqrd) 8x +15 / 2x(sqrd)  7x  15
I don't get it will someone please give me an answer to this and some explanation PLEASE?
If you factorise both the numerator and denominator, what do you get?
In these questions, the likelihood is that when you factorise, one of the brackets in the numerator will cancel with one of the brackets in the denominator 
(Original post by Qu33n)
Hey, I did a past paper yesterday and i got this question:
Simplify fully x(sqrd) 8x +15 / 2x(sqrd)  7x  15
I don't get it will someone please give me an answer to this and some explanation PLEASE?
2x^27x15 factorises to (2x+3)(x5)
because
2x^27x15
= 2x^210x+3x15
= 2x(x5)+3(x5)
= (2x+3)(x5)
So answer is (x3)/(2x+3) 
(Original post by dugdugdug)
x^28x+15 factorises to (x3)(x5)
2x^27x15 factorises to (2x+3)(x5)
because
2x^27x15
= 2x^210x+3x15
= 2x(x5)+3(x5)
= (2x+3)(x5)
So answer is (x3)/(2x+3)
Right! Now, you may think I'm daft, but, I still don't understand how you got the bottom line lool! I got the top line the same as you, but the bottom one has to add up to 7 not 8 so I'm confused how you used the same numbers, 3 and 5.
Thanks for helping btw 
(Original post by RajPopat94)
If you factorise both the numerator and denominator, what do you get?
In these questions, the likelihood is that when you factorise, one of the brackets in the numerator will cancel with one of the brackets in the denominator 
(Original post by Qu33n)
the bottom one has to add up to 7 not 8 
(Original post by TGH1)
10x + 3x = 7x 
(Original post by Qu33n)
But doesn't it have to times together to make 15 too loool am confused
Given an equation ax^2+bx+c = 0
Case 1: a = 1
You need to find the factors of c that add up to b.
In the example x^28x+15, you need to find the factors of 15 that add up to 8. These factors are 3 and 5.
Hence (x3)(x5) = 0
Case 2: a>1
You need to find the factors of ac that add up to b.
In the example 2x^27x15, you need to find the factors of 2 times 15, ie 30 that add up to 7. These factors are 10 and 3.
Now rewrite original expression with the middle term (the x term) split up.
It was 7x but rewrite it as 10x+3x
So 2x^27x15
= 2x^210x+3x15
Consider 2x^210x which factorises to 2x(x5)
Consider 3x15 which factorises to 3(x5)
Both have a common (x5)
so 2x^210x+3x15 becomes 2x(x5)+3(x5)
= (2x+3)(x5) by the DISTRIBUTIVE LAW 
(Original post by dugdugdug)
Here's some general theory behind factorising.
Given an equation ax^2+bx+c = 0
Case 1: a = 1
You need to find the factors of c that add up to b.
In the example x^28x+15, you need to find the factors of 15 that add up to 8. These factors are 3 and 5.
Hence (x3)(x5) = 0
Case 2: a>1
You need to find the factors of ac that add up to b.
In the example 2x^27x15, you need to find the factors of 2 times 15, ie 30 that add up to 7. These factors are 10 and 3.
Now rewrite original expression with the middle term (the x term) split up.
It was 7x but rewrite it as 10x+3x
So 2x^27x15
= 2x^210x+3x15
Consider 2x^210x which factorises to 2x(x5)
Consider 3x15 which factorises to 3(x5)
Both have a common (x5)
so 2x^210x+3x15 becomes 2x(x5)+3(x5)
= (2x+3)(x5) by the DISTRIBUTIVE LAW
Ahh!! Thanks I think I finally get it
Reply
Submit reply
Register
Thanks for posting! You just need to create an account in order to submit the post Already a member? Sign in
Oops, something wasn't right
please check the following: