Struggling with Algebraic Fractions

[grishmabeli6]

Hi everyone. I am revisiting GCSE Maths and am encountering difficulty understanding the process for solving the following problem. The question is: (a) Factorise: x2 - x - 2

(b) Simplify: 3x+2/x2 -x-2 + 1/x+1

Apologies for the formatting, I have no idea how to make the fractions look more conventional.

Ok, so I worked out the answer to (a), which is (x+1)(x-2).

I know the answer to (b) is 4x/(x+1)(x-2). I looked this up in the hope I could reverse engineer the answer after 3 hours of trying.

Many pages of scribbling later, the problem I am having is I don’t understand how to come to that answer. I have tried multiplying them as you do when adding fractions but I keep ending up with the same result and I’m not sure why or whether I am even on the right path.

The stage I keep getting to is: 4x2 +4x / (x2 -x-2)(x+1)

I feel as though I should be using the (x+1)(x-2) answer from (a) in my working for (b) but again I am not sure where this should be and why. I don’t even think the stage I’m getting to is correct either so any assistance or advice will be greatly appreciated.

Thank you all for your time.

[Guest]

Hi everyone. I am revisiting GCSE Maths and am encountering difficulty understanding the process for solving the following problem. The question is: (a) Factorise: x2 - x - 2

You mean x^2- x- 2, using "^" to indicate an exponent. The only time it is really simple to factor is if only integers involved. The only ways to factor -2 are (-1)(2) or (1)(-2) so TRY (x- 2)(x+ 1) and (x+ 2)(x- 1).

(x- 2)(x+ 1)= x^2+ x- 2x- 2= x^2- x- 2. Yes, that works!
(If we had first tried, instead, (x+ 2)(x- 1) we would have got x^2- x+ 2x- 2= x^2+ x- 2 which has the wrong sign pn "x" so would have gone to (x- 2)(x+ 1).)

(b) Simplify: 3x+2/x2 -x-2 + 1/x+1

Apologies for the formatting, I have no idea how to make the fractions look more conventional.

You could at least have used parentheses to make it clear. What you wrote is 3x+ (2/x^2)- x- 2+ (1/x)+ 1
but I am sure you meant (3x+ 2)/(x^2- x- 2)+ 1/(x+ 1) which is quite different.

To add fractions you need to get "common denominators". What are the denominators?
They are x^2- x- 2 and x+ 1. You already know that x^2- x- 2= (x- 2)(x+ 1).

Since there is already "x+ 1" in the denominator of the second fraction you just need to get "x- 2" there.
You do that by multiplying both numerator and denominator by x- 2.

(3x+ 2)/(x^2- x- 2)+ (x- 2)/(x+1)(x- 2)= (3x+ 2)/(x^2- x- 2)+ (x-2)/(x^2- x- 2)
= (3x+ 2+ x- 2)/(x^2- x- 2)= 4x/(x^2- x- 2).

Ok, so I worked out the answer to (a), which is (x+1)(x-2).

I know the answer to (b) is 4x/(x+1)(x-2). I looked this up in the hope I could reverse engineer the answer after 3 hours of trying.

Many pages of scribbling later, the problem I am having is I don’t understand how to come to that answer. I have tried multiplying them as you do when adding fractions but I keep ending up with the same result and I’m not sure why or whether I am even on the right path.

The stage I keep getting to is: 4x2 +4x / (x2 -x-2)(x+1)

It looks like you are multiply the numerator and denominator of 4x/(x^2- x- 2) by x+ 1 but there is no reason to do that. It already has "x+ 1" in the denominator, as a factor or x^2- x- 2. It is the fraction 1/(x+ 1) that you need to multiply numerator and denominator by x- 2 to get x^2- x- 2 as its denominator.

I feel as though I should be using the (x+1)(x-2) answer from (a) in my working for (b) but again I am not sure where this should be and why. I don’t even think the stage I’m getting to is correct either so any assistance or advice will be greatly appreciated.

Thank you all for your time.

Frankly it looks like you have memorized a lot of "methods" without understanding why you do them!

[Angus53]

Hello! Let's break down the problem step by step.

For part (a), you correctly factorized \( x^2 - x - 2 \) into \( (x + 1)(x - 2) \).

Now, moving to part (b), where you have \( \frac{3x+2}{x^2 -x-2} + \frac{1}{x+1} \):

First, express the fractions with a common denominator. For the second fraction, multiply the numerator and denominator by \( (x - 2) \) to get a common denominator of \( (x + 1)(x - 2) \):

\[ \frac{3x+2}{x^2 -x-2} + \frac{1}{x+1} = \frac{3x+2}{(x+1)(x-2)} + \frac{(x-2)}{(x+1)(x-2)} \]

Combine the fractions:

\[ = \frac{3x+2 + (x-2)}{(x+1)(x-2)} \]

\[ = \frac{4x}{(x+1)(x-2)} \]

So, you were indeed on the right track! The answer is \( \frac{4x}{(x+1)(x-2)} \). You got the numerator correct, but it seems there was a slight error in the denominator. Make sure to check the arithmetic carefully. Keep up the good work!

[Guest]

Hello! Let's break down the problem step by step.

For part (a), you correctly factorized
x^2−x−2 into (x+1)(x−2)

Now, moving to part (b), where you have

3x+2/x^2−x−2+1/x+1

First, express the fractions with a common denominator. For the second fraction, multiply the numerator and denominator by (x−2) to get a common denominator of (x+1)(x−2):

3x+2/x^2−x−2+1/x+1=3x+2/(x+1)(x−2)+(x−2)/(x+1)(x−2)

Combine the fractions:

=3x+2+(x−2)/(x+1)(x−2)
=4x/(x+1)(x−2)

So, you were indeed on the right track! The numerator is correct, but it seems there was a slight error in the denominator. Make sure to check the arithmetic carefully. Keep up the good work!

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