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Year 12s: what will be the first way you research your future options? >>

C3 Algebraic fractions

In the attachment below this is given
(in the "where things can go ugly" section)
Blindly multiplying the two denominators when there might be a common factor, e.g. (𝑥+3)(𝑥−3) and (𝑥+3) should become (𝑥+3)(𝑥−3) rather than (𝑥+3)²(𝑥−3).

But surely multiplying
(x+3)(x-3) by (x+3) would give (x-3)²(x-3)????

(edited 7 years ago)

C3CheatSheet.pdf428.5KB

If you're adding 2 fractions, say

\frac{A}{(x+3)(x-3)} + \frac{B}{x+3}

then you're going to end up with

\frac{A(x+3)+B(x+3)(x-3)}{x+3}{x-3}{x+3}

which is unnecessary because there's a common factor of (x+3) in the fraction. So it just ends up as

Unparseable latex formula:

\frac{A+B(x-3)}{(x+3)(x-3)

fml.. Latex is broken at the moment I think.

edit:

(edited 7 years ago)

OP

Original post by Parallex

If you're adding 2 fractions, say

\frac{A}{(x+3)(x-3)} + \frac{B}{x+3}

then you're going to end up with

\frac{A(x+3)+B(x+3)(x-3)}{x+3}{x-3}{x+3}

which is unnecessary because there's a common factor of (x+3) in the fraction. So it just ends up as

\frac{A+B(x-3)}{(x+3)(x-3)}

fml.. Latex is broken at the moment I think.

edit:

But the question is ....
oh i forgot the definition of denominator xD silly me thanks anyway :smile:

:smile:

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