At algebraic fractions are fractional algebraic expressions that have at least one unknown in the denominator. Often, there are factors that appear in both the numerator and denominator of these fractions, leaving the possibility of simplifying them. What many ignore is that there are some rules, studied since the beginning of Elementary School, that guide this simplification process. Therefore, any simplification who breaks these rules has great potential to be wrong. Therefore, we list below the three most frequent errors in simplifying algebraic fractions and the correct way to perform these procedures.
Before proceeding, we recommend reading the article Algebraic fraction simplification for those who still have questions about this matter.
1 – Cut elements equal in numerator and denominator
This is the most common mistake. At the beginning of learning, students want to "cut" all the same elements in the numerator and denominator of a algebraic fraction. However, they are not equal elements that must be "cut", but, yes, factors equals.
The rule is as follows: If there is equal factors in the numerator and denominator, these factors can be cut. Remember: the division between them will give 1, which does not influence a division or multiplication. As these factors simply disappear, this process has become known as “cutting”. Also remember that the numbers in a multiplication are called factors.
Elements being added or subtracted you can not be cut, because its division does not result in 1. Thus, taking the example below that involves a sum, we will see the correct and incorrect way to perform the simplification.
Example: Simplify the following algebraic fraction.
4x + 4y
x + y
Incorrect:
4x + 4y = 4 + 4 = 8
x + y
Note that the unknown numbers that have been cut off (highlighted in red) are not factors of a multiplication, but rather parts of an addition. Therefore, the cut made above is wrong.
Right:
4x + 4y
x + y
making the process of polynomial factorization by common factor, we will have:
4(x + y) = 4
x + y
In the numerator of the algebraic fraction, we find a multiplication where the factors are 4 and x + y. In the denominator, we find only x + y. Note that x + y is a factor as it is not being added or subtracted by any other number or unknowns. For a better view, just put parentheses:
4(x + y) = 4
(x + y)
If, instead of x + y, there were only the number 4 in the denominator, it would also be possible to simplify, cutting the number 4 only.
Now look at a case where there could not be simplification:
4(x + y)
x + y + k
*k is any number, unknown or monomial.
2 – Factoring the perfect square trinomial using the common factor process in evidence
Almost whenever a polynomial in a algebraic fraction, it must be factored. After that, the factors present in the numerator and denominator must be compared in search of those that can be simplified (another word for “cut”).
What happens is that students are faced with a perfect square trinomial and forget that it is the result of a remarkable product, just returning to this product to perform the factorization. So the attempt is made to put common factors in evidence.
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People who make this kind of attempt often make the above mistake.
Note the following example, which also shows the correct form and the most frequent incorrect form of resolution.
Example: Simplify the following algebraic fraction.
4x2 + 8xy + 4y2
x + y
Incorrect:
4x2 + 8xy + 4y2
x + y
4(x2 + 2xy + y2)
x + y
or
4(x + 2y) + 4y2
x + y
Note that it is not even possible to simplify, precisely because the factoring process was not carried out properly.
Right:
4x2 + 8xy + 4y2
x + y
(2x + 2y)2
x + y
(2x + 2y)(2x + 2y)
x + y
In this step, note that the number 2 is common to all elements of the two numerator factors. In this situation, it is necessary to factor by factor common to the two factors. We will have as a result:
2·(x + y)·2·(x + y)
x + y
2·2·(x + y)(x + y)
x + y
4·(x + y)(x + y)
x + y
Now, yes, we can cut the factor that repeats itself in both the numerator and the denominator.
4·(x + y)(x + y)= 4·(x + y)
x + y
3 – Confuse the remarkable products
Note the list of notable products below which involves squares or product of sum for difference.
(x+y)2 = x2 + 2xy + y2
(x - y)2 = x2 –2xy + y2
(x+ y)(x – y) = x2 - y2
Every time a polynomial takes the form of a perfect square trinomial or two square difference - found in right side of the equalities above -, it is possible to replace them by the remarkable product that generated them (left side corresponding).
At simplification of algebraic fractions, forgetting that remarkable product corresponds to the perfect square trinomial is a very recurrent error - especially when it comes to the two square difference. When it appears, it is common to imagine that it is already factored or that exponent 2 can be put “in evidence” (and, of course, it is not possible to do this).
Note the following example involving two square difference:
Example: Simplify the following algebraic fraction.
4x2 – 4y2
x + y
Correct:
Remember that the numerator is a two-square difference and can be replaced by:
(2x - 2y)(2x + 2y)
x + y
The simplification will be done by placing the 2 in evidence, once again, in the two factors.
2·(x - y)·2·(x + y)
x + y
2·2·(x – y)·(x + y)
x + y
4·(x - y)·(x + y) = 4·(x – y)
x + y
Note that, in the difference of two squares, in one of the factors there is an addition and, in the other, a subtraction.
Incorrect:
Use one of the other two notable product cases:
4x2 – 4y2
x + y
(2x + 2y)(2x + 2y)
x + y
Or "put the exponent 2 in evidence":
4x2 – 4y2
x + y
4(x - y)2
x + y
To avoid these last two errors, we suggest reading the text sum square, Common factor in evidence and Potentiation.
Good studies!
By Luiz Paulo Moreira
Graduated in Mathematics
