At the end of last week’s lesson on fractions, I left our readers with a challenge problem to solve:
X2 + 3X + 2
X2 + 4X + 4
Simplify.
This week, we will discuss in detail how to solve this and other problems involving the simplification of rational expressions.
Factoring Algebraic Expressions
As you may recall, one of the first things we covered in our discussion of basic fractions was factorization. When dealing with algebraic rational expressions – which are really fractions in disguise – you must become comfortable with an analogous process. This process we will now discuss.
Suppose, for example, that you have been asked to factor the expression 2X2 + 4X. How would you go about it?
First, you must break down each term into its basic pieces:
2*X*X + 2*2*X
Now, you must identify which pieces the two terms have in common:
2*X*X + 2*2*X
Then, you must divide each term by the common pieces:
(2*X*X)/2*X + (2*2*X)/2*X
Write what is left in parentheses. Put the common pieces outside:
2X(X + 2)
This process that we have just completed is called factoring out the greatest common factor. Many problems on the SAT and elsewhere can be solved in this fashion. But there are other types of factoring problems we must worry about. For example, consider the following expression:
2X2 + 5X + 3.
If we break the terms in this expression down into their component pieces, we discover that they have no pieces in common:
2*X*X + 5*X + 3
This expression – a trinomial of the form AX2 +/- BX +/- C – thus requires a different approach. The steps for factoring such an expression are as follows:
1) Multiply A and C. A is the coefficient of the X2 term. C is the constant at the end.
A*C = 2*3 = 6
2) Write the factors of A*C.
6: 1*6 or 2*3
3) Decide which pair of factors add up to the middle term B. In this step, you may have to make one or both of the factors in your selected pair negative. (We’ll go over examples involving negatives in a moment.)
B=5, 2+3=5
4) Rewrite your expression with the middle term broken down into its two pieces.
2X2 + 2X + 3X + 3
5) Group the first two and the last two terms using parentheses.
(2X2 + 2X) + (3X + 3)
6) Factor out the greatest common factor for each group using the process described above.
(2*X*X + 2*X) + (3*X + 3)
2X(X+1) + 3(X+1)
7) Write the binomial that appears in both sets of parentheses as one factor. Then write what is left as the other factor.
(X+1)(2X+3)
Here’s are some example problems that involve negatives:
3X2 – 19X + 6
A*C = 3*6 = 18
18: 1*18, 2*9, 3*6
B = -19, -1 + -18 = -19
3X2 – 1X – 18X +6
(3X2 – 1X) + (-18X + 6)
(3*X*X – 1*X) + (-2*3*3*X + 2*3)
X(3X – 1) – 6(3X – 1)
(3X – 1)(X – 6)
X2 + X – 12
A*C = 1*-12 = -12
-12: 1*-12, -1*12, 2*-6, -2*6, 3*-4, -3*4
B = 1, -3 + 4 = 1
X2 – 3X + 4X – 12
(X2 – 3X) + (4X – 12)
(X*X – 3*X) + (2*2*X – 2*2*3)
X(X – 3) + 4(X – 3)
(X – 3)(X + 4)
There is also one special pattern you absolutely must know for the college entrance exams: the difference of squares pattern. For example, what if you were asked to factor the following:
X2 – 16
Both X2 and 16 are perfect squares. Thus, this expression can be factored using the difference of squares pattern.
First, you need to take the square root of each piece. The square root of X2 is X. The square root of 16 is 4.
Next, you write the square roots in two factors as follows:
(X – 4)(X + 4)
All expressions that feature a difference of squares can be factored in this way. Thus, X2 – 9 = (X – 3)(X + 3); 4X2 – 25 = (2X – 5)(2X + 5); 9X2 – 121 = (3X – 11)(3X + 11); and so on.
Simplifying Rational Expressions
So what of last week’s challenge problem? How do we simplify a rational expression? It’s really quite simple once you have mastered the art of factoring algebraic expressions.
First, we need to break the top and the bottom expressions down into their component factors.
X2 + 3X + 2
A*C = 1*2 = 2
2: 1*2
B = 3, 1 + 2 = 3
X2 + X + 2X + 2
(X2 + X) + (2X + 2)
(X*X + X) + (2*X + 2)
X(X + 1) + 2(X + 1)
(X + 1)(X + 2)
X2 + 4X + 4
A*C = 1*4 = 4
4: 1*4, 2*2
B = 4, 2 + 2 = 4
X2 + 2X + 2X + 4
(X2 + 2X) + (2X + 4)
(X*X + 2*X) + (2*X + 2*2)
X(X + 2) + 2(X + 2)
(X + 2)(X + 2)
With the numerator and denominator factored, our rational expression looks like this:
(X + 1)(X + 2)
(X + 2)(X + 2)
Your next step is to cross out the factor(s) the numerator and the denominator have in common:
(X + 1)(X + 2)
(X + 2)(X + 2)
What’s left is our simplified expression!
(X + 1)
(X + 2)
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Free algebra worksheets (with answer keys) can be found at the Kuta Software site if you would like some extra practice.
