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Mastering algebra is an important requisite of understanding calculus. The motivation for this review is to refresh algebraic concepts that might not have been encountered for some time, and to try to correct any algebraic misconceptions before delving into calculus. Inside this review are numerous examples, as well as some practice problems at the very end.
The Distributive Property
The distributive property states that
a(b + c) = ab + ac,
which implies (b + c)a = ab + ac,
by commutativity of multiplication. We can use the distributive property to either multiply terms through, or factor terms out. For instance, if we are confronted with 13x + 4x, we can use the distributive property to factor out the x, and find that
13 x + 4x = (13 + 4)x = 17x.
This works exactly the same for a more complicated example, such as 13yx + 4x. We can apply the distributive property to find that
13 yx + 4x = (13y + 4)x.
Similarly, if we have 12(x + 2), the distributive property tells us that we can multiply through by 12, and find that 12(x + 2) = 12x + 24.
We can also apply the distributive property multiple times; for instance, to arrive at the important result of factoring a quadratic polynomial.
(a + b)(c + d) = a(c + d) + b(c + d) = ac + ad + bc + bd.
In a similar fashion, by writing c = d + e we find that
a(b + d + e) = ab + ad + ae.
Using this idea we can apply the distributive property to expressions with an arbitrary number of terms.
Common Pitfalls
- Not Applying the Distributive Property: The simplest mistake one can make is to expand powers in the following way (x + y)^2 = x^2 + y^2. This analysis completely ignores the distributive property - we should find that
(x + y)^2 = x^2 + 2xy + y^2.
It is noteworthy that these two expressions are only equal if either x or y is zero, in which the problem is reduced to one in which we do not need the distributive property.
- Not Distributing Negative Signs: It is important to remember to carry through negative signs when using the distributive property. For instance, −(x + 1) = −x − 1. Note that the negative sign is “carried” to both terms. Similarly,
−h(x + 1) = −hx − h, where once again the second term is also a negative h. Similarly, we find that h(x − 1) = hx − h,
where the second term is negative, because we had a negative 1.
- Distributing nth^ Powers: It may seem tempting to directly apply the distributive property to expressions such as 2(x + 1)^2 , by writing that 2(x + 1)^2 = (2x + 2)^2. This, however, is incorrect. If we expand both expressions we find that 2(x + 1)^2 = 2(x + 1)(x + 1) = 2(x^2 + 2x + 1) = 2x^2 + 4x + 2, and (2x + 2)^2 = (2x + 2)(2x + 2) = 4(x + 1)(x + 1) = 4x^2 + 8x + 4, so clearly the two expressions are not equal. The above analysis does, however, suggest how we could properly rewrite the expression using the distributive property. 2(x + 1)^2 = 2(x + 1)(x + 1) =
2(x + 1)(x + 1) = (
2 x +
2 x +
2 x +
2)^2
Although we are able to rewrite the expression in this way, it is generally not worth the effort required to do so. One is better off simply leaving an expression of the form 2(x + 1)^2 as it is. Similarly, factoring coefficients out of such an expression such as (2x + 2)^2 is generally not worthwhile.
- Improperly Separating Fractions: Suppose we have the fraction
x + 1 x − 3 One might be tempted to say that x + 1 x − 3
x + 1 x
x + 1 3
but this would not be correct, because we cannot separate the terms in the denominator. As a simple example, note that 1 4
Nevertheless, we can separate the terms in the numerator using the distributive property, so x + 1 x − 3
x − 3
(x + 1) = x x − 3
x − 3
is a valid statement.
then one of the following is true: a = 0, b = 0, or a = b = 0. Thus, when one is faced with an equation like x(x + 2) = 0 there are two solutions: x = 0, and x + 2 = 0, which implies x = −2.
- Not Performing an Operation to All Terms of an Equation: Suppose that
η(x + 4) − x = γ.
In this case, one might try to divide both sides of the equation by η to find
(x + 4) − x = γ η
This is incorrect, because dividing both sides of the equation by η does not simply move η to the right side of the equation. Performing such an operation would actually yield
x + 4 −
x η
γ η
which is not extremely useful in solving the equation. A better strategy would be to multiply through by η using the distributive property to find
ηx + 4η − x = γ.
From there the x terms could be combined, yielding
x(η − 1) + 4η = γ.
Next we would subtract 4η from both sides and find
x(η − 1) = γ − 4 η.
Finally, we would divide by (η − 1) completing the solution for x
x = γ − 4 η η − 1
Multiplying by 1
The multiplicative identity is 1, which means that for any number a, we find
a · 1 = a
Thus, unlike the arithmetic operators, which we must apply to both sides of the equation, we can multiply a single entity by 1 without changing our equation at all. For instance, if we need to solve
x 7
x 2
we can multiply both of the fraction terms by 1, in order to create a common denominator. Thus
x 7
x 2
x 7
x 2
x 7
x 2
2 x 14
7 x 14
2 x + 7x 14
Up to this point, we’ve used the trick of multiplying by 1 in order to create a common denominator. Now we can apply the distributive property to factor out the x, and then perform arithmetic operations to find a solution. We find that 2 x + 7x 14
(2 + 7)x 14
9 x 14
9 x = 14 x =
Common Pitfalls
- Improperly Canceling Terms of a Fraction: Suppose you aer faced with a fraction containing polynomial terms, such as x x + 1
In this situation, one might see an x in the numerator and denominator of the function, and feel tempted to cancel them, yielding either x x^ + 1
or x x^ + 1
Both of these are incorrect. In order to cancel an x from both the numerator and the denomi- nator, it would have to be multiplying all of the terms in both the numerator and denominator, as follows 2 x x^2 + x
Above we can factor x from the terms in the denominator, and then cancel a factor of x from the numerator and denominator as follows 2 x x^2 + x
2 x x(x + 1)
x x
x + 1
x + 1
Note that the expression 2 x x^2 + x is not defined for x = 0. When we write x x
we are assuming that x 6 = 0. Thus, the expressions 2 x x^2 + x and
x + 1 are only equal when x 6 = 0. Similarly, if we have a fraction multiplied by a constant, we need to distribute the constant to all terms within the numerator of the fraction, so
2 ·
x + 1 x
2(x + 1) x
2 x + 2 x
Similarly, if we have a term consisting of multiple variables such as
x = p(2x + 1)x,
we can manipulate it as a single entity yielding
x − p(2x + 1)x = 0.
Now applying the distributive property we find
x(1 − p(2x + 1)) = 0.
This is an equation with two solutions x = 0, and 1 − p(2x + 1) = 0, or x = 12 ( (^1) p − 1). It is left to the reader to verify these are indeed solutions.
General Strategy
In order to solve algebraic equations, one must apply the above tools in the proper order, and avoid the common pitfalls along the way. The general strategy to solving an algebraic equation is as follows
- Separate the variable and constant terms. The first step is to apply arithmetic operators in order to move all terms containing the variable being solved for to one side of the equation, and the remaining constant terms to the other. After this process is completed, if possible, simplify the expressions (such as canceling a 1 and −1 on the same side of the equation).
- Factor the variable from constant terms, if necessary. If there are multiple terms containing the variable, use the distributive property to factor out the variable from the constants. Depending on the form of the equation, more complicated factoring schemes may need to be used (such as (a + b)(c + d) = ac + ad + bc + bd).
- Divide by any constant terms that resulted from factoring, to isolate the variable. If you were able to factor out all instances of the variable from the constant terms in the previous step, then divide by the constant in order to isolate the variable.
- If the equation has multiple terms multiplied together which equal 0, check for solutions to each of the terms equalling 0. It may turn out that your equation looks something like x(x − 2) = 0 after it is simplified. Remember that setting either of the terms on the left equal to zero provides a solution, so there are 2 solutions, not just one (x = 0 and x = 2).
It should be emphasized that this is a general strategy, but it is not an algorithm that may be followed blindly in order to solve an equation. Thus, it is only with practice, and an understanding of how to utilize the tools presented above that one can master solving algebraic equations. Having read the above examples, try to solve the following equations for x.
p^2 (2x + 4) + eατ^ = 1
p(2x − 17) = 4q + x
x =
x x + p(1 − x)
