Note that, while the basic knowledge of the order of operations makes it possible to simplify most basic expressions, specialized techniques are needed to simplify many variable expressions, including nearly all polynomials. See Method Two below for more information.
As an example, let’s try to simplify the expression 2x + 4(5 + 2) + 32 - (3 + 4/2). In this expression, we would solve the terms in parentheses, 5 + 2 and 3 + 4/2, first. 5 + 2 = 7. 3 + 4/2 = 3 + 2 = 5. The second parenthetical term simplifies to 5 because, owing to the order of operations, we divide 4/2 as our first act inside the parentheses. If we simply went from left to right, we might instead add 3 and 4 first, then divide by 2, giving the incorrect answer of 7/2. Note - if there are multiple parentheses nested inside one another, solve the innermost terms first, than the second-innermost, and so on.
The second parenthetical term simplifies to 5 because, owing to the order of operations, we divide 4/2 as our first act inside the parentheses. If we simply went from left to right, we might instead add 3 and 4 first, then divide by 2, giving the incorrect answer of 7/2.
The second parenthetical term simplifies to 5 because, owing to the order of operations, we divide 4/2 as our first act inside the parentheses. If we simply went from left to right, we might instead add 3 and 4 first, then divide by 2, giving the incorrect answer of 7/2.
After dealing with the parentheses, our example expression is now 2x + 4(7) + 32 - 5. The only exponent in our example is 32, which equals 9. Add this back into the equation in the place of 32 to get 2x + 4(7) + 9 - 5.
There are two instances of multiplication in our problem: 2x (2x is 2 × x) and 4(7). We don’t know the value of x, so let’s leave 2x as it is. . 4(7) = 4 × 7 = 28. We can rewrite our equation as 2x + 28 + 9 - 5.
Because we already solved a division problem (4/2) when we tackled the terms in parentheses, our example no longer has any division in it, so we will skip this step. This brings up an important point - you don’t have to perform every operation in the PEMDAS acronym when simplifying an expression, just the ones that are present in your problem.
Our example expression has been partially simplified to “2x + 28 + 9 - 5”. Now, we must add what we can - let’s look at each addition problem from left to right. We can’t add 2x and 28 because we don’t know the value of x, so let’s skip it. 28 + 9 = 37, so let’s rewrite or expression as “2x + 37 - 5”.
In our expression, “2x + 37 - 5”, there is only one subtraction problem. 37 - 5 = 32
Our final answer is “2x + 32”. We can’t address this final addition problem until we know the value of x, but when we do, this expression will be much easier to solve than our initial lengthy-expression.
This rule also extends to terms with multiple variables. For instance, 2xy2 can be added to -3xy2, but not -3x2y or -3y2. Let’s look at the expression x2 + 3x + 6 - 8x. In this expression, we can add the 3x and -8x terms because they are like terms. Simplified, our expression is x2 - 5x + 6.
For example, let’s consider the fraction 36/60. If we have a calculator handy, we can divide to get an answer of . 6. If we don’t, however, we can still simplify by removing common factors. Another way to think of 36/60 is (6 × 6)/(6 × 10). This can be rewritten as 6/6 × 6/10. 6/6 = 1, so our expression is actually 1 × 6/10 = 6/10. However, we’re not done yet - both 6 and 10 share the factor 2. Repeating the above procedure, we are left with 3/5. In other words, you have to divide both the numerator and denominator by their greatest common factor (which in the example above is 12: (12 × 3)/(12 × 5)).
Let’s consider the expression (3x2 + 3x)/(-3x2 + 15x). This fraction can be rewritten as (x + 1)(3x)/(3x)(5 - x), 3x appears both in the numerator and in the denominator. Removing these factors from the equation leaves (x + 1)/(5 - x). Similarly, in the expression (2x2 + 4x + 6)/2, since every term is divisible by 2, we can write the expression as (2(x2 + 2x + 3))/2 and thus simplify to x2 + 2x + 3. Note that you can’t cancel just any term - you can only cancel multiplicative factors that appear both in the numerator and denominator. For instance, in the expression (x(x + 2))/x, the “x” cancels from both the numerator and denominator, leaving (x + 2)/1 = (x + 2). However, (x + 2)/x does not cancel to 2/1 = 2.
For instance, the expression 3(x2 + 8) can be simplified to 3x2 + 24, while 3x(x2 + 8) can be simplified to 3x3 + 24x. Note that, in some cases, such as in variable fractions, the constant adjacent to the parentheses gives an opportunity for cancellation and thus shouldn’t be multiplied through the parentheses. In the fraction (3(x2 + 8))/3x, for instance, the factor 3 appears both in the numerator and the denominator, so we can cancel it and simplify the expression to (x2 + 8)/x. This is simpler and easier to work with than (3x3 + 24x)/3x, which would be the answer we would get if we had multiplied through.
Let’s consider the expression x2 - 5x + 6 once more. This expression can factor to (x - 3)(x - 2). So, if x2 - 5x + 6 is the numerator of a certain expression with one of these factor terms in the denominator, like is the case with the expression (x2 - 5x + 6)/(2(x - 2)), we may want to write it in factored form so that we can cancel it with the denominator. In other words, with (x - 3)(x - 2)/(2(x - 2)), the (x - 2) terms cancel, leaving us with (x - 3)/2. As hinted at above, another reason you may want to factor your expression has to do with the fact that factoring can reveal answers to certain equations, especially when those equations are written as expressions equal to 0. For instance, let’s consider the equation x2 - 5x + 6 = 0. Factoring gets us (x - 3)(x - 2) = 0. Since any number times zero equals zero, we know that if we can get either of the terms of parentheses to equal zero, the whole expression on the left side of the equals sign will equal zero as well. Thus, 3 and 2 are two answers to the equation.
