For example, 3x2 and 4x2 are like terms because each contains the variable x raised to the second power. However, x and x2 are not like terms because each term has x raised to a different power. Similarly, -3yx and 5xz are not like terms because each term has a different set of variables.
For example, if we wanted to factor 20, we might write it as 4 × 5. Note that variable terms can also be factored - 20x, for instance, can be written as 4(5x). Prime numbers can’t be factored because they are only evenly divisible by themselves and 1.
Parentheses Exponents Multiplication Division Addition Subtraction
As an example problem, for the next few steps, let’s consider the expression 1 + 2x - 3 + 4x.
For example, let’s identify like terms in our equation 1 + 2x - 3 + 4x. 2x and 4x both have the same variable raised to the same exponent (in this case, the x’s aren’t raised to any exponent at all). In addition, 1 and -3 are like terms, as neither has any variables. So, in our equation, 2x and 4x and 1 and -3 are like terms.
Let’s add the like terms in our example. 2x + 4x = 6x 1 + -3 = -2
In our example, our simplified terms are 6x and -2, so our new expression is 6x - 2. This simplified expression is equal to the original (1 + 2x - 3 + 4x), but is shorter and easier to manage. It’s also easier to factor, which, as we’ll see below, is another important simplifying skill.
For example, let’s consider the equation 5(3x-1) + x((2x)/(2)) + 8 - 3x. It would be incorrect to immediately identify 3x and 2x as like terms and combine them because the parentheses in the expression dictate that we’re supposed to do other operations first. First, let’s perform the arithmetic operations in the expression in accordance with the order of operations to obtain terms we can use. See below: 5(3x-1) + x((2x)/(2)) + 8 - 3x 15x - 5 + x(x) + 8 - 3x 15x - 5 + x2 + 8 - 3x. Now, since the only operations left are addition and subtraction, we can combine like terms. x2 + (15x - 3x) + (8 - 5) x2 + 12x + 3
Let’s use the equation 9x2 + 27x - 3. Notice that every term in this equation is divisible by 3. Since the terms aren’t all evenly divisible by any larger number, we can say that 3 is our expression’s greatest common factor.
Let’s factor our equation by its greatest common factor, 3. To do so, we’ll divide each term by 3. 9x2/3 = 3x2 27x/3 = 9x -3/3 = -1 Thus, our new expression is 3x2 + 9x - 1.
For our example expression, 3x2 + 9x - 1, we would enclose the expression in parentheses and multiply by the greatest common factor of the original equation to get 3(3x2 + 9x - 1). This equation is equal to the original, 9x2 + 27x - 3.
Let’s say our original example expression, 9x2 + 27x - 3, is the numerator of a larger fraction with 3 in the denominator. This fraction would look like this: (9x2 + 27x - 3)/3. We can use factoring to simplify this fraction. Let’s substitute the factored form of our original expression for the expression in the numerator: (3(3x2 + 9x - 1))/3 Notice that now, both the numerator and the denominator share the coefficient 3. Dividing the numerator and denominator by 3, we get: (3x2 + 9x - 1)/1. Since any fraction with “1” in the denominator is equal to the terms in the numerator, we can say that our original fraction can be simplified to 3x2 + 9x - 1.
Let’s tackle an example that doesn’t necessarily require drawn-out factoring. For the fraction (5x2 + 10x + 20)/10, we may want to divide every term in the numerator by the 10 in the denominator to simplify, even though the “5” coefficient in 5x2 isn’t bigger than 10 and thus can’t have 10 as a factor. Doing so gets us ((5x2)/10) + x + 2. If we like, we may want to rewrite the first term as (1/2)x2 to get (1/2)x2 + x + 2.
Let’s tackle a simple example - √(90). If we think of the number 90 as the product of two of its factors, 9 and 10, we can take the square root of 9 to give the whole number 3 and remove this from the radical. In other words: √(90) √(9 × 10) (√(9) × √(10)) 3 × √(10) 3√(10)
For example, let’s consider the expression 6x3 × 8x4 + (x17/x15). In each occasion where it’s necessary to multiply or divide by exponents, we’ll subtract or add the exponents, respectively, to quickly find a simplified term. See below: 6x3 × 8x4 + (x17/x15) (6 × 8)x3 + 4 + (x17 - 15) 48x7 + x2 For an explanation of why this works, see below: Multiplying exponential terms is essentially like multiplying long strings of non-exponential terms. For example, since x3 = x × x × x and x 5 = x × x × x × x × x, x3 × x5 = (x × x × x) × (x × x × x × x × x), or x8. Similarly, dividing exponential terms is like dividing long strings of non-exponential terms. x5/x3 = (x × x × x × x × x)/(x × x × x). Since each term in the numerator can be canceled out by a matching term in the denominator, we’re left with two x’s in the numerator and none in the bottom, giving us an answer of x2
