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How to Expand and Simplify an Expression

Lesson 1: Overview

Expansion is done using the distributive law of multiplication given below:

a(b+c)=ab+ac

This law states that when we have a factor outside a pair of parentheses, we distribute the factor by multiplying it by each of the term in the parentheses and add the resulting terms together.

If we have more than one term in the first factor, we multiply every term in the first factor with every term in the second factor and add the resulting terms together. For instance,

\begin{aligned} (a+b)(c+d) &=ac+ad+bc+bd\end{aligned}

In the example above, we multiply a with the two terms in the second factor (c and d) to get ac+ad. We also multiply b with the two terms in the second factor (c and d) to get bc+bd. Finally, we add the results together to get ac+ad+bc+bd.

Next, when multiplying negative terms, we use the rules below,

\begin{aligned} (+)(-)&=(-)\\(-)(+)&=(-) \\(-)(-)&=(+)\end{aligned}

We determine whether a term is positive or negative based on the sign in front of it. For instance, if we have the expression 6x-3+5y

  • The first term is positive (i.e., 6x)
  • The second term is negative (i.e., -3)
  • The third term is positive (i.e., 5y)

Finally, adding a negative term is equivalent to doing a subtraction. In other words

p+(-q)=p-q

The remaining lessons provide examples of the following topics:

  1. Distributing a Positive Factor
  2. Distributing a Negative Factor
  3. Expansion with Multiple Terms in the First Factor
  4. Combining Expansions

Lesson 2: Distributing a Positive Factor

Distributing a positive factor is straightforward. Let’s look at some examples.

Example 1

\begin{aligned}5(2x+9)&=(5)(2x)+(5)(9)\\&=10x+45\end{aligned}

Here, we distribute 5 to both terms (2x and 9) in the parentheses and add the resulting terms together.

Example 2

\begin{aligned} 3x(x-2)&=(3x)(x)+(3x)(-2)\\&=3x^2+(-6x)\\&=3x^2-6x \end{aligned}

Here, we distribute 3x to both terms (x and -2) in the parentheses and add the resulting terms together.

Next, we simplify +(-6x) to -6x.

Note that 3x\times x=3x^2 .

Example 3

\begin{aligned} 3y(x+9y-5)&=(3y)(x)+(3y)(9y)+(3y)(-5)\\&=3yx+27y^2+(-15y)\\&=3yx+27y^2-15y \end{aligned}

For this example, we distribute 3y to the three terms (x, 9y and -5) in the parentheses and add the resulting terms together.

Next, we simplify +(-15y) to -15y.

Note that 3y\times 9y=27y^2.

Lesson 3: Distributing a negative factor

Distributing a negative factor is similar to distributing a positive one, except that a negative factor toggles the signs of the terms that it is distributed to. Let’s illustrate this with examples.

Example 1

\begin{aligned} -(3x-9+6y)=-3x+9-6y\end{aligned}

For this example, we have a negative sign outside the pair of parentheses. Distributing a negative sign simply toggles the signs of the terms in the parentheses.

  • 3x becomes -3x
  • -9 becomes 9
  • 6y becomes -6y

Example 2

\begin{aligned} -3(6x+5)&=(-3)(6x)+(-3)(5)\\&=-18x+(-15)\\&=-18x-15 \end{aligned}

Here, we distribute -3 to the two terms (6x and 5) in the parentheses and add the resulting terms together.

  • Multiplying -3 with 6x gives us -18x.
  • Multiplying -3 with 5 gives us -15.

Next, we simply +(-15) to -15.

Note that we initially had two positive terms in the parentheses (6x and 5). After distribute the negative factor, we end up with two negative terms.

Example 3

\begin{aligned} -7(3x-2)&=(-7)(3x)+(-7)(-2)\\&=-21x+14\end{aligned}

Here, we distribute -7 to the two terms (3x and -2) in the parentheses and add the resulting terms together.

  • Multiplying -7 with 3x gives us -21x.
  • Multiplying -7 with -2 gives us 14.

Example 4

\begin{aligned} -5x(2x-6+9y)&=(-5x)(2x)+(-5x)(-6)+(-5x)(9y)\\&=-10x^2+30x+(-45xy)\\&=-10x^2+30x-45xy \end{aligned}

For this example, we distribute -5x to the three terms (2x, -6 and 9y) in the parentheses and add the resulting terms together.

Next, we simplify +(-45xy) to -45xy.

Lesson 4: Expansion with Multiple Terms in the First Factor

If the first factor has more than one term, we need to multiply all the terms in the first factor with all the terms in the second.

Example 1

\begin{aligned} (3x+y)(x+2y)&=(3x)(x)+(3x)(2y)+(y)(x)+(y)(2y)\\&=3x^2+6xy+yx+2y^2\\&=3x^2+7xy+2y^2\end{aligned}

Here, we first multiply 3x with the two terms (x and 2y) in the second factor. Next, we multiply y with the same two terms.

After multiplying, we add the terms together.

Finally, we simplify the two like terms (6xy+xy) in the second last step to get 7xy in the last step.

Note that yx=xy.

Example 2

\begin{aligned} &(2x-3)(x+5-y)\\&=(2x)(x)+(2x)(5)+(2x)(-y)+(-3)(x)+(-3)(5)+(-3)(-y)\\&=2x^2+10x+(-2xy)+(-3x)+(-15)+3y \\&=2x^2+10x-2xy-3x-15+3y \\&=2x^2+7x-2xy-15+3y\end{aligned}

Here. we first multiply 2x with the three terms (x, 5 and -y) in the second factor.

Next, we multiply -3 with the same three terms.

After multiplying, we add the terms together.

Next, we simplify +(-2xy), +(-3x) and +(-15) to -2xy, -3x and -15, respectively.

Finally, we simplify the two like terms 10x-3x in the second last step to get 7x in the last step.

Lesson 5: Combining Products

Sometimes, an expression may have more than one product. To simplify the expression, we expand the products first and add (or subtract) the resulting terms. Let’s look at two examples.

Example 1

\begin{aligned}5(3x-2)+6(x+7)&=(5)(3x)+(5)(-2)+(6)(x)+(6)(7)\\&=15x+(-10)+6x+42\\&=15x-10+6x+42\\&=21x+32\end{aligned}

For this example, we have two products, 5(3x-2) and 6(x+7). We need to expand the two products and add the resulting terms together.

To expand 5(3x-2), we distribute 5 to the two terms in (3x-2).

To expand 6(x+7), we distribute 6 to the two terms in (x+7).

After distributing, we add the resulting terms together.

Next, we simplify +(-10) to -10.

Finally, we simplify the like terms 15x+6x and -10+42 in the second last step to get 21x and 32 in the last step.

Example 2

\begin{aligned} 3x(x-4)-6(x+5)&=(3x^2-12x)-(6x+30)\\&=3x^2-12x-6x-30\\&=3x^2-18x-30\end{aligned}

Here, we have two products: 3x(x-4) and 6(x+5); the second product is subtracted from the first.

To expand the first product, we distribute 3x to the two terms in (x-4). That gives us 

\begin{aligned}(3x)(x)+(3x)(-4)&=3x^2+(-12x)\\&=3x^2-12x\end{aligned}

To expand the second product, we distribute 6 to the two terms in (x+5). That gives us 

\begin{aligned}(6)(x)+(6)(5)&=6x+30\end{aligned}

After expanding the two products, we subtract the result of the second product from the first product.

As we need to do a subtraction, it is important that we enclose the results of the second product in a pair of parentheses, as we have to subtract both terms (6x and 30) from the result of the first product. Without the parentheses, only the first term 6x is subtracted, which is incorrect.

An alternative method to expand 3x(x-4)-6(x+5) is to interpret is as 3x(x-4)+[-6(x+5)].

When interpreted as such, we have a sum of two products: 3x(x-4) and -6(x+5).

We can expand the two products and add the resulting terms together.

The alternative solution is as follows:

\begin{aligned} 3x(x-4)-6(x+5)&=(3x)(x)+(3x)(-4)+(-6)(x)+(-6)(5)\\&=3x^2+(-12x)+(-6x)+(-30)\\&=3x^2-12x-6x-30\\&=3x^2-18x-30\end{aligned}

Lesson 6: Presenting Our Answers

In the preceding examples, we provided a step-by-step breakdown of the expansion process, which can be particularly helpful when you’re new to this concept.

However, once you are familiar with the concept, you can skip most of the steps.

For instance, for the last example above, you can just present your answer as

\begin{aligned} 3x(x-4)-6(x+5)&=3x^2-12x-6x-30\\&=3x^2-18x-30\end{aligned}

You do not have to write (3x)(-4) as +(-12x) since you know that it simplifies to -12x.

Similarly, you do not have to write (-6)(x) as +(-6x) or (-6)(5) as +(-30).

Lesson 7: Expansion Formulas

Last but not least, there are some expansion formulas that are useful to know.

The commonly used ones are:

\begin{aligned}(a+b)^2&=a^2+2ab+b^2&&(1)\\ (a-b)^2&=a^2-2ab+b^2&&(2)\\(a+b)(a-b)&=a^2-b^2&&(3)\\(a+b)^3&=a^3+3a^2b+3ab^2+b^3 &&(4)   \end{aligned}

We can prove the formulas by expanding the left-hand side. For instance, for the first formula, we have

\begin{aligned}  (a+b)^2&=(a+b)(a+b)\\&=(a)(a)+(a)(b)+(b)(a)+(b)(b)\\&=a^2+ab+ba+b^2\\&=a^2+2ab+b^2  \end{aligned}

Let’s look at how we can apply the above formulas.

Example 1

\begin{aligned} (2x-1)^2&=(2x)^2-2(2x)(1)+(1)^2\\&= 4x^2-4x+1 \end{aligned}

Here, we use formula (2) to expand (2x-1)^2, by letting a=2x and b=1.

Note that (2x)^2=(2x)(2x)=4x^2.

Example 2

\begin{aligned} (2+5x)^3&=(2)^3+3(2)^2(5x)+3(2)(5x)^2+(5x)^3\\&=8+3(4)(5x)+3(2)(25x^2)+125x^3\\&=8+60x+150x^2+125x^3   \end{aligned}

Here, we use formula (4) to expand (2+5x)^3, by letting a=2 and b=5x.

Note that (5x)^2=(5x)(5x)=25x^2 and (5x)^3=(5x)(5x)(5x)=125x^3.