Multiplying Two Binomials -

When multiplying a binomial (an expression with two terms) by another binomial, we end up with four terms..

Example 1: Integers – we can multiply integers using brackets.

$27 \times 34=(20+7)(30+4)$

Let’s suppose the numbers are actually the length and width of a rectangle.

The total area of this rectangle is the sum of these four calculations:

$\begin{align*}7 \times 30 &= 210\\[10pt]7 \times 4 &= 28\\[10pt]20\times 30&=600\\[10pt]20\times 4 &= 80\end{align*}$

Which is: $210+28+600+80=918$ .

In other words, $(20+7)(30+4)=600+80+210+28$ .

A binomial multiplied by a binomial yields four terms.

Example 2: In general.

To multiply a binomial by binomial we are multiplying

$(a+b)(c+d)$

…where the $a, b, c$ and $d$ can be any number or variable or mixture of numbers and variables.

Using a rectangle diagram, let the width be $a+b$ units; the length be $c+d$ units.

The total area of the rectangle is $ac+ad+bc+bd$ .

Example 3:

Multiply $(x+4)(x+5)$ .

Here is a rectangle model of this expression:

Add the four terms:

$x^2+4x+5x+20$ .

Gather the like terms: $4x+5x=9x$

Therefore,

$(x+4)(x+5)=x^2+9x+20$

Example 4: Expand and simplify without drawing a rectangle.

Suppose we wish to expand and simplify:

$(x+2)(x+6)$

Everything in the first bracket multiplies everything in the second bracket. We can write this as follows:

$\begin{align*}&(x+2)(x+6)\\[10pt]=&x(x+6)+2(x+6)\\[10pt]=&x^2+6x+2x+12\\[10pt]=&x^2+8x+12\end{align*}$

This technique can be used when multiplying out any brackets. See examples 9 & 10.

Example 5: The algorithm FOIL: First Outside Inside Last

This algorithm helps to calculate all four terms efficiently. It is very nice but can only be used when multiplying two terms by two terms.

We multiply the first terms in each bracket to get $x^2$

We multiply the outside terms in each bracket to get $6x$

We multiply the inside terms in each bracket to get $2x$

We multiply the last terms in each bracket to get $12$ .

Then we add those terms together and simplify to $x^2+8x+12$ .

Example 6: Watch out!

Remember that a square number means multiply the number by the number? So…

$(x-5)^2$

is expanded as follows:

$\begin{align*}&(x-5)^2\\[10pt]=&(x-5)(x-5)\\[10pt]=&x^2-5x-5x+25\\[10pt]=&x^2-10x+25\end{align*}$

Notice that the answer is not simply $x^2+5^2$ – exponents do not distribute over addition.

Example 7: When the first factor is a whole number:

$\begin{align*}&3(x-10)(2x-1)\\[10pt]=&3(2x^2-x-20x+10)\\[10pt]=&3(2x^2-21x+10)\\[10pt]=&6x^2-63x+30\end{align*}$

alternatively, multiply the three into the first bracket:

$\begin{align*}&3(x-10)(2x-1)\\[10pt]=&(3x-30)(2x-1)\\[10pt]=&6x^2-3x-60x+30\\[10pt]=&6x^2-63x-30\end{align*}$

Example 8: When the first factor is negative one:

$\begin{align*}&-(x-10)(2x-1)\\=&-(2x^2-x-20x+10)\\[10pt]=&-(2x^2-21x+10)\\[10pt]=&-2x^2+21x-10\end{align*}$

Practice

Use paper and pencil before entering your answer.

Applet link: https://www.geogebra.org/m/udx9xygp

Practice Expanding brackets of all kinds on tranum maths: https://www.transum.org/software/SW/Starter_of_the_day/Students/Brackets.asp?Level=9

Example 9: Extending the principle

In example 4 we said that ‘everything in the first bracket multiplies everything in the second.’

The method used in example four can be used for any bracket multiplication. Let’s expand

$\begin{align*}&(x+y+3)(x+2)\\[10pt]=&x(x+2)+y(x+2)+3(x+2)\\[10pt]=&x^2+2x+yx+2y+3x+6\\[10pt]=&x^2+5x+yx+2y+6\end{align*}$

Example 10: Three brackets

For three brackets, we simply multiply out the first two then we have two brackets. Etc.

$\begin{align*}&(x+2)(x+3)(x+6)\\[10pt]=&(x^2+3x+2x+6)(x+6)\\[10pt]=&(x^2+5x+6)(x+6)\\[10pt]=&x^2(x+6)+5x(x+6)+6(x+6)\\[10pt]=&x^3+6x^2+5x^2+30x+6x+36\\[10pt]=&x^3+11x^2+36x+36\end{align*}$

Try the ten questions at the bottom of this MathisFun page here.