Introduction
We've already come across binomial expressions when we looked at how to expand brackets using the distributive law . Expressions such as 2\left(x-3\right) are the product of a term (outside the brackets) and a binomial expression (the sum or difference of two terms). So a binomial is a mathematical expression in which two terms are added or subtracted. They are usually surrounded by brackets or parentheses, such as \left(x+7\right).
Recall that to expand 2\left(x-3\right) we use the distributive law: A\left(B+C\right)=AB+AC
Now we want to look at how to multiply two binomials together, such as \left(ax+b\right) \left(cx+d \right).
Multiply two binomials
When we multiply binomials of the form \left(ax+b\right) \left(cx+d \right) we can treat the second binomial \left(cx+d \right) as a constant term and apply the distributive property in the form \left(B+C\right) \left(A \right) = BA + CA. The picture below shows this in action:
As you can see in the picture, we end up with two expressions of the form A\left(B+C\right).
We can expand these using the distributive property again to arrive at the final answer:
\begin {aligned} ax\left(cx+d\right) + b\left(cx+d\right) &= acx^{2} +adx+bcx+bd \\ &= acx^{2} + \left(ad+bc \right)x + bd \end {aligned}
Examples
Example 1
Expand and simplify \left(x+6\right)\left(x-12\right).
Example 2
Expand and simplify the following: 3\left(y+3\right)\left(y+5\right)
Example 3
Fill in the blanks to make the expression true. (t+6)(⬚+⬚)=t^{2} +17t+66
To multiply two binomials together we apply the distributive law twice:
\begin {aligned} (ax+b)(cx+d) &= ax\left(cx+d\right) + b\left(cx+d\right) \\ &= acx^{2} +adx+bcx+bd \\ &= acx^{2} + \left(ad+bc \right)x + bd \end {aligned}