1.08 Binomial expansions
The distributive law says that for any numbers $A,B,$A,B, and $C$C, $A\left(B+C\right)=AB+AC$A(B+C)=AB+AC. However $A$A can also be an expression in brackets, and the distributive law still holds.
Consider the expression $\left(A+B\right)\left(C+D\right)$(A+B)(C+D). If we want to expand this using the distributive law we get $A\left(C+D\right)+B\left(C+D\right)$A(C+D)+B(C+D). If we then expand the brackets in both terms we get $AC+AD+BC+BD$AC+AD+BC+BD. That is, $\left(A+B\right)\left(C+D\right)=AC+AD+BC+BD$(A+B)(C+D)=AC+AD+BC+BD.
Worked examples
Example 1
Expand $\left(x+4\right)\left(x-8\right)$(x+4)(x−8)
Think: We can use the rule $\left(A+B\right)\left(C+D\right)=AC+AD+BC+BD$(A+B)(C+D)=AC+AD+BC+BD.
Do:
| $\left(x+4\right)\left(x-8\right)$(x+4)(x−8) | $=$= | $x\times x+x\times\left(-8\right)+4\times x+4\times\left(-8\right)$x×x+x×(−8)+4×x+4×(−8) |
Using the rule $\left(A+B\right)\left(C+D\right)=AC+AD+BC+BD$(A+B)(C+D)=AC+AD+BC+BD |
| $=$= | $x^2-8x+4x-32$x2−8x+4x−32 |
Simplifying the products |
|
| $=$= | $x^2-4x-32$x2−4x−32 |
Collecting the $x$x terms |
Reflect: After using the rule we can then simplify the expression using any of the algebraic rules that we have learned.
Example 2
Expand $\left(x+4\right)^2$(x+4)2
Think: Since $\left(x+4\right)^2=\left(x+4\right)\left(x+4\right)$(x+4)2=(x+4)(x+4) we can still use the rule $\left(A+B\right)\left(C+D\right)=AC+AD+BC+BD$(A+B)(C+D)=AC+AD+BC+BD.
Do:
| $\left(x+4\right)^2$(x+4)2 | $=$= | $\left(x+4\right)\left(x+4\right)$(x+4)(x+4) |
Since $\left(x+4\right)^2=\left(x+4\right)\left(x+4\right)$(x+4)2=(x+4)(x+4) |
| $=$= | $x\times x+x\times4+4\times x+4\times4$x×x+x×4+4×x+4×4 |
Using the rule $\left(A+B\right)\left(C+D\right)=AC+AD+BC+BD$(A+B)(C+D)=AC+AD+BC+BD |
|
| $=$= | $x^2+4x+4x+16$x2+4x+4x+16 |
Simplifying the products |
|
| $=$= | $x^2+8x+16$x2+8x+16 |
Collecting the $x$x terms |
Reflect: This method will work with any expression squared. In fact, if we generalise to $\left(A+B\right)^2$(A+B)2, then we can see that $\left(A+B\right)^2=A^2+AB+BA+B^2=A^2+2AB+B^2$(A+B)2=A2+AB+BA+B2=A2+2AB+B2. We call this perfect square expansion.
Example 3
Expand $\left(x+4\right)\left(x-4\right)$(x+4)(x−4)
Think: We can use the rule $\left(A+B\right)\left(C+D\right)=AC+AD+BC+BD$(A+B)(C+D)=AC+AD+BC+BD.
Do:
| $\left(x+4\right)\left(x-4\right)$(x+4)(x−4) | $=$= | $x\times x+x\times\left(-4\right)+4\times x+4\times\left(-4\right)$x×x+x×(−4)+4×x+4×(−4) |
Using the rule $\left(A+B\right)\left(C+D\right)=AC+AD+BC+BD$(A+B)(C+D)=AC+AD+BC+BD |
| $=$= | $x^2-4x+4x-16$x2−4x+4x−16 |
Simplifying the products |
|
| $=$= | $x^2-16$x2−16 |
Collecting the $x$x terms |
Reflect: This method will work with any expression in this form. In fact, if we generalise to $\left(A+B\right)\left(A-B\right)$(A+B)(A−B), then we can see that $\left(A+B\right)\left(A-B\right)=A^2-AB+BA-B^2=A^2-B^2$(A+B)(A−B)=A2−AB+BA−B2=A2−B2. We call this difference of two squares expansion.
We can expand the product of two binomial expressions using the rule $\left(A+B\right)\left(C+D\right)=AC+AD+BC+BD$(A+B)(C+D)=AC+AD+BC+BD.
There are two special cases of expanding binomials:
- $\left(A+B\right)^2=A^2+2AB+B^2$(A+B)2=A2+2AB+B2 (called a perfect square)
- $\left(A+B\right)\left(A-B\right)=A^2-B^2$(A+B)(A−B)=A2−B2 (called a difference of two squares)
Practice questions
Question 1
$\left(3+4\right)^2=3^2+4^2$(3+4)2=32+42
Is this statement true or false?
False
ATrue
BWhat whole number needs to be added to the right-hand side to make the statement true?
$\left(3+4\right)^2=3^2$(3+4)2=32$+$+$\editable{}$$+$+$4^2$42
Question 2
Expand the following perfect square:$\left(x+10\right)^2$(x+10)2
Question 3
Expand the following:
$\left(u+5\right)\left(u-5\right)$(u+5)(u−5)