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iGCSE (2021 Edition)

3.06 Binomial expansions (Extended)

Lesson

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)(x8)

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)(x8) $=$= $x\times x+x\times\left(-8\right)+4\times x+4\times\left(-8\right)$x×x+x×(8)+4×x+4×(8)

Using the rule

  $=$= $x^2-8x+4x-32$x28x+4x32

Simplifying the products

  $=$= $x^2-4x-32$x24x32

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)

  $=$= $x\times x+x\times4+4\times x+4\times4$x×x+x×4+4×x+4×4

Using the rule

  $=$= $x^2+4x+4x+16$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)(x4)

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)(x4) $=$= $x\times x+x\times\left(-4\right)+4\times x+4\times\left(-4\right)$x×x+x×(4)+4×x+4×(4)

Using the rule

  $=$= $x^2-4x+4x-16$x24x+4x16

Simplifying the products

  $=$= $x^2-16$x216

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)(AB), 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)(AB)=A2AB+BAB2=A2B2. We call this difference of two squares expansion.

 
Example 4

Expand $\left(x+2\right)\left(x-4\right)\left(x+1\right)$(x+2)(x4)(x+1)

Think:  First we will multiply the first two brackets together, then we will multiply the result with the final bracket.

Do:

$\left(x+2\right)\left(x-4\right)\left(x+1\right)$(x+2)(x4)(x+1) $=$= $\left(x\times x+x\times\left(-4\right)+2\times x+2\times\left(-4\right)\right)\left(x+1\right)$(x×x+x×(4)+2×x+2×(4))(x+1)

 

  $=$= $\left(x^2-4x+2x-8\right)\left(x+1\right)$(x24x+2x8)(x+1)

 

  $=$= $\left(x^2-2x-8\right)\left(x+1\right)$(x22x8)(x+1)

 

  $=$= $x^2\times x-2x\times x-8\times x+x^2\times1-2x\times1-8\times1$x2×x2x×x8×x+x2×12x×18×1

 

  $=$= $x^3-2x^2-8x+x^2-2x-8$x32x28x+x22x8

 

  $=$= $x^3-x^2-10x-8$x3x210x8

 

Summary

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)(AB)=A2B2 (called a difference of two squares)

Practice questions

Question 1

$\left(3+4\right)^2=3^2+4^2$(3+4)2=32+42

  1. Is this statement true or false?

    False

    A

    True

    B
  2. What 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)(u5)

Outcomes

0607E2.7

Expansion of brackets, including the square of a binomial.

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