IB MYP 3 2021 Edition

2.04 Expansion

Lesson

Normally, when an expression has a multiplication and an addition or subtraction, for example $5+8\times9$5+8×9, we evaluate the multiplication first. The exception is when the addition or subtraction is in brackets, for example, $\left(5+8\right)\times9$(5+8)×9.

It will help to visualise a rectangle with a height of $9$9 cm and a width of $5+8$5+8 cm.

The rectangle has an area of $\left(5+8\right)\times9$(5+8)×9 cm^{2}. We can work this out as follows.

$\left(5+8\right)\times9$(5+8)×9 | $=$= | $13\times9$13×9 |
Evaluate the addition in the brackets first |

$=$= | $117$117 cm^{2} |
Evaluate the multiplication |

However, we can see that the rectangle is made up of two smaller rectangles, one with area $5\times9$5×9 cm^{2} and the other with area $8\times9$8×9 cm^{2}. So we can also work out the total area like this.

$5\times9+8\times9$5×9+8×9 | $=$= | $45+72$45+72 |
Evaluate the multiplications |

$=$= | $117$117 cm^{2} |
Evaluate the addition |

So $\left(5+8\right)\times9=5\times9+8\times9$(5+8)×9=5×9+8×9. This can be extended to any other numbers.

If $A,B$`A`,`B`, and $C$`C` are any numbers then $A\left(B+C\right)=AB+AC$`A`(`B`+`C`)=`A``B`+`A``C`. This is known as the distributive law.

The distributive law is particularly useful for algebraic expressions where we can't evaluate the expression in the brackets.

Expand $7\left(x-12\right)$7(`x`−12).

**Think:** Expand means to write an algebraic expression without brackets. We can expand this expression using the distributive law.

**Do:**

$7\left(x-12\right)$7(x−12) |
$=$= | $7\times x+7\times\left(-12\right)$7×x+7×(−12) |
Use the distributive law, $A\left(B+C\right)=AB+AC$ Here, $A=7,B=x,$ |

$=$= | $7x-84$7x−84 |
Evaluate the multiplication |

**Reflect:** Because of the distributive law we know that both sides of the equation are equal. But now we have a way to write an equal expression without brackets.

We had to be careful of the negative sign here. Because $A$`A` is positive and $C$`C` is negative, $AC$`A``C` is negative.

To solve the previous example we could also use a slightly different version of the rule that accounts for the negative sign: $A\left(B-C\right)=AB-AC$`A`(`B`−`C`)=`A``B`−`A``C`. Notice that in this case we are assuming $C$`C` is positive, but we are taking away $AC$`A``C`.

Summary

We can use the distributive law to expand an algebraic expression brackets like so:

$A\left(B+C\right)=AB+AC$`A`(`B`+`C`)=`A``B`+`A``C`,

and if the second term in the brackets is negative:

$A\left(B-C\right)=AB-AC$`A`(`B`−`C`)=`A``B`−`A``C`

Expand the expression $9\left(5+w\right)$9(5+`w`).

Expand the expression $\left(y+8\right)\times7$(`y`+8)×7.

Expand the expression $-8\left(c-5\right)$−8(`c`−5).