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2 Algebra
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Australia
Year 9

2.03 Binomial expansion

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Introduction

We've already come across binomial expressions when we looked at how to expand brackets using the  distributive law  . Expressions such as 2(x−3)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 (x+7)\left(x+7\right).

Recall that to expand 2(x−3)2\left(x-3\right) we use the distributive law: A(B+C)=AB+ACA\left(B+C\right)=AB+AC

Now we want to look at how to multiply two binomials together, such as (ax+b)(cx+d)\left(ax+b\right) \left(cx+d \right).

Multiply two binomials

When we multiply binomials of the form (ax+b)(cx+d)\left(ax+b\right) \left(cx+d \right) we can treat the second binomial (cx+d)\left(cx+d \right) as a constant term and apply the distributive property in the form (B+C)(A)=BA+CA\left(B+C\right) \left(A \right) = BA + CA. The picture below shows this in action:

This image shows how the distributive property applies to multiplying binomials. Ask your teacher for more information.

As you can see in the picture, we end up with two expressions of the form A(B+C)A\left(B+C\right).

We can expand these using the distributive property again to arrive at the final answer:

ax(cx+d)+b(cx+d)=acx2+adx+bcx+bd=acx2+(ad+bc)x+bd\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 (x+6)(x−12)\left(x+6\right)\left(x-12\right).

Worked Solution
Create a strategy

Use the distributive law to expand the expression.

Apply the idea
(x+6)(x−12)\displaystyle \left(x+6\right)\left(x-12\right)=\displaystyle =x(x−12)+6(x−12)\displaystyle x\left(x-12\right) + 6 \left(x-12\right)Use the distributive property
=\displaystyle =x2−12x+6x−72\displaystyle x^{2} -12x+6x-72Distribute both sets of brackets
=\displaystyle =x2−6x−72\displaystyle x^{2} -6x-72Simplify like terms

Example 2

Expand and simplify the following: 3(y+3)(y+5)3\left(y+3\right)\left(y+5\right)

Worked Solution
Create a strategy

Use the distributive law to expand the expression.

Apply the idea
3(y+3)(y+5)\displaystyle 3\left(y+3\right)\left(y+5\right)=\displaystyle =(3y+9)(y+5)\displaystyle \left(3y+9\right) \left(y+5\right)Multiply (y+3)\left(y+3\right) by 33
=\displaystyle =3y(y+5)+9(y+5)\displaystyle 3y\left(y+5\right) + 9\left(y+5\right)Use the distributive law
=\displaystyle =3y2+15y+9y+45\displaystyle 3y^{2} +15y +9y +45Expand both sets of brackets
=\displaystyle =3y2+24y+45\displaystyle 3y^{2} +24y +45Simplify like terms

Example 3

Fill in the blanks to make the expression true. (t+6)(⬚+⬚)=t2+17t+66(t+6)(⬚+⬚)=t^{2} +17t+66

Worked Solution
Create a strategy

Consider what the terms in the first bracket need to be multiplied by to get the terms on the right hand side.

Apply the idea

On the right hand side of the equation, the t2t^{2} term tells us that we need to multiply tt by something to get t2.t^2.

But t×t=t2t\times t=t^2 so the first missing term is tt: (t+6)(t+⬚)=t2+17t+66(t+6)(t+⬚)=t^{2} +17t+66

The constant term of 6666 tells us that we need to multiply 66 by something to get 66.66.

But 6×11=666\times 11=66 so the second missing term is 1111: (t+6)(t+11)=t2+17t+66(t+6)(t+11)=t^{2} +17t+66

Idea summary

To multiply two binomials together we apply the distributive law twice:

(ax+b)(cx+d)=ax(cx+d)+b(cx+d)=acx2+adx+bcx+bd=acx2+(ad+bc)x+bd\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}

Outcomes

ACMNA213

Apply the distributive law to the expansion of algebraic expressions, including binomials, and collect like terms where appropriate

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