Distributive laws and the difference of squares

Some prod­ucts come up so often in alge­bra that it pays to know their answers by heart, the way you know $7 \times 8 = 56$ with­out work­ing it out. These ready-made results are called alge­braic iden­ti­ties. Every one of them comes from a sin­gle tool: the dis­trib­u­tive law. This les­son sets out the dis­trib­u­tive laws in all their forms, then proves and uses the first iden­tity, $(a + b)(a - b) = a^2 - b^2$.

The dis­trib­u­tive laws#

The dis­trib­u­tive law says that a quan­tity out­side a bracket mul­ti­plies every term inside it. Here are its eight forms.

$$\begin{aligned} a(b + c) &= ab + ac \\ a(b - c) &= ab - ac \\ (a + b)c &= ac + bc \\ (a - b)c &= ac - bc \end{aligned}$$

When both fac­tors are brack­ets, each term of the first mul­ti­plies each term of the sec­ond, so there are four prod­ucts:

$$\begin{aligned} (a + b)(c + d) &= ac + ad + bc + bd \\ (a + b)(c - d) &= ac - ad + bc - bd \\ (a - b)(c + d) &= ac + ad - bc - bd \\ (a - b)(c - d) &= ac - ad - bc + bd \end{aligned}$$

Look at the signs in the last line. $(-b) \times (-d) = +bd$, because a neg­a­tive times a neg­a­tive is pos­i­tive. Every sign in these four lines fol­lows from the sign rules for inte­gers, so you do not need to learn them sep­a­rately. Just mul­ti­ply term by term and keep each sign with its term.

Exam­ple 1#

Find $(2x + 3)(x - 5)$.

$$\begin{aligned} (2x + 3)(x - 5) &= 2x \times x - 2x \times 5 + 3 \times x - 3 \times 5 \\ &= 2x^2 - 10x + 3x - 15 \\ &= 2x^2 - 7x - 15 \end{aligned}$$

What an iden­tity is#

An equa­tion such as $2x + 1 = 7$ is true for only one value of $x$, namely $x = 3$. An iden­tity is an equal­ity that is true for every value of the let­ters in it. For exam­ple, $2(x + 3) = 2x + 6$ is true what­ever num­ber $x$ is. Iden­ti­ties are proved by alge­bra, and once proved they can be used as short­cuts.

Iden­tity 1: the prod­uct of a sum and a dif­fer­ence#

$$(a + b)(a - b) = a^2 - b^2$$

Proof. Use the dis­trib­u­tive law:

$$\begin{aligned} (a + b)(a - b) &= a^2 - ab + ab - b^2 \\ &= a^2 - b^2 \end{aligned}$$

The two mid­dle terms, $-ab$ and $+ab$, can­cel each other. That is why only two terms are left.

In words: (first + sec­ond)(first − sec­ond) = first squared − sec­ond squared. The first and sec­ond terms can be any­thing: num­bers, let­ters, or whole terms such as $2x$ or $3y$.

Check with num­bers: take $a = 5$ and $b = 3$. Then $(5 + 3)(5 - 3) = 8 \times 2 = 16$, and $5^2 - 3^2 = 25 - 9 = 16$. They agree.

When a whole term is squared, every fac­tor in it is squared, because $(ab)^n = a^n b^n$. So $(2x)^2 = 4x^2$ and $(3y)^2 = 9y^2$.

Exam­ple 2#

Find $(2x + 3y)(2x - 3y)$.

Here the first term is $2x$ and the sec­ond is $3y$.

$$\begin{aligned} (2x + 3y)(2x - 3y) &= (2x)^2 - (3y)^2 \\ &= 4x^2 - 9y^2 \end{aligned}$$

Exam­ple 3#

Find $(2a^3 + b^3)(2a^3 - b^3)$.

$$\begin{aligned} (2a^3 + b^3)(2a^3 - b^3) &= (2a^3)^2 - (b^3)^2 \\ &= 4a^6 - b^6 \end{aligned}$$

Squar­ing $a^3$ dou­bles its power: $(a^3)^2 = a^3 \times a^3 = a^6$.

Exam­ple 4#

Find $\displaystyle \left(\frac{x}{2} + \frac{1}{3}\right)\left(\frac{x}{2} - \frac{1}{3}\right)$.

$$\displaystyle \left(\frac{x}{2}\right)^2 - \left(\frac{1}{3}\right)^2 = \frac{x^2}{4} - \frac{1}{9}$$

Using the iden­tity for num­bers#

The iden­tity also makes some mul­ti­pli­ca­tions of num­bers quick. Write the num­bers as a sum and a dif­fer­ence of the same two num­bers.

Exam­ple 5#

Find $103 \times 97$ with­out long mul­ti­pli­ca­tion.

$103 = 100 + 3$ and $97 = 100 - 3$.

$$103 \times 97 = (100 + 3)(100 - 3) = 100^2 - 3^2 = 10000 - 9 = 9991$$

Exam­ple 6#

Find $52^2 - 48^2$.

Read the iden­tity the other way round: $a^2 - b^2 = (a + b)(a - b)$.

$$52^2 - 48^2 = (52 + 48)(52 - 48) = 100 \times 4 = 400$$

This is much quicker than work­ing out $2704 - 2304$.

Com­mon mis­takes#

• Squar­ing only the let­ter: $(3y)^2$ is $9y^2$, not $3y^2$.
• Writ­ing $(a + b)(a - b) = a^2 + b^2$. The answer is a dif­fer­ence, $a^2 - b^2$.
• Using the iden­tity when the two brack­ets do not have the same two terms. $(2x + 3)(3x - 2)$ is not of this form; mul­ti­ply it out term by term.

Prac­tice#

1. Find $(3x + 2)(x - 4)$.
2. Find $(a - 2b)(c - 3d)$.
3. Find $(x + 7)(x - 7)$.
4. Find $(5m + 4n)(5m - 4n)$.
5. Find $(3p^2 + 2q)(3p^2 - 2q)$.
6. Find $\displaystyle \left(\frac{2a}{3} + \frac{b}{4}\right)\left(\frac{2a}{3} - \frac{b}{4}\right)$.
7. Find $105 \times 95$ using an iden­tity.
8. Find $63^2 - 37^2$ using an iden­tity.
9. Find $(x^2 + y^2)(x + y)(x - y)$.

Answers#