# How do you multiply (1+3a)(1-3a)?

Aug 12, 2016

The answer to this expression would be $1 - 9 {a}^{2}$ .

#### Explanation:

When multiplying $\left(1 + 3 a\right) \left(1 - 3 a\right)$ you are multiplying binomials. In addition, you can use the strategy FOIL (which stands for First, Outers, Inners and Last) to solve this problem.
For instance, in the example above (letter C), you start by multiplying $a \cdot a$ which gives you ${a}^{2}$ (the Firsts). After that you multiply $a$ with 5 which would give you $5 a$ (the Outers).

Later on, you multiply $3$ with $a$, which would give you $3 a$ (the Inners).

Finally, you multiply $3$ with $5$ which would give you $15$ (the Lasts) and then all you have to do is combine like terms so, that the final answer is ${a}^{2} + 8 a + 15$.

Now apply this new knowledge to the problem $\left(1 + 3 a\right) \left(1 - 3 a\right)$.

1.Multiply $1$ with $1$ (the Firsts) $= 1$
2.Multiply $1$ with $- 3 a$ (the Outers). $= - 3 a$
3.Multiply $3 a$ with $1$ (the Inners). $= 3 a$
4.Multiply $3 a$ with $- 3 a$ (the Lasts). $= - 9 {a}^{2}$

Now combine like terms.

$1 + \left(- 3 a\right) + 3 a + \left(- 9 {a}^{2}\right)$
$1 - 9 {a}^{2}$

The answer is $1 - 9 {a}^{2}$.

Aug 12, 2016

$\left(1 + 3 a\right) \left(1 - 3 a\right) = 1 - 9 {a}^{2}$
Difference of squares

#### Explanation:

If you look at the 2 factors carefully you should recognise them as the factors to the difference of squares.

If ${x}^{2} - {y}^{2} = \left(x + y\right) \left(x - y\right)$ , then if you recognise

$\left(x + y\right) \left(x - y\right)$ you will be able to write ${x}^{2} - {y}^{2}$ without having to multiply out in full.

Similarly here.

$\left(1 + 3 a\right) \left(1 - 3 a\right) = 1 - 9 {a}^{2}$

Aug 18, 2016

${1}^{2} - {\left(3 a\right)}^{2}$

#### Explanation:

The objective of this question is to teach you that
$\left({a}^{2} - {b}^{2}\right) = \left(a - b\right) \left(a + b\right)$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b l u e}{\left(1 + 3 a\right) \textcolor{b r o w n}{\left(1 - 3 a\right)}}$

Multiply everything in the right hand side (RHS) bracket by everything inside the left hand side (LHS) bracket.

Note that the + from $+ 3 a$ follows the $3 a$

$\textcolor{b r o w n}{\textcolor{b l u e}{1} \left(1 - 3 a\right) \text{ } \textcolor{b l u e}{+ 3 a} \left(1 - 3 a\right)}$

$\text{ "1-3a" } + 3 a - 9 {a}^{2}$

But $- 3 a + 3 a = 0$ giving:

$1 + 0 - 9 {a}^{2}$

$1 - 9 {a}^{2}$

But $1 = {1}^{2}$ and $9 {a}^{2} = {\left(3 a\right)}^{2}$ giving:

${1}^{2} - {\left(3 a\right)}^{2} \text{ "->" "1-(3a)^2 " }$ if you prefer!