# How does determining the area of rectangle relate to binomial multiplication?

So using the strategy to determine the area of the large rectangle I simply did

$10\times10, 10\times2, 10\times4, 2\times4$ to get $168\mathrm{cm}$ total.

The next question goes on to ask how this strategy relates to binomial multiplication.

$14\times12 = 10\times10 + 10\times2 + 10\times4 + 2\times4$

I'm confused as to how this question relates to binomial multiplication because it doesn't look like it can be put in a for such as $(5x+2)(3x-1)$.

Take your second diagram and replace $10$ cm and $2$ cm along the top with $5x$ and $2$, respectively, and similarly replace $10$ cm and $4$ cm on the side with $3x$ and $-1$. Now, this is a little strange since $-1$ is negative, but the principle remains the same.

The total area of the rectangle is $(5x+2)(3x-1)$, which you can see as the area of four smaller rectangles $5x \cdot 3x$, etc.

$10 \times 10 + 10 \times 2 = 10(10 + 2)$
Do the same for the other two terms $10 \times 4 + 2 \times 4$ (pull out common factor). You should then have a common binomial to factor out leading to 2 binomials multiplied together.
You may be expecting to see a variable like $x$ (based on your last sentence) in your binomials but you won't in this case. Should just have numbers in each binomial.
Binomial multiplication is really anything that can be written in the form $(a+b)\times(c+d)$. We already know that the area of the rectangle is given by $12\times14$, so you just need to split $12$ into the right $a$ and $b$ and split $14$ into the right $c$ and $d$.