# How do you find the product (3/4k+8)^2?

Nov 5, 2017

${\left(\frac{3}{4} k + 8\right)}^{2} = \frac{9}{36} {k}^{2} + 12 k + 64$

#### Explanation:

${\left(x + y\right)}^{2} = {x}^{2} + 2 x y + {y}^{2}$

Nov 5, 2017

$\frac{9 {k}^{2}}{16} + 12 k + 64$

#### Explanation:

The mistake most people make is simply squaring both terms individually, so you to avoid this error, you should start by writing out the binomial like this:
$\left(\frac{3}{4} k + 8\right) \left(\frac{3}{4} k + 8\right)$

This is because it was squared; if it was cubed, you'd write it thrice, and so on.

This makes it easier to visualise what exactly you're doing, therefore you're less likely to make mistakes when solving it.

Now, I don't know if you've heard of FOIL method but I'll link a video here so you can understand it better before continuing with the question: enter link description here

So, using the FOIL method, we need to multiply:
the first terms of each binomial. i.e. $\frac{3}{4} k \cdot \frac{3}{4} k$ which is $\frac{9 {k}^{2}}{16}$
Then the outer terms, i.e. $\frac{3}{4} k \cdot 8$ which gives $\frac{24 k}{4}$
Then the inside terms, i.e $8 \cdot \frac{3}{4} k$ which is $\frac{24 k}{4}$
And finally the last terms, i,e, $8 \cdot 8$ which is $64$

Now you'll need to add these terms:
$\frac{9 {k}^{2}}{16} + \frac{24 k}{4} + \frac{24 k}{4} + 64$

[you can add the terms with the same variable; but remember: you can't add ${k}^{2}$ and $k$ because they have different powers.]

So,
$\frac{9 {k}^{2}}{16} + \frac{48 k}{4} + 64$

You can also now simplify the individual fractions, which is:
$\frac{9 {k}^{2}}{16} + 12 k + 64$