Find the product using the distributive law of multiplication.
 $ \left( x-8 \right)\left( x-2 \right) $

Hint: In order to solve this problem, we need to understand the distributive law of multiplication. The distributive property of multiplication over addition or subtraction can be used when a number is multiplied by a summation or subtraction of numbers. In the case of two brackets of multiplication, we can expand any bracket first. The answer will not change.

Complete step-by-step answer:
Let’s first understand what distributive law of multiplication means.
The distributive property of multiplication over addition or subtraction can be used when a number is multiplied by a summation or subtraction of numbers.
It is expressed mathematically as follows,
 $ a\left( b+c \right)=ab+bc $
Where, a, b, c are just the constants.
In this sum, we have two brackets with subtraction.
We need to open any bracket and perform distributive law.
Opening the first bracket we get,
 $ \left( x-8 \right)\left( x-2 \right)=x\left( x-2 \right)-8\left( x-2 \right) $
Now, we can again use the distributive law of multiplication for each of the brackets,
We can simplify bt,
 $ x\left( x-2 \right)-8\left( x-2 \right)=\left( x\times x \right)+\left( x\times -2 \right)+\left( -8\times x \right)+\left( -8\times -2 \right) $
Simplifying further we get,
 $ \left( x-8 \right)\left( x-2 \right)=\left( x\times x \right)+\left( x\times -2 \right)+\left( -8\times x \right)+\left( -8\times -2 \right) $
Multiplication of two same terms is the square of the same term.
Therefore, $ x\times x={{x}^{2}} $ .
Then the product of a negative and a positive number is always a negative number.
Therefore, $ \left( -8\times x \right)=-8x $ .
And the product of a negative and a negative number is always a positive number.
Therefore, $ \left( -8\times -2 \right)=+16 $ .
 $ \begin{align}
  & \left( x-8 \right)\left( x-2 \right)={{x}^{2}}-2x-8x+16 \\
 & ={{x}^{2}}-10x-16
\end{align} $
Therefore the product using the distributive law of multiplication is $ {{x}^{2}}-10x-16 $

Note: We will get the same answer expanding the second bracket first. We can show that as follows,
\[\begin{align}
  & \left( x-8 \right)\left( x-2 \right)=x\left( x-8 \right)-2\left( x-8 \right) \\
 & =\left( x\times x \right)+\left( x\times -8 \right)+\left( -2\times x \right)+\left( -8\times -2 \right) \\
 & ={{x}^{2}}-8x-2x+16 \\
 & ={{x}^{2}}-10x+16 \\
\end{align}\]
As we can see that the answer does not change irrespective of which bracket we are opening.