Question

For 5x+20 and the multiplicative inverse of 5, how do you write the product and then write the expression in standard form?

Answer 1

Hint: We first find the multiplicative inverse of 5. Then we take the multiplication of $5x+20$ and $\dfrac{1}{5}$. We have to find the standard form for which we take the factored form of $5x+20$.
We find a linear expression as a solution.

Complete step-by-step solution:
There are two terms given. One is $5x+20$ and the other one is a multiplicative inverse of 5.
We first find the multiplicative inverse of 5.
If $a$ be the multiplicative inverse of $b$ then $b=\dfrac{1}{a}$.
For 5, multiplicative inverse would be $\dfrac{1}{5}$.
Now we have to take the product of these two to find the solution.
Multiplication of $5x+20$ and $\dfrac{1}{5}$ gives $\left( 5x+20 \right)\times \dfrac{1}{5}=\dfrac{5x+20}{5}$.
We now have to find the standard form of $\dfrac{5x+20}{5}$.
We now find the factored form of $5x+20$.
The only process that is available for this equation to factorise is to take a common constant out of the terms 20 and $5x$.
Now we are actually taking the constant from 20 and 5x. We are finding the maximum possible constant to take out. It will be the greatest common divisor of the numbers 20 and 5.
Now, we need to find the GCD of 20 and 5.
$\begin{align}
  & 5\left| \!{\underline {\,
  5,20 \,}} \right. \\
 & 1\left| \!{\underline {\,
  1,4 \,}} \right. \\
\end{align}$
Therefore, the GCD will be 5.
Therefore, the factorisation is $5x+20=5\left( x+4 \right)$.
The standard form will be $\dfrac{5x+20}{5}=\dfrac{5\left( x+4 \right)}{5}=\left( x+4 \right)$.

Note: We can verify the result of the factorisation by taking an arbitrary value of x where $x=2$.
We put $x=2$ on the left-hand side of the equation $5x+20$ and get
$5x+20=5\times 2+20=10+20=30$
Now we put $x=2$ on the right-hand side of the equation $5\left( x+4 \right)$ and get
$5\left( x+4 \right)=5\left( 2+4 \right)=5\times 6=30$.