# What is the LCD of 7(y+2) and y?

Jan 25, 2018

$7 {y}^{2} + 14 y$

#### Explanation:

To find the LCD of regular numbers, you use the following steps:

$\text{Write out the prime factorizations of all of the numbers}$

$\text{For each prime factor, determine which}$
$\text{number has the highest power of that factor}$

$\text{Multiply together all of the}$ "$\text{highest}$" $\text{powers of factors to get the LCD}$

Working with polynomials like this is not much different. The only real difference you'll see here is that some of our prime factors have variables in them, but they're still prime factors because they're as simple as we can get them.

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So, let's find the LCD. Our two numbers are $y$ and $7 \left(y + 2\right)$

Prime factorizations:

$7 \times \left(y + 2\right)$
$y$

The factor $\textcolor{b l u e}{7}$ occurs the most in the first term, where it occurs $\textcolor{red}{1}$ time, so we will multiply ${\textcolor{b l u e}{7}}^{\textcolor{red}{1}}$ into our LCD.

The factor $\textcolor{\mathmr{and} a n \ge}{y}$ occurs the most in the second term, where it occurs $\textcolor{red}{1}$ time, so we will multiply ${\textcolor{\mathmr{and} a n \ge}{y}}^{\textcolor{red}{1}}$ into our LCD.

The factor $\textcolor{\lim e g r e e n}{\left(y + 2\right)}$ occurs the most in the first term, where it occurs $\textcolor{red}{1}$ time, so we will multiply ${\textcolor{\lim e g r e e n}{\left(y + 2\right)}}^{\textcolor{red}{1}}$ into our LCD.

Therefore, our LCD is:

${7}^{1} \times {y}^{1} \times {\left(y + 2\right)}^{1}$

$7 y \left(y + 2\right)$

$7 {y}^{2} + 14 y$