# How do you find the greatest common factor of 45y^{12}+30y^{10}?

Jun 11, 2018

$15 {y}^{10}$

#### Explanation:

The largest number that goes into 45 and 30 is 15

${y}^{10}$ is the biggest term that goes into ${y}^{12} \mathmr{and} {y}^{10}$

Jun 11, 2018

The greatest common factor is $15 {y}^{10}$.

#### Explanation:

I suppose you're asking for the greatest common factor between $45 {y}^{12}$ and $30 {y}^{10}$, since the greatest common factor is computed between two number.

We need to find the greatest common factor for both the numeric and literal part.

For the numeric part, we can use the prime factorization of the two numbers:

$45 = 9 \cdot 5 = {3}^{2} \cdot 5$

$30 = 3 \cdot 10 = 2 \cdot 3 \cdot 5$

So, what's the biggest number that "fits" inside both $45$ and $30$, given their prime factorizations?

Well, we can't choose $2$, because it fits in $30$ but not in $45$. $3$ appears in both factorization, but we can pick it only once, since two three's (i.e. $9$) fit inside $45$, but not inside $30$. Finally, we can pick $5$ once since it appears in both factorizations.

So, the answer is $3 \cdot 5 = 15$

As for the literal part, we have a similar way to proceed: since "there are" $12$ $y$'s in the first term and only $10$ in the second, we can take at most $10$ $y$'s from both.

So, the greatest common factor is $15 {y}^{10}$. In fact, you can factor it from both terms to get

$45 {y}^{12} + 30 {y}^{10} = 15 {y}^{10} \left(3 {y}^{2} + 2\right)$

and there is nothing else to factor between $3 {y}^{2}$ and $2$.

Jun 11, 2018

$15 {y}^{10} \left(3 {y}^{2} + 2\right)$

Broken down into steps

#### Explanation:

$\textcolor{b r o w n}{\text{If you are not sure break it down into stages.}}$

We know that 5 is a factor of both 45 and 30

$5 \left(9 {y}^{12} + 6 {y}^{10}\right)$

We know that 3 is a factor of both 9 and 6

$5 \left[3 \left(3 {y}^{12} + 2 {y}^{10}\right)\right]$

$15 \left[3 {y}^{12} + 2 {y}^{10}\right]$

We know that ${y}^{12} \to \textcolor{red}{{y}^{10}} \textcolor{g r e e n}{\times {y}^{2}}$ giving:

$15 \left[\textcolor{w h i t e}{\frac{2}{2}} \textcolor{w h i t e}{\text{dddd")3y^12color(white)("ddd.d}} + 2 {y}^{10} \textcolor{w h i t e}{\frac{2}{2}}\right]$

$\textcolor{g r e e n}{15 \left[\textcolor{w h i t e}{\frac{2}{2}} \overbrace{\left(3 \textcolor{red}{\times {y}^{10}} \times {y}^{2}\right)} + 2 \textcolor{red}{{y}^{10}} \textcolor{w h i t e}{\frac{2}{2}}\right]}$

Factor out the $\textcolor{red}{{y}^{10}}$ giving:

$15 {y}^{10} \left(3 {y}^{2} + 2\right)$