# (6p^2q^2) - (9pq^3)?

## I honestly don't know how to factorise and I really need my homework done for tomorrow, thank you for your help if you may.

May 22, 2018

$\left(6 {p}^{2} {q}^{2}\right) - \left(9 p {q}^{3}\right) = 3 p {q}^{2} \left(2 p - 3 q\right)$

#### Explanation:

It helps to use the script function in Socrates - makes it easier to read. I read your expression as $\left(6 {p}^{2} {q}^{2}\right) - \left(9 p {q}^{3}\right)$

First, you don't need the parentheses here, since you have only one term in each parenthesis. Therefore you can write it as:

$\left(6 {p}^{2} {q}^{2}\right) - \left(9 p {q}^{3}\right) = 6 {p}^{2} {q}^{2} - 9 p {q}^{3}$

Next we need to ask: What is common in the two terms?

Firstly: 6 and 9 are both multiples of 3: $6 = 2 \cdot 3$ and $9 = 3 \cdot 3$. We can, therefore, draw it out from both terms.

Secondly: ${p}^{2} {q}^{2}$ and $p {q}^{3}$ both have $p$ and ${q}^{2}$ in common, so we can also draw them out from both terms.

So from the 1st term: $6 {p}^{2} {q}^{2} = 3 p {q}^{2} \cdot 2 p$
2nd term: $9 p {q}^{3} = 3 p {q}^{2} \cdot 3 q$

If we combine these, we get:
$3 p {q}^{2} \cdot 2 p - 3 p {q}^{2} \cdot 3 q = 3 p {q}^{2} \left(2 p - 3 q\right)$ where we put together those parts that are not equal in the two terms.

This gives us:
$\left(6 {p}^{2} {q}^{2}\right) - \left(9 p {q}^{3}\right) = 3 p {q}^{2} \left(2 p - 3 q\right)$

I hope the step by step example was suffiently clear so that you could see what we do.