Question

Simplify: \[15xy - 6x + 5y - 2\]

Answer 1

Hint:
We will rearrange the terms in the given polynomial in order to take out common factors. Then, we will find the factors of the polynomial. Factorization or factoring is defined as the breaking or decomposition of an entity which may be a number, a matrix, or a polynomial into a product of another entity, or factors, which when multiplied together give the original number or a matrix.

Complete step by step solution:
The equation is given as \[15xy - 6x + 5y - 2\]
We usually start with taking the term having \[x\] variables together and the term having \[y\] variables together.
In this case we will take the term having \[xy\] variable and \[x\] variable together.
\[15xy - 6x + 5y - 2 = \left( {15xy - 6x} \right) + \left( {5x - 2} \right)\]
Taking\[3x\]common in first bracket and 1 common in second bracket, we get
\[ \Rightarrow 15xy - 6x + 5y - 2 = 3x\left( {5y - 2} \right) + 1\left( {5x - 2} \right)\]
As we can see both the bracket have same value we will take it common, so we will get
\[ \Rightarrow 15xy - 6x + 5y - 2 = \left( {3x + 1} \right)\left( {5x - 2} \right)\]

So our simplified form of \[15xy - 6x + 5y - 2\] is\[\left( {3x + 1} \right)\left( {5x - 2} \right)\].

Note:
Another way in which the above value can be simplified is by taking term having variable \[xy\] with term having variable y as
\[15xy - 6x + 5y - 2 = \left( {15xy + 5y} \right) - \left( {6x + 2} \right)\]
Now taking \[5y\]common from first bracket and 2 common from second bracket we get,
\[ \Rightarrow 15xy - 6x + 5y - 2 = 5y\left( {3x + 1} \right) - 2\left( {3x + 1} \right)\]
As we can see that both brackets has same term we will take it common and get,
\[\left( {5y - 2} \right)\left( {3x + 1} \right)\] This is the same as the previous answer.
So we can state that no matter what term we take together we will always get the same simplified form if the equation can be simplified. Bracket plays a very important role when we simplify and equation as it gets clear what term is coming common.