Question

How do you factor $ 2{{x}^{2}}+7xy-15{{y}^{2}} $ ?

Answer 1

Hint: We are given an expression consist of 2 variable and consist of 3 terms,
We start our solution by learning about the trinomial, bivariable trinomial, then we learn about how to faster bi variable trinomial, use those steps, we will split the middle term of our given expression and factor it to get the required solution.

Complete step by step answer:
We are given that we have $ 2{{x}^{2}}+7xy-15{{y}^{2}} $ . We are asked to factor it.
Before we start, we should know about we are having an expression $ 2{{x}^{2}}+7xy-15{{y}^{2}} $ in this we have 3 expressions, so it is a trinomial.
Now we have different kinds of trinomial one are those in which these are just one variable.
While others are one which has more than one variable.
Our expression $ 2{{x}^{2}}+7xy-15{{y}^{2}} $ has two variables ‘x’ and ‘y’. These are called bi variable trinomials.
Now to factor the trinomial with 2 variable we follow the following steps:
1) Multiply the leading coefficient by the last number
2) find the sum of the two number that adds to the middle number
3) Split the middle term and group in two’s by taking common from each group
4) And then write In factored form
Now we have –
 $ 2{{x}^{2}}+7xy-15{{y}^{2}} $
Now product of leading coefficient with last term is $ 2\times 15=30 $
Now, we can write $ 30=3\times 10 $ and also we can write 7 using these number as –
 $ 7=10-3 $
So, we use this $ 7=10-3 $ to split our middle term so we get –
 $ 2{{x}^{2}}+7xy-15{{y}^{2}} $
Become
 $ 2{{x}^{2}}+\left( 10-3 \right)xy-15{{y}^{2}} $
Opening brackets, we get –
 $ 2{{x}^{2}}+10xy-3xy-15{{y}^{2}} $
We make pair of two
 $ 2{{x}^{2}}+10xy-3xy-15{{y}^{2}} $
Now, as 2x is common in $ 2{{x}^{2}} $ and 10xy, so we take it out.
Similarly, 3y is common in 3xy and $ 15{{y}^{2}} $, so take it out.
We get –
 $ 2x\left( x+5y \right)-3x\left( x+5y \right) $
Now as $ \left( x+5y \right) $ is common to both, we take it and simplify
So, we get –
 $ \left( 2x-3x \right)\left( x+5y \right) $
So, finally we get –
 $ 2{{x}^{2}}+7xy-15{{y}^{2}}=\left( 2x-3x \right)\left( x+5y \right) $
So, we get our factor.

Note:
 A key point to remember is that if the sign of the coefficient of leading and endpoint are same then we add a term to split middle term if their sign is opposite like our lane where we have +2 and -15, so we subtract term to get the middle term. So, we use $ 7=10-3 $ . This is one of the ways to check you are moving in the right direction.