91Ó°ÊÓ

We have an equation- $30x^2-53xy-21y^2$. We will first find out all the factors of the first and the last term.
Then by trying them with each other, we will find the correct set of factors which will result into the given equation when multiplied.

The first term is $30$.

The factors of it are- $(1×30),(2×15),(3×10)$

The last term is $-21$.

The factors of it are-role="math" localid="1644571044294" $(1×-21),(3×-7)$.

The possibilities are-

"Not possible options" are declared so, because as there is no common factor in the equation, there cannot be any common factor in any of the factors also.

As we can see from the above table, we can conclude that the correct factors are those whose product matches with the given equation.

So the correct factors are $(3x+y)(10x-21y)$.

We will check the solution by multiplying. If we get the same given equation, our calculations are right.

So,

$$\begin{gathered} =(3x+y)(10x-21y) \\ =30x^2-63xy+10xy-21y^2 \\ =30x^2-53xy-21y^2 \end{gathered}$$