A polynomial is given in terms of two variables $x$ and $y$, and it consists of three terms.

$2x^2y^2-7xy-30$

Let us learn how to factorise this algebraic expression by splitting its middle term.

It is essential to check whether the given expression in either descending or ascending order. The given algebraic expression is in descending order. So, no need to think about this step.

$2x^2y^2-7xy-30$

In this polynomial, $2x^2y^2$ is the first term and $-30$ is the last term. Now, calculate the product of the two terms with their signs.

$(2x^2y^2)(-30) \,=\, -60x^2y^2$

$-7xy$ is the middle and try to split it as either sum or difference of two terms but their product should be equal to the product of the first and last terms $-60x^2y^2$.

- $-7xy$ $\,=\,$ $-12xy+5xy$
- $(-12xy)(5x) \,=\, -60x^2y^2$

The above two statements reveal that the factorization by splitting the middle term can be used to factorise the given polynomial.

The trinomial is successfully expanded as four-term polynomial. It is time to factorise the algebraic expression by taking out the common factors from the terms.

$=\,\,\,$ $2x^2y^2-12xy+5xy-30$

There is a common factor in the first two terms. Similarly, there is a common factor in the remaining two terms. So, the four terms can be written as two groups and it is useful to factorize the expression by grouping.

$=\,\,\,$ $(2x^2y^2-12xy)$ $+$ $(5xy-30)$

Now, express the terms in each group in factor form to separate the common factor from them.

$=\,\,\,$ $(2xy \times xy-6 \times 2xy)$ $+$ $(5 \times xy-5 \times 6)$

It is time to take out the factor common from each group in the above expression.

$=\,\,\,$ $2xy \times (xy-6)$ $+$ $5 \times (xy-6)$

In the above expression, $xy-6$ is a common factor in both terms. Hence, it can be taken out common from both terms to complete the process of factorization in this problem.

$=\,\,\,$ $(xy-6)(2xy+5)$

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