The product of $9$ and variable $x$ is subtracted from $15$ times the product of variables $x$ and $y$, $25$ times variable $y$ is subtracted from the difference and then the number $15$ is added to them to represent an unknown quantity in the following algebraic expression.
The given algebraic expression consists of four terms and they can be arranged as two groups for separating the common factor from the terms.
$=\,\,\,$ $(15xy-9x)$ $+$ $(-25y+15)$
Let’s learn how to factorise or factorize the expression by grouping the terms.
It is right time to explore the common factors from the terms in each group and it can be done by the factorization (or factorisation) of the terms in the groups.
$=\,\,\,$ $(3x \times 5y-3 \times 3x)$ $+$ $\big(5 \times (-5y)+5 \times 3\big)$
There is a factor common in the terms of each group. So, the common factor can be taken out from the terms of every group by factorisation.
$=\,\,\,$ $3x \times (5y-3)$ $+$ $5 \times (-5y+3)$
Closely observe the factors $5y-3$ and $-5y+3$ in the terms. There is a similarity between them. One factor can be converted into second factor by multiplying it with $-1$. This technique helps us to complete the procedure for factoring the expression.
According to the multiplication of numbers, $5$ times $1$ is equal to $5$. So, the factor $5$ can be written as a product of $5$ and $1$ in the second term of the expression.
$=\,\,\,$ $3x \times (5y-3)$ $+$ $5 \times 1 \times (-5y+3)$
Similarly, the factor $1$ can be written as a product of two negative ones in the second term.
$=\,\,\,$ $3x \times (5y-3)$ $+$ $5 \times (-1) \times (-1) \times (-5y+3)$
Now, multiply the factor $5$ by another factor $-1$ to get their product in the second term.
$=\,\,\,$ $3x \times (5y-3)$ $-$ $5 \times (-1) \times (-5y+3)$
Finally, distribute the factor $-1$ over the subtraction by multiplication in second term as per the distribute property.
$=\,\,\,$ $3x \times (5y-3)$ $-$ $5 \times \big((-1) \times (-5y)+3 \times (-1)\big)$
It is time to calculate the products of the factors in the second factor of the second term.
$=\,\,\,$ $3x \times (5y-3)$ $-$ $5 \times (5y-3)$
In this expression, $5y-3$ is a common factor in both terms. So, take the common factor out from the terms to finish the factorization.
$=\,\,\,$ $(5y-3) \times (3x-5)$
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