91Ó°ÊÓ

When dealing with algebraic expressions, it’s crucial to group like terms together. Like terms are terms in an expression that have the same variables raised to the same powers. For example, in the expression \( r x^{2}-r x+5 r+s x^{2}-s x+5 s \) we can see like terms that involve \( x^{2}, x, \text{and constant terms (terms without variables)} \).By grouping them together, the expression becomes easier to manage and factor. So, we rearrange it as \((r x^{2} + s x^{2}) + (-r x - s x) + (5 r + 5 s)\). This grouping helps in the next steps, where we factor common terms out of each group.

After grouping like terms, the next step is to look for common factors within each group. A common factor is a term that can be evenly divided from each term in the group. This is similar to finding the greatest common divisor in a set of numbers.For the grouped expression \((r x^{2} + s x^{2}) + (-r x - s x) + (5 r + 5 s)\), each group has a common factor. For the first group \(r x^{2} + s x^{2}\), \(x^{2} \) is common. For the second, \(-r x - s x\), \(x\) is common, and for the third group \(5 r + 5 s\), \( 5\) is common.Factoring these out, we get \((r + s) x^{2} - (r + s) x + 5 (r + s)\). Now, each term within these groups is simpler, which helps in the overall simplification of the expression.

The final step is factoring out the common binomial. This involves identifying a binomial term (a term with two parts) that is common across the entire expression. In our problem, \(r + s \) appears in each term of \((r + s) x^{2} - (r + s) x + 5 (r + s)\).When a common binomial is identified, you can factor it out in much the same way you would with a singular common factor. For our expression, we factor \(r + s \) out, resulting in:i.e., write(r + s) (x^{2} - x + 5)This final step simplifies the expression significantly. The expression is now in its factored form, clearly showing the binomial \(r + s\) and another polynomial\(x^{2} - x + 5\)nested within.