Problem 5
Question
1–8 ? Factor out the common factor. $$ y(y-6)+9(y-6) $$
Step-by-Step Solution
Verified Answer
Factor out (y-6) to get (y-6)(y+9).
1Step 1: Identify the Common Factor
Examine each term in the expression \(y(y-6) + 9(y-6)\). Notice that both terms share a common factor of \((y-6)\).
2Step 2: Rewrite the Expression
Rewrite the expression by factoring out the common term \((y-6)\) across both terms. This results in: \((y-6)(y+9)\).
3Step 3: Confirm the Factored Expression
Verify the factored expression is correct by expanding it: \((y-6)(y+9) = y(y-6) + 9(y-6)\), which confirms the factoring was done correctly.
Key Concepts
common factorfactored expressionexpand
common factor
A common factor is an algebraic term shared by two or more terms in an expression. Identifying a common factor is the foundational step in factoring. In the expression \( y(y-6)+9(y-6) \), the common factor is \( (y-6) \). Both terms, \( y(y-6) \) and \( 9(y-6) \), include this expression. Why is finding the common factor so important?
- It simplifies the expression by allowing you to "extract" the shared term.
- This process makes further algebraic manipulations, like solving equations, easier.
factored expression
A factored expression results when you rewrite an original expression using its common factors. This process transforms a sum of terms into a product of terms. In our exercise, the expression \( y(y-6) + 9(y-6) \) is changed to \( (y-6)(y+9) \) after factoring. The benefit of having a factored expression is clear:
- It represents the expression in a more simplified and compact form.
- It reveals important characteristics, like roots or solutions, easier.
expand
Expanding is the reverse operation of factoring. It involves distributing terms to return a product back into a sum. In our example, if you take the factored expression \( (y-6)(y+9) \) and expand it, you end up back at the original expression \( y(y-6) + 9(y-6) \).Here’s how you do it:
- Apply the distributive property: multiply each term inside the first bracket by each term inside the second bracket.
- This yields \( y(y-6) + 9(y-6) \), confirming our factoring was correct.
Other exercises in this chapter
Problem 5
\(1-6=\) An expression is given. (a) Evaluate it at the given value. (b) Find its domain. $$ \sqrt{x+3}, \quad x=6 $$
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\(3-10=\) State the property of real numbers being used. $$ (x+2 y)+3 z=x+(2 y+3 z) $$
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