An electric device delivers a current of [tex]$15.0 A$[/tex] for 30 seconds. How many electrons flow through it?
Answer (2)
• We identify the coefficients of the quadratic expression. • We find two numbers that multiply to a c and add up to b .
• We rewrite the middle term using these two numbers and factor by grouping. • The factored form of the quadratic expression is ( 5 z + 1 ) ( z + 5 ) .
Explanation
Understanding the Problem We are asked to factor the quadratic expression 5 z 2 + 26 z + 5 .
Identifying Coefficients The quadratic expression is of the form a z 2 + b z + c , where a = 5 , b = 26 , and c = 5 .
Finding the Right Numbers We need to find two numbers that multiply to a c = 5 × 5 = 25 and add up to b = 26 .
Identifying the Numbers The two numbers are 25 and 1 because 25 × 1 = 25 and 25 + 1 = 26 .
Rewriting the Middle Term Rewrite the middle term using these two numbers: 5 z 2 + 25 z + z + 5 .
Factoring by Grouping Factor by grouping: 5 z ( z + 5 ) + 1 ( z + 5 ) .
Factoring out the Common Factor Factor out the common factor ( z + 5 ) : ( 5 z + 1 ) ( z + 5 ) .
Final Factored Form The factored form of the quadratic expression is ( 5 z + 1 ) ( z + 5 ) .
Examples
Factoring quadratic expressions is a fundamental skill in algebra, with applications in various fields. For instance, engineers use factoring to analyze the stability of structures, economists use it to model supply and demand curves, and computer scientists use it to optimize algorithms. Imagine you're designing a rectangular garden with an area represented by the expression 5 z 2 + 26 z + 5 . By factoring this expression into ( 5 z + 1 ) ( z + 5 ) , you determine the dimensions of the garden, ensuring it fits perfectly in your backyard. This skill is not just theoretical; it's a practical tool for problem-solving in many real-world scenarios.
To find how many electrons flow through an electric device delivering 15.0 A for 30 seconds, first, calculate the total charge using Q = I ⋅ t , which gives 450 C. Then, by dividing this charge by the charge of a single electron (approximately 1.6 × 1 0 − 19 C ), we find that about 2.81 × 1 0 21 electrons flow through the device. This calculation shows the large number of electrons involved in electric current.
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