Factor completely.
[tex]$v^3-125$
Answer (2)
The expression v 3 − 125 can be factored as ( v − 5 ) ( v 2 + 5 v + 25 ) using the difference of cubes formula. The quadratic part v 2 + 5 v + 25 cannot be factored further due to its negative discriminant. Thus, the complete factorization is ( v − 5 ) ( v 2 + 5 v + 25 ) .
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Recognize the expression as a difference of cubes: v 3 − 125 = v 3 − 5 3 .
Apply the difference of cubes formula: a 3 − b 3 = ( a − b ) ( a 2 + ab + b 2 ) .
Substitute a = v and b = 5 into the formula: ( v − 5 ) ( v 2 + 5 v + 25 ) .
The quadratic v 2 + 5 v + 25 cannot be factored further, so the complete factorization is ( v − 5 ) ( v 2 + 5 v + 25 ) .
Explanation
Recognizing the Problem Structure We are asked to factor the expression v 3 − 125 completely. Notice that this is a difference of cubes.
Recalling the Difference of Cubes Formula Recall the difference of cubes factorization formula: a 3 − b 3 = ( a − b ) ( a 2 + ab + b 2 ) .
Identifying a and b In our case, we have v 3 − 125 = v 3 − 5 3 . So, a = v and b = 5 .
Applying the Formula Substitute a = v and b = 5 into the formula: v 3 − 5 3 = ( v − 5 ) ( v 2 + 5 v + 5 2 ) = ( v − 5 ) ( v 2 + 5 v + 25 ) .
Checking the Quadratic Factor Now, we need to check if the quadratic v 2 + 5 v + 25 can be factored further. To do this, we can calculate the discriminant, which is given by Δ = b 2 − 4 a c . In this case, a = 1 , b = 5 , and c = 25 . So, Δ = 5 2 − 4 ( 1 ) ( 25 ) = 25 − 100 = − 75 .
Final Factorization Since the discriminant is negative, the quadratic v 2 + 5 v + 25 has no real roots and cannot be factored further using real numbers. Therefore, the complete factorization of v 3 − 125 is ( v − 5 ) ( v 2 + 5 v + 25 ) .
Examples
The difference of cubes factorization is useful in various engineering and physics applications. For example, when analyzing the stress distribution in a material, you might encounter an expression like x 3 − 8 , which represents the difference in volumes. Factoring this expression helps simplify the analysis and understand the material's behavior under different conditions. Similarly, in fluid dynamics, expressions involving the difference of cubes can arise when calculating flow rates or pressure differences, and factoring them simplifies the calculations.
