Algebra help please?

Question

how do i reduce to lowest terms

a^2+2a-35
_________ (division)
a^3-125

^ means to the # power
thank you

Answer 1

Factor

(a+7)(a-5) / (a-5)(a^2 +5a + 25)

Cancel

(a+7) / (a^2 + 5a + 25)

Answer 2

Start by factoring both the numerator and denominator.

Numerator------a^2 + 2a - 35
factored----(a +7) (a - 5)

Denominator-----a^3 - 125
factored-----(a - 5) (a - 5) (a - 5) then........

(a - 5) in both numerator and denominator cancel out, leaving
you with a + 7
__________
a^2 - 10a + 25

Hope that helps, or maybe shows you the way.

Answer 3

Factor each expression, and cancel like terms.

First, you want to factor the expression in the numerator to the form (a+b)*(a-c). Multiply that out, and you get a^2+a*b-a*c-b*c, or a^2+a*(b-c)-b*c. Look at b and c. You want to choose values for them that make (b-c)=2 and b*c=35. By inspection, b=7 and c=5.

To factor the denominator, remember that a^3-b^3=(a-b)*(a^2+a*b+b^2). Because 5^3=125, substitute 5 for b, and you have your factors.

Finally, cancel out the similar terms.

Answer 4

numerator:
a^2 + 2a - 35 = (a-5)(a+7)

denominator:
a^3 - 125 = (a-5)(a^2 + 5a + 25)

the (a-5) terms cancel so

ANSWER = (a+7) / (a^2 + 5a + 25)

Answer 5

you need to factor the numerator and also the denominator
(a - 5 ) ( a + 7 ) is the numerator
-------------------
( a - 5 ) ( a^2 +5a + 25) is the denominator

now take out common factors
(a + 7 )
----------
a^2 + 5a +25
that is reduced to lowest terms. and you need to restrict the value of a to all real numbers except 5.

Answer 6

The numerator and denominator can be factored:

(a + 7) (a - 5)
-----------------
(a - 5)^3


and one instance of (a - 5) can be cancelled from the top and bottom to leave:

(a + 7)
---------
(a - 5)^2

Answer 7

(a^2 + 2a - 35)/(a^3 - 125)
((a + 7)(a - 5))/((a - 5)(a^2 + 5a + 25))
(a + 7)/(a^2 + 5a + 25)