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Factor completely:   5x^2y^3 - 180y

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gudeapp | Student, College Freshman | eNoter

Posted June 9, 2011 at 2:11 AM via web

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Factor completely:

  5x^2y^3 - 180y

4 Answers | Add Yours

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted June 10, 2011 at 12:10 PM (Answer #2)

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We have to factor 5x^2y^3 - 180y

5x^2y^3 - 180y

=> 5y(x^2y^2 - 36)

use x^2 - y^2 = (x - y)(x + y)

=> 5y(xy - 6)(xy + 6)

The factors of 5x^2y^3 - 180y are 5y(xy - 6)(xy + 6)

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giorgiana1976 | College Teacher | Valedictorian

Posted June 13, 2011 at 3:22 AM (Answer #3)

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We notice that 180 = 10 * 18 = 5*2*18

According to this relation between coefficients, we can factorize by 5y first:

5x^2y^3 - 180y=5y(x^2y^2 - 36)

As we can notice, we've obtained a difference of squares:

a^2 - b^2 = (a-b)(a+b)

We'll put a = xy and b = 6

x^2y^2 - 6^2 = (xy - 6)(xy + 6)

The result of factorization is:

5x^2y^3 - 180y = 5y(xy - 6)(xy + 6)

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kanishkporwal | Student, College Freshman | eNoter

Posted September 20, 2011 at 8:16 PM (Answer #4)

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the last term is 180 so any number with 0 or 5 in last is divisible by 5 so take 5y common

5x^2y^3 - 180y=5y(x^2y^2 - 36)

now appy the formula

a^2 - b^2 = (a-b)(a+b)

x^2y^2 - 6^2 = (xy - 6)(xy + 6)

The result of factorization is:

5x^2y^3 - 180y = 5y(xy - 6)(xy + 6)

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted December 27, 2011 at 3:50 AM (Answer #5)

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You need to factor 5x^2y^3 - 180y.

Writing 180 = 18*2*5 yields:

 5x^2y^3 - 180y =  5x^2y^3 - 18*2*5y

The common factor that arrises is 5y:

5x^2y^3 - 18*2*5y = 5y(x^2y^2 - 36) => 5x^2y^3 - 18*2*5y = 5y(xy - 6)(xy+6)

Answer: 5x^2y^3 - 180y = 5y(xy - 6)(xy+6)

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