Post a New Question

Algebra

posted by on .

Factor the numerator and denominator of each fraction if neceesary. Rewrite each one as a product. Then look for ones and simplify. The denominator is not zero.

x^2+6x+9 2x^2-x-10 28x^2-x-15
-------- --------- ----------
x^2-9 3x^2+7x+2 28x^2-x-15


x^2+4x
-------
2x+8

  • Algebra - ,

    very hard to figure out what you typed.
    Please use brackets next time, building fractions in this format does not work.

    I think you meant:
    (x^2+6x+9)/(x^2-9) * (2x^2-x-10)/(x^2+7x+2) * (28x^2-x-15)/(28x^2-x-15)
    I assume you know how to factor, since this is usually the type of exercise with that topic

    = (x+3)(x+3)/[(x-3)(x+3)] * (2x-5)(x+2)/[(3x+1)(x+2)] * (28x^2-x-15)/(28x^2-x-15)
    = (x+3)(2x-5)/[(x+3)(3x+1) , x not equal to -3,-1/3

    I did not attempt to factor the last part, since it just canceled.

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question