Perform the indicated operations and simplify

[LotinElite]

[(6x^2-13xy-5y^2) * (2x-7)] /
[(3x+y) * (2x^2+3xy-20y)]

 

[mmm4444bot]

[(6x^2-13xy-5y^2) * (2x-7)] / [(3x+y) * (2x^2+3xy-20y)] <<< Is this term correct?

I'm thinking that factor should be 2x^2 + 3xy - 20y^2, instead, because these sorts of exercises with algebraic fractions are usually cooked up with factorable expressions, so that some nice cancellations take place.

Both of the following two expressions can be factored.

6x^2 -13xy - 5y^2

2x^2 + 3xy - 20y^2

Are you familiar with factoring by grouping?

I'll factor the first one, and then you can try the second one.

Multiply (6)(-5) = -30

Look for two integers whose product is -30 and whose sum is -13

(2)(-15) = -30
2 - 15 = -13

The integers are -15 and 2, so rewrite the middle term (-13xy): -15xy + 2xy

(we could also rewrite it as 2xy - 15xy ; either way works)

6x^2 - 15xy + 2xy - 5y^2

"Group" the first two terms, and factor them

(3x)(2x - 5y) + 2xy - 5y^2

"Group" the last two terms, and factor them

(3x)(2x - 5y) + (y)(2x - 5y)

There is a factor of (2x - 5y) on each side of the plus sign; factor it out

(2x - 5y)(3x + y)

Now, in the denominator of your exercise, factor 2x^2 + 3xy - 20y^2. Then finish the simplification by cancelling the common factors.

If you would like more help with this exercise, then please show whatever work that you can accomplish, and try to say something about WHY you're stuck, so that people might determine where to continue helping you.

 

[Denis]

Mark is correct; proper way to simplify is factoring; you'll get:
(???????)(???????)(2x - 7) / [(???????)(???????)(3x + y)]

2 different terms will cancel out, leaving (2x - 7) / (???????)
YES, the (3x + y) will cancel out! And that's a huge hint.