am i doing rationals right?

Question

Reduce the rational expression to lowest terms. If it is already in lowest terms, enter the expression in the answer box.

 

2r³-6r²-20r/(r²-r-6)

 

and i got 2r(r²-3r-10)

and got 2r(r-5) (r+2)/(r-3)(r+2)

 

so i cancelled out (r+2) and was left with 2r(r-5)/r-3)

 

is this correct? 

 

Answer 1

(2r³-6r²-20r)/(r²-r-6)

 

As you noted, you can factor 2r out of

each term in the denominator giving

2r(r² -3r -10)  for the numerator

 

However your polynomial r² - 3r -10 is also

further factorable.

Look for factors of -10 that add to -3...

-10 = (-5)(2) and -5 + 2 = -3  so this factors to:

(r-5)(r+2)

 

Your numerator is:  2r(r-5)(r+2)

 

Now look at the denominator:

r²-r-6

Again, this is factorable.  The factors of -6 that

add to -1 are (-3)(2) so this factors to:

(r-3)(r+2) for the denominator

 

Putting this back together we have:

 

[2r(r-5)(r+2)]/[(r-3)(r+2)]

 

We can cancel out the (r+2) leaving:

 

2r(r-5)/(r-3)

for the final answer.

 

For restrictions on the variable, look for values

of r that would make the denominator equal zero

since dividing by zero is undefined.

 

(r-3)(r+2) = 0

The values that would make the denominator

be zero are r=3, -2

 

Final final answer:

 

(2r³-6r²-20r)/(r²-r-6) = [2r(r-5)]/(r-3),  r≠3, r≠-2

 

 

Note:  Candice, you did an excellent job factoring this

expression down to lowest terms.  Fantastic!!

But, remember to check for restrictions on the variable,

and also, you left the grouping parenthesis off of the

polynomial in the numerator.  Again, GREAT JOB.