(125t^3+8) / (5t+2)

Question

how do you solve this?

Answer 1

Hi Julia,

 

First, you must remind yourself that you have to use (0t² and 0t) (not written but exists) inside the long division.  You must also switch the signs as you go through this long division:

 

             ____________________

(5t + 2) | (125t³ + 0t² + 0t + 8)

 

 

                          25t² - 10t  +4

             ____________________
(5t + 2) | (125t³ + 0t² + 0t + 8)

             - (125t³  +50t²)              ---> (5t + 2) (25t²)

              ----------------------

                         -50t² + 0t

                      - (-50t² -20t)         ---> (5t +2) (-10t)

              ----------------------------

                                 +20t + 8

                              - (+20t + 8)  ---> (5t + 2) (+4)

              ----------------------------

                                           0

 

 

(25t² -10t + 4) will be the answer.

 

Let's check it:

 

(25t² -10t + 4) (5t + 2)

 

First      (25t² -10t + 4) (5t) = 125t³ -50t² + 20t

second  (25t² -10t + 4) (2)  = +50t² -20t +8

 

all the t²  (-50 & +50) cancels out and all the ts (+20 & -20) cancels out as well.

 

And it becomes 125t³ + 8.  It checks out!!!

 

There you have it.  Good luck.

 

P.S.  It was sum of cubes (125t³ + 8) = (25t² -10t +4) (5t + 2)  =  (a² + ab +b2) (a + b) 

                                      where a = 5t & b= 2.

 

Answer 2

Julie, I hope you are still up to get this!

 

So, in the numerator, we have the sum of cubes.  The cube root of 125t³ is 5t, right?  The cube root of 8 is 2, right?

 

Remember that a³ + b³ = (a + b) (a² - ab + b²)

 

So, 125t³ + 8 = (5t + 2) (25t² - 10t + 4)

 

Add the denominator and we have [(5t + 2) (25t² - 10t + 4)] / (5t + 2).

 

Since (5t +2) / (5t + 2) = 1, your answer is 25t² - 10t + 4.

 

Hope that helps!