SOLUTION: Can you help me answer this question and explain the process of what you have done, so I can complete the remaining questions. The questions is : Determine the remainder, r,

Algebra ->  Equations -> SOLUTION: Can you help me answer this question and explain the process of what you have done, so I can complete the remaining questions. The questions is : Determine the remainder, r,      Log On


   

Question 135106This question is from textbook Advanced Functions & Introductory Calculus
: Can you help me answer this question and explain the process of what you have done, so I can complete the remaining questions.
The questions is :
Determine the remainder, r, so that the division statement is true.
a) (2x-3)(3x-5)+ r = 6x^2 + x + 5 This question is from textbook Advanced Functions & Introductory Calculus

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
First of all, you have an error in the statement of the problem:
You show: %282x-3%29%28red%283x-5%29%29%2B+r+=+6x%5E2+%2B+x+%2B+5

And it should be %282x-3%29%28green%283x%2B5%29%29%2B+r+=+6x%5E2+%2B+x+%2B+5

Use polynomial long division or synthetic division to divide 6x%5E2+%2B+x+%2B+5
by either 2x-3 or 3x%2B5.

2x goes into 6x%5E2 3x times. So 3x is the first term of your quotient.

3x times 2x-3 is 6x%5E2-9x

Subtract 6x%5E2-9x from 6x%5E2%2Bx resulting in 10x (remember, change the sign and add)

Bring down the 5

2x goes into 10x 5 times. So 5 is the second term of your quotient.

5 times 2x-3 is 10x-15

Subtract 10x-15 from 10x%2B5 resulting in 20 which is your remainder.

So: %282x-3%29%283x%2B5%29%2B20=6x%5E2%2Bx%2B5

Check the answer:
%282x-3%29%283x%2B5%29%2B20=6x%5E2%2Bx-15%2B20=6x%5E2%2Bx%2B5 Checks.