Can anyone solve this problem

[isuranja]

1) 6x²+6+7x ÷ 2x+1

2) 3x³-5ab+2b² ÷ 3x-2



Regards,
Isuranja

 

[glenjoy]

6x²+6+7x
__________
2x+1

and

3x³-5ab+2b²
____________
3x-2

Regards,
Isuranja

Solve it first and show use your solution. Why too lazy to study and do it yourself?

 

[heo83]

however, I did not understand much about the isuranja's problem. There are two fractions :?
and no explanation!

 

[isuranja]

however, I did not understand much about the isuranja's problem. There are two fractions :?
and no explanation!

I rearranged it. I think you can undersatand now.
Regards,
Isuranja

 

[glenjoy]

1) 6x²+6+7x ÷ 2x+1

2) 3x³-5ab+2b² ÷ 3x-2



Regards,
Isuranja

Use synthetic division.

OR if you want to get the roots:

 

[lokeyh]

1) 6x²+6+7x ÷ 2x+1

2) 3x³-5ab+2b² ÷ 3x-2



Regards,
Isuranja

Do you mean (6x²+6+7x ÷ 2x+1) or (6x²+6+7x) ÷ (2x+1)? (3x³-5ab+2b² ÷ 3x-2) or (3x³-5ab+2b²) ÷ (3x-2)?

 

[isuranja]

1) (6x²+6+7x) ÷ (2x+1 )

2) (3x³-5ab+2b²) ÷ (3x-2 )

Regards
Isuranja

 

[steve10]

Solve for what? You want to do factorization? That is impossible as there are no common factors.

 

[isuranja]

Solve for what? You want to do factorization? That is impossible as there are no common factors.

There is a way to do this even not common factors.The given answer to 1st one as follows,


Quetioant - (3x+2)
remainder - 4

I want to know the step that get this answer.

Regards,
Isuranja

 

[tkbits]

Have you ever seen polynomial long division?

2x - 1 ) 4x² + 7x + 1 ( 2x + 4.5
      - (4x² - 2x)
         --------
               9x + 1
            - (9x - 4.5)
               --------
                    5.5

 

[glenjoy]

Solve for what? You want to do factorization? That is impossible as there are no common factors.

There is a way to do this even not common factors.The given answer to 1st one as follows,


Quetioant - (3x+2)
remainder - 4

I want to know the step that get this answer.

Regards,
Isuranja

ok, I will give you a short cut for number 1:

Here it is to get the remainder (shortcut),

6x² + 6 + 7x = remainder <--- equation 1

2x + 1 = 0
1 = - 2x

x = - 1/2 <--- equation 2

Substitute equation 2 to equation 1

6(-1/2)² + 6 + 7(1/2) = remainder

remainder = 4

 

[IanP]

(6x²+7x+6)2x+1)

First step:
(6x²+7x)2x+1)=3x
-(6x²+3x)
--------(4x+6) <---remainder from 1-st step

Second step
(4x+6)2x+1)=2
-(4x+2)
------4 <-----remainder from 2nd step and from whole division

Quotient= sum from the first and the second steps= 3x+2 ... remainder 4

Following this example you can try to do the second division ..

Good luck ..

 

[heo83]

After having a trip on the beach (05/01 is Labour Day, you know) , I’m back and now I understand what you want to. IanP, glenjoy and tkbits gave you their explanations and I will give you mine. You can follow any that easy to understand for you.
There are two ways to get the same result.
Firstly, using the polynomial division:
(I use the first division for example)
(6x²+6+7x) ÷ (2x+1)
a) take (6x² ÷ 2x), we have 3x
b) using 3x multiply (2x+1) back we have (6x² + 3x)
c) (6x²+6+7x) - (6x² + 3x) then we have (4x +6)
d) then taking 4x divide 2x we have 2
e) 2 multiply (2x+1) back we have (4x + 2)
f) (4x +6) - (4x + 2) = 4 this is the reminder.
g) And (3x + 2) is the quotient

Sorry for using multiple posts! But this illustrates my idea.
Using this method you can solve more complex division like this

 

[heo83]

I think these two examples are enough to illustrate the idea.
The next method is very quick and easy to do with two hands () but it can only be used if the degree of polynomial divisor is ONE. I want to mention to Hornor’s scheme (it sounds similarly to Hornor, please check this spelling for me, and say sorry to Mr. Hornor because I can’t write exactly his name)
You only use the coefficients of the polynomial function.


See explanation below:

 

[heo83]

See explanation below:

a) pull down the first number (in this example we have 6)
b) get the -1/2 multiply to the number we pull down earlier (6) then add to the next coefficient of the devided multi-nomial (we have 7 here) to have 4: (-1/2)*6+7=4
c) use the same pattern from now on. Get -1/2 multiply to the number you found earlier step (4) then add to the next coefficient: (-1/2)*4+6=4
Remember, the missing coefficients in the multinomial are replaced by zeros. Thus,
f(x) = 4x^4+3x+2
f(x) | 4 0 0 3 2
Please reply if there is something that is not clear.

 

[glenjoy]

I think for problem number 2, if it is required to find the values of a and b, I think the remainder must be given.

 

[heo83]

This is the reminder of the problem 2

8/9 - 5ab + 2b^2

 

[IanP]

Well, ... almost right ..

Don't you think the remainder of problem 2 should be:
4/9+2b²-5ab

 

[heo83]

This is my work!

 

[IanP]

I was jus playing devil's advocate ...

Your answers are 100+++++ % correct ..