SOLUTION: I have the answer to this. Division of polynomial but I think I am messing up the steps. (#1)1-3x^2/x+2 answer -3x+6-11/x+2. (#2) the product of twice a number and three is the sam

Algebra ->  Expressions -> SOLUTION: I have the answer to this. Division of polynomial but I think I am messing up the steps. (#1)1-3x^2/x+2 answer -3x+6-11/x+2. (#2) the product of twice a number and three is the sam      Log On


   



Question 458319: I have the answer to this. Division of polynomial but I think I am messing up the steps. (#1)1-3x^2/x+2 answer -3x+6-11/x+2. (#2) the product of twice a number and three is the same as the diffrence of five times the number and 3/4. Find the number. (#3). Solution or equation, I have the answer need help setting it up. x-3/7=-2: -11 thank you very much. A greatful mom.
Answer by Edwin McCravy(20056) About Me  (Show Source):
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I have the answer to this. Division of polynomial but I think I am messing up the steps. (#1)1-3x^2/x+2 answer -3x+6-11/x+2. (#2) the product of twice a number and three is the same as the diffrence of five times the number and 3/4. Find the number. (#3). Solution or equation, I have the answer need help setting it up. x-3/7=-2: -11 thank you very much. A greatful mom.
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I have the answer to this. Division of polynomial but I think I am messing up the steps.



(#1)  1 - 3x²
      ———————
       x + 2

You must do two things with the numerator.
1. Arrange it in descending powers of x.
   Change it from 1 - 3x² to -3x² + 1 
2. Insert a placeholder term of 0x to hold
   the place for that term so you'll have
the highest power term first, then the first
power of x-term and then a constant number
on the end like this -3x² + 0x + 1  

So we put this in an array with x + 2 as the
divisor, and -3x² + 0x + 1 as the dividend.
You say "x+2 divided into -3x²+0x+1"
 
                         
x + 2)-3x² + 0x +  1
    
The x on the far left goes into -3x² and 
gives -3x because %28-3x%5E2%29%2Fx=-3x. So
we write -3x above the 0x, like this
             

            -3x     
x + 2)-3x² + 0x +  1
      
Now multiply the -3x by both of the terms of x + 2.
-3x(x + 2) = -3x² - 6x, so we write -3x² - 6x underneath
the first two terms of the dividend, like this


            -3x     
x + 2)-3x² + 0x +  1   
      -3x² - 6x
            
Now you draw a line underneath:

            -3x     
x + 2)-3x² + 0x +  1
      -3x² - 6x


Now you subtract those:

    (-3x² - 0x) - (-3x² - 6x) = -3x² - 0x + 3x² + 6x = 6x.

So you put 6x under the - 6x:

            -3x     
x + 2)-3x² + 0x +  1
      -3x² - 6x
             6x 
             
Now we do something like the first step. The x on
the far left goes into 6x and gives 6 because %286x%29%2Fx=6. So
we write + 6 above the + 1, like this
                 


            -3x +  6
x + 2)-3x² + 0x +  1
      -3x² - 6x
             6x 

Next we bring down the + 1 to the bottom, like this



            -3x +  6
x + 2)-3x² + 0x +  1
      -3x² - 6x
             6x +  1
             
Now like in the 2nd step, we now multiply the 6 by both of 
the terms of x + 2.  6(x + 2) = 6x + 12, so we write 6x + 12 
underneath the two terms at the bottom, like this



            -3x +  6
x + 2)-3x² + 0x +  1
      -3x² - 6x
             6x +  1
             6x + 12 
         
Next we draw a line underneath:

            -3x +  6
x + 2)-3x² + 0x +  1
      -3x² - 6x
             6x +  1
             6x + 12

        
Now like in the 3rd step, we subtract those: 

(6x + 1) - (6x + 12) = 6x + 1 - 6x - 12 = -11 So you put -11 under the + 12:

            -3x +  6
x + 2)-3x² + 0x +  1
      -3x² - 6x
             6x +  1
             6x + 12 
                 -11

-11 is the remainder. And you're done with the division 
algorithm but that's not all you do.  You have to interpret 
a division of this form:

        QUOTIENT   
DIVISOR)DIVIDEND
        ...
            ....
        REMAINDER

as 
           REMAINDER 
QUOTIENT + —————————  
            DIVISOR           
           
or
           -11
-3x + 6 + ————— 
          x + 2 

And since the numerator of the fraction is negative,
you can just change the plus sign before the fraction
to a minus, and get:

            11
-3x + 6 - ————— 
          x + 2

----------------------------------------------
 
(#2) the product of twice a number and three is the same as the diffrence of five times the number and ¾. 

Replace the words "a number" and "the number" with a letter variable,
say N, and rewrite it:

(#2) the product of twice N and three is the same as the difference of five times N and ¾.

Now we replace the words "twice N" with "2N" and the words "five times N"
by "5N", and the word "three" by "3" and rewrite it:

(#2) the product of 2N and 3 is the same as the difference of 5N and ¾.

"Product" means to multiply and "difference" means to subtract. Therefore
we replace the words "The product of 2N and 3" with "(2N)·(3)" and we
replace the words "the difference of 5N and ¾" by "5N-¾"

     (2N)·(3) is the same as 5N-¾

Finally we replace the words "is the same as" by an equal sign "=":

     (2N)·(3) = 5N-¾

That's the equation.  So to solve it we replace (2N)·(3) by 6N:

     6N = 5N-¾

We clear of fractions by multiplying every term by the LCD of 4

    24N = 20N - 3

Subtract 20N from both sides:

     4N = -3

Divide both sides by 4

      N = -¾  

That's the answer.

--------------------------------
         x - 3 
(#3).    ————— = -2
           7

Clear of fractions by multiplying every term on
both sides by LCD = 7:

         x - 3 = -14

Add 3 to both sides:
         
             x = -11 


Edwin