Question 146421
Can someone please help me with a couple of problems on polynomials finding the quotient and the remainder? 
1) {{{p(x)= 2x3-3x2+x-1}}}
{{{d(x) = x-3}}}
<pre>
By long division you would start with
     ___________________
x - 3) 2x<sup>3</sup> - 3x<sup>2</sup> + x - 1 

But with synthetic division

Change the sign of the -3 to +3
and write this:

   3 |
     |__________ 

Then list the coefficients of the terms:

   3 |2 -3  1 -1
     |__________   
     

Start by bringing down the 2

   3 |2 -3  1 -1
     |__________  
      2 

Multiply the 2 at the bottom by the 3 at the left, 
getting 6. Write that 6 above and to the right of 
the 2, under the next term -3:

   3 |2 -3  1 -1
     |<u>   6      </u>  
      2  

Combine the -3 and the 6, getting 3, and write this
below the line under the 6:

   3 |2 -3  1 -1
     |<u>   6      </u>  
      2  3  

Multiply that 3 on the bottom by the 3 at the left,
getting 9, and write this 9 above and to the right of the
3, above the line below the 1: 

   3 |2 -3  1 -1
     |<u>   6  9   </u> 
      2  3  

Combine the 1 and the 9, getting 10, and write this
below the line under the 9:

   3 |2 -3  1 -1
     |<u>   6  9   </u>
      2  3 10 

Multiply 10 on the bottom by the 3 at the left, getting 30,
and write this 30 above and to the right of the 10, above 
the line below the -1:

   3 |2 -3  1 -1
     |<u>   6  9 30</u>
      2  3 10 

Combine the -1 and the 30, getting 29, and write this
29 below the line under the 30:

   3 |2 -4  1 -1
     |<u>   6  9 30</u>
      2  3 10 29

Now we must interpret that row of numbers 2  3 10 29
across the bottom of the synthetic division.

The largest power of x in the original polynomial is 3.
So the 2 is multiplied by a power of x which is 1 lower 
than the degree of the original polynomial. One lower 
than 3 is 2, so we write x<sup>2</sup> after the 2. Then we 
write x (first power) after the 3. Then 10 is the 
constant term.  So the quotient only is

     2x<sup>2</sup> + 3x + 10

and the last number on the bottom right is the remainder.  So
we put the remainder 29 over the divisor x - 3  and we have
as the final answer:

    {{{2x^2+3x+10+29/(x-3)}}}

==========================================================

2) {{{P(x) = x^3 - 9x^2 + 15x + 25}}}
d(x) = {{{x-5}}} 


By long division you would start with
     _____________________
x - 5) x<sup>3</sup> - 9x<sup>2</sup> + 15x + 25 

But with synthetic division

Change the sign of the -5 to +5
and write this:

   5 |
     |__________ 

Then list the coefficients of the terms:

   5 |1 -9 15 25
     |__________   
     

Start by bringing down the 1

   5 |1 -9 15 25
     |__________  
      1 

Multiply the 1 at the bottom by the 5 at the left, 
getting 5. Write that 5 above and to the right of 
the 1, under the next term -9:

   5 |1 -9 15 25
     |<u>   5      </u>  
      1  

Combine the -9 and the 5, getting -4, and write this
below the line under the 5:

   5 |1 -9 15 25
     |<u>   5      </u>  
      1 -4  

Multiply that -4 on the bottom by the 5 at the left,
getting -20, and write this -20 above and to the right of the
-4, above the line below the 15: 

   5 |1 -9 15 25
     |<u>   5-20   </u> 
      1 -4  

Combine the 15 and the -20, getting -5, and write this
below the line under the -20:

   5 |1 -9 15 25
     |<u>   5-20   </u>
      1 -4 -5 

Multiply -5 on the bottom by the 5 at the left, getting -25,
and write this -25 above and to the right of the -5, above 
the line below the -20:

   5 |1 -9 15 25
     |<u>   5-20-25</u>
      1 -4 1-5  

Combine the 25 and the -25, getting 0, and write this
0 below the line under the -25:

   5 |1 -9 15 25
     |<u>   5-20-25</u>
      1 -4 -5  0

Now we must interpret that row of numbers 1 -4 -5  0
across the bottom of the synthetic division.

The largest power of x in the original polynomial is 3.
So the 1 is multiplied by a power of x which is 1 lower 
than the degree of the original polynomial. One lower 
than 3 is 2, so we write x<sup>2</sup> after the 1. Then we 
write x (first power) after the -4. Then -5 is the 
constant term.  So the quotient is

      x<sup>2</sup> - 4x - 5

and the last number on the bottom right is the remainder. 
Since that number is 0, that is the final answer.

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

3) P(x) = x<sup>4</sup> + 6x<sup>3</sup>
d(x) = x-1

Nere we must use placeholders for x<sup>2</sup>, x<sup>1</sup>, and the constant
term

By long division you would start with
     _______________________
x - 1) x<sup>4</sup> + 6x<sup>3</sup> + 0x + 0 + 0 

But with synthetic division

Change the sign of the -1 to +1
and write this:

   1 | 1   6   0   0   0
     |__________________
       
Start by bringing down the 1

   1 | 1   6   0   0   0
     |__________________
       1
 
Multiply the 1 at the bottom by the 1 at the left, 
getting 1. Write that 1 above and to the right of 
the 1 at the bottom, under the next term 6:

   1 | 1   6   0   0   0
     |_____1____________
       1

Combine the 6 and the 1, getting 7, and write this
7 below the line under the 1:

   1 | 1   6   0   0   0
     |_____1____________
       1   7

Multiply the 7 at the bottom by the 1 at the left, 
getting 7. Write that 7 above and to the right of 
the 7 at the bottom, under the next term 0:

   1 | 1   6   0   0   0
     |_____1___7________
       1   7

Combine the 0 and the 7, getting 7, and write this
7 below the line under the 7:

   1 | 1   6   0   0   0
     |_____1___7________
       1   7   7

Multiply the second 7 at the bottom by the 1 at the 
left, getting 7. Write that 7 above and to the right of 
the second 7 at the bottom, under the next term 0:

   1 | 1   6   0   0   0
     |_____1___7___7____
       1   7   7

Combine the 0 and the 7, getting 7, and write this
7 below the line under the 7:

   1 | 1   6   0   0   0
     |_____1___7___7____
       1   7   7   7

Multiply the third 7 at the bottom by the 1 at the 
left, getting 7. Write that 7 above and to the right of 
the third 7 at the bottom, under the last term 0:

   1 | 1   6   0   0   0
     |_____1___7___7___7
       1   7   7   7

Combine the 0 and the 7, getting 7, and write this
7 below the line under the 7:

   1 | 1   6   0   0   0
     |_____1___7___7___7
       1   7   7   7   7

Now we must interpret that row of numbers 1   7   7   7   7
across the bottom of the synthetic division.

The largest power of x in the original polynomial is 4.
So the 1 is multiplied by a power of x which is 1 lower 
than the degree of the original polynomial. One lower 
than 4 is 3, so we write x<sup>3</sup> after the 1. Then we 
write x<sup>2</sup> after the first 7. Then x after the second 7.
The third 7 on the bottom is the constant term.  The
quotient only is

     x³ + 7x² + 7x + 7

and the last number on the bottom right is the remainder.  So
we put the remainder 7 over the divisor x - 1  and we have
as the final answer:

    {{{x^3+7x^2+7x+7+7/(x-1)}}}

Edwin</pre>