Be taught polynomial lengthy division with two examples which might be straightforward to observe and straight to the purpose.

Explaining in additional particulars the polynomial lengthy division within the determine above

The polynomial lengthy division above is a abstract displaying how one can do lengthy division with polynomials. Learn instance #1 to totally perceive it!

Instance #1

Divide x2 + 3x – 10 by x – 2

Divide the main time period of x2 + 3x – 10 by the main time period of x – 2

x2 ÷ x = x

Write x as the primary time period of the quotient 

            x                         
 x – 2)  x2 + 3x – 10

Multiply the primary time period of the quotient by the divisor

x(x – 2) = x2 – 2x

Subtract x2 – 2x from the dividend

            x                         
 x – 2)  x2 + 3x – 10
          -( x2 – 2x)

            x                     
 x – 2)  x2 + 3x – 10
          -x2 + 2x
       ______________
                   5x  

Carry down -10

          x                         
 x – 2)  x2 + 3x – 10
          -x2 + 2x
       ______________
                   5x  – 10

Divide the main time period of 5x – 10 by the main time period of x – 2

5x ÷ x = 5

Write 5 because the second time period of the quotient. Since 5 is constructive, you’ll be able to put a + signal between the primary time period and the second time period.

              x  +  5               
 x – 2)  x2 + 3x – 10
          -x2 + 2x
       ______________
                   5x  – 10

Multiply the second time period of the quotient by the divisor

5(x – 2) = 5x – 10

Subtract 5x – 10 from 5x – 10

             x  +  5                     
 x – 2)  x2 + 3x – 10
          -x2 + 2x
        ______________
                   5x  – 10
                  -5x + 10
              __________
                        0

(x2 + 3x – 10) ÷ (x – 2) = x + 5

One other easy instance displaying polynomial lengthy division

Instance #2

Divide x2 – 5x + 1 by x + 3

Divide the main time period of x2 – 5x + 1 by the main time period of x + 3

x2 ÷ x = x

Write x as the primary time period of the quotient 

          x                         
 x + 3)  x2 – 5x + 1

Multiply the primary time period of the quotient by the divisor

x(x + 3) = x2 + 3x

Subtract x2 + 3x from the dividend

          x                         
 x + 3)  x2 – 5x + 1
          -( x2 + 3x)

          x                         
 x + 3)  x2 – 5x + 1
          -x2 – 3x
       ______________
                -8x  

Carry down 1

          x                         
 x + 3)  x2 – 5x + 1
          -x2 – 3x
       ______________
                   -8x  + 1

Divide the main time period of -8x  + 1 by the main time period of x + 3

-8x ÷ x = -8

Write -8 because the second time period of the quotient. Discover that placing a + signal between the primary time period and the second time period doesn’t change the issue.

            x + -8                     
 x + 3)  x2 – 5x + 1
          -x2 – 3x
       ______________
                   -8x  + 1

Multiply the second time period of the quotient by the divisor

-8(x + 3) = -8x – 24

Subtract -8x – 24 from -8x + 1

          x  + -8                     
 x + 3)  x2 – 5x + 1
         -x2 – 3x
        ______________
                   -8x  + 1
                    8x + 24
              __________
                           25       

(x2 – 5x + 1) ÷ (x + 3) = x + -8 with a the rest of 25

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