Synthetic Division
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Synthetic Division : This method is used to find remainder as well as used in factorization.When you are dividing by a linear factor, you don't "have" to use long polynomial division; instead, you can use synthetic-division, which is much quicker.
Example : Find the remainder of x 2 + 5x + 6 and x-1.
Write the coefficients ONLY inside an upside-down division symbol: | ![]() |
Make sure you leave room inside, underneath the row of coefficients, to write another row of numbers later. Put the test zero, x -1=0 ⇒x = 1, at the left: | ![]() |
Take the first number inside, representing the leading coefficient, and carry it down, unchanged, to below the division symbol: | ![]() |
Multiply this carry-down value by the test zero, and carry the result up into the next column: | ![]() |
Add down the column: | ![]() |
Multiply the previous carry-down value by the test zero, and carry the new result up into the last column: | ![]() |
Add down the column: This last carry-down value is the remainder. | ![]() |
3) x 3 +7x 2 +7x -15
Solution :
As sum of all the coefficient ( 1 + 7 + 7 -15 )= 0 so (x -1 ) is one of the factor
After getting one factor find the other factors by Synthetic - Division.

As at the bottom of the line, there are 3 numbers so degree of the polynomial will be (3 -1 =2) and the numbers are 1,8,15.So the polynomial will be
x 2 + 8x + 15
Now find the factors of 15 in such a way that the addition and subtraction of that factors will be 8x.
So the factors of 15 = 5 and 3
x 3 +7x 2 +7x -15 = (x-1)(x + 5)(x + 3) are the factors.
Polynomial
• Degree of the Polynomial
• Zeros of Polynomial
• Remainder Theorem
• Find remainder by Synthetic Division
• Rational root test in Polynomial
• Solved Examples on Polynomial
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