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(Synthetic Substitution)

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(Synthetic Substitution)

When compared to Polynomial Long Division, Synthetic Division reduces the number of steps to divide a polynomial by a binomial divisor of form x − r; where r is a real or complex number.

Let P(x) = x4 + 6x3 + 5x2 − 4x + 2, the dividend.

Let F(x) = x + 2, the binomial divisor.

P(x) is a polynomial of degree four and we know that if x4 is divided by x the result will be x3 [ i.e., x4 / x = x3 ], one degree less. The same idea continues as we divide each term of P(x) having an x. This tells us that dividing a polynomial of highest degree xn by x results in a quotient with xn − 1 highest degree and that the other terms having an x are also reduced by one magnitude. By understanding this fundamental concept we can eliminate x from the binomial divisor during the polynomial divide process and only focus on r of x − r. This is exactly what we do to synthetically divide a polynomial.

To divide P(x) by F(x) synthetically first note that the binomial divisor F(x) is x + 2; the binomial divisor needs to be of form x − r, so the +2 of x + 2 will be changed to −2 for the division process, its negative (if the divisor was x − 2 instead of x + 2 the sign of −2 would be changed to +2 for the synthetic division). The setup follows.

First remove the x from all terms of P(x) = x4 + 6x3 + 5x2 − 4x + 2 and write as:

P(x) with each x removed and the divisor −2

Blank line (for tabulation)

Blank line (for tabulation)

1 + 6 + 5 − 4 + 2 (−2

____________

The coefficient of the first term of the quotient will be the same as that of the dividend so just write this coefficient on the third line:

1 + 6 + 5 − 4 + 2 (−2

_____________

1

Next, multiply the divisor −2 by the first coefficient on line 3 and write that result on line two beneath the second coefficient of the first line.

1 + 6 + 5 − 4 + 2 (−2

− 2

_____________

1 + 4

The divisor −2 is multiplied by the first coefficient on line three.

The result +4 is the addition of line one +6 and second line −2.

Repeat this process for each coefficient…

The divisor −2 is multiplied by the second coefficient on line three.

Result −3 is the addition of +5 (line one) and −8 (second line).

1 + 6 + 5 − 4 + 2 (−2

− 2 − 8

_____________

1 + 4 − 3

The divisor −2 is multiplied by the third coefficient on line three.

Result +2 is the addition of −4 (line one) and +6 (second line).

1 + 6 + 5 − 4 + 2 (−2

− 2 − 8 + 6

_____________

1 + 4 − 3 + 2

The divisor −2 is multiplied by the fourth coefficient on line three.

Result −2 is the addition of +2 (line one) and −4 (second line).

1 + 6 + 5 − 4 + 2 (−2

− 2 − 8 + 6 − 4

_____________

1 + 4 − 3 + 2 − 2

The coefficient on line three, farthest right, is the remainder, −2. All other coefficients on line three to the left of the remainder are coefficients of the quotient (1 + 4 − 3 + 2). The result of the synthetic division can now be written as:

x3 + 4x2 − 3x + 2 with remainder −2.

Note that if a term is missing from a polynomial, that term would be a +0 coefficient during synthetic dividing. If P(x) = x4 + 6x3 − 4x + 2 we would setup the synthetic division as:

1 + 6 + 0 − 4 + 2 (−2

_____________

The x2 term is missing from the divisor P(x) and has +0 for synthetic division.

One more example:

P(x) = x4 − 4x3 +3x + 10

F(x) = x − 3

1 − 4 + 0 + 3 + 10 (+3

+ 3 − 3 − 9 − 18

_____________

1 − 1 − 3 − 6 − 8

The quotient is x3 − x2 − 3x − 6 with a remainder of −8.

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