Synthetic Division is only applicable if in \(\frac{{P(x)}}{{g(x)}}\) g(x) is in the form x - c; where c is a constant.

\(\frac{{P(x)}}{{g(x)}}\) ==> P(x) = g(x) q(x) + R(x)

Where:

q(x) is the quotient

R(x) is the remainder

\(P(x) = {c_1}{x^n} + {c_2}{x^{n - 1}} + {c_3}{x^{n - 2}} + ...\)

\(q(x) = {b_1}{x^{n - 1}} + {b_2}{x^{n - 2}} + {b_3}{x^{n - 3}} + ...\)

So, to be able to find the quotient q(x), we just need to calculate the values of b's.

---

\(P(x) = q(x)g(x) + R(x) = (x - c)g(x) + R(x)\)

\( = (x - c)({b_1}{x^{n - 1}} + {b_2}{x^{n - 2}} + ...) + P(r)\)

\( = {b_1}x{x^{n - 1}} - c{b_1}{x^{n - 1}} + {b_2}x{x^{n - 2}} - c{b_2}{x^{n - 2}} + ... + {b_n}x - c{b_n} + P(r)\)

\( = {b_1}{x^n} - c{b_1}{x^{n - 1}} + {b_2}{x^{n - 1}} - c{b_2}{x^{n - 2}} + {b_3}{x^{n - 2}} + ... + - c{b_n} + P(r)\)

\( = {b_1}{x^n} + \left( { - c{b_1} + {b_2}} \right){x^{n - 1}} + \left( { - c{b_2} + {b_3}} \right){x^{n - 2}} + ... + \left( { - c{b_n} + P(r)} \right)\)

\(P(x) = {c_1}{x^n} + {c_2}{x^{n - 1}} + {c_3}{x^{n - 2}} + ...\)

\({c_1}{x^n} + {c_2}{x^{n - 1}} + {c_3}{x^{n - 2}} + ... = (x - c)g(x) + R(x)\)

\({c_1}{x^n} + {c_2}{x^{n - 1}} + {c_3}{x^{n - 2}} + ... = {b_1}{x^n} + \left( { - c{b_1} + {b_2}} \right){x^{n - 1}} + \left( { - c{b_2} + {b_3}} \right){x^{n - 2}} + ... + \left( { - c{b_n} + P(r)} \right)\)

By equating coefficients of equal exponents.

Then we solve the value of b's.

Maybe in this form it's a lot harder to see that this is a simplified form of polynomial division. But if arranged in a table form it would be easier. See the table below.

c	\({x^n}\)	\({x^{n - 1}}\)	\({x^{n - 2}}\)	...	\({c_n}\)
	\({c_1}\)	\({c_2}\)	\({c_3}\)	...	\({c_n}\)
	\({c_1}\)	\({c_2} + c{b_1}\)	\({c_3} + c{b_2}\)	...	\({c_n} + c{b_n}\)
	\({b_1}\)	\({b_2}\)	\({b_3}\)	...	\(P(r)\)