### Math Notes

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#### Algebra Solutions

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Derivation of Synthetic Division

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.

$${c_1} = {b_1}$$

$${c_2} = {b_2} - c{b_1}$$

$${c_3} = {b_3} - c{b_2}$$

:
.

$${c_n} = P(r) - c{b_n}$$

Then we solve the value of b's.

$${b_1} = {c_1}$$

$${b_2} = {c_2} + c{b_1}$$

$${b_3} = {c_3} + c{b_2}$$

:
.

$$P(r) = {c_n} + c{b_n}$$

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