# Algebra

### Remainder and Factor Theorem

###### Remainder Theorem and Factor Theorem

If a polynomial $$F(x)$$ divided by $$p(x)$$, the quotient is $$H(x)$$ and the remainder is $$S(x)$$, then the polynomial can be written as:

$$F(x) = p(x) \:.\: H(x) + S(x)$$

Example 01

Find the quotient and the remainder of $$2x^3 + 5x^2 - 6x + 3$$ divided by $$(x + 2)$$

\begin{array}{r} \bbox[5px, border: 2px solid red] {2x^2 + x - 8}\\ x + 2\enclose{longdiv}{2x^3 + 5x^2 - 6x + 3}\\ \underline{2x^3 + 4x^2}\hspace{4em}\\ x^2 - 6x + 3\hspace{.33em}\\ \underline{x^2 + 2x}\hspace{2em}\\ -8x + 3\hspace{.33em}\\ \underline{-8x - 16}\\ \bbox[5px, border: 2px solid blue] {19} \end{array}

Quotient:  $${\color {red} 2x^2 + x - 8}$$

Remainder: $${\color {blue} 19}$$

$$2x^3 + 5x^2 - 6x + 3 = (x + 2) \:.\: (2x^2 + x - 8) + 19$$

Remainder Theorem

If a polynomial $$F(x)$$ divided  by $$(x - a)$$, the remainder is $$F(a)$$.

Factor Theorem

If $$(x - a)$$ is a factor of a polynomial $$F(x)$$, then $$F(a) = 0$$.

Example 02

Find the remainder of $$F(x) = x^3 + 4x^2 - 6x + 1$$ divided by $$(x - 2)$$

$$F(x) = x^3 + 4x^2 - 6x + 1$$

$$F(2) = 2^3 + 4 \:.\: 2^2 - 6 \:.\: 2 + 1$$

$$F(2) = 13$$

We can write the polynomial as

$$x^3 + 4x^2 - 6x + 1 = (x - 2) \:.\: H(x) + 13$$

Example 03

Prove that $$(x + 3)$$ is a factor of $$F(x) = x^3 - x^2 - 10x + 6$$

$$F(-3) = (-3)^3 - (-3)^2 - 10(-3) + 6 = 0$$

Since $$F(-3) = 0$$, then $$(x + 3)$$ is a factor of $$F(x)$$.

##### Exercise

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