#### Perform division of polynomials

• Find the quotient and the remainder of $$(3x^2 – 6x + 2) ÷ (x – 1)$$
• $$\begin{array}{c}\hspace{3cm}{\rm{ }}3x{\rm{ }} - 3\\x - 1\left){\vphantom{1\begin{array}{l}3{x^2} - 6x + 2\\\underline {3{x^2} - 3x{\rm{ }}} \\{\rm{ }} - 3x + 2\\{\rm{ }}\underline { - 3x + 3} \\{\rm{ }} - 1\end{array}}}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}3{x^2} - 6x + 2\\\underline {3{x^2} - 3x{\rm{ }}} \\{\rm{ }} - 3x + 2\\{\rm{ }}\underline { - 3x + 3} \\{\rm{ }} - 1\end{array}}}\end{array}$$

∴ Quotient $$= \underline{\underline {3x - 3}}$$
Remainder $$= \underline{\underline { - 1}}$$

#### Understand the remainder theorem

• Consider $$f(x) = 2{x^3} + 6{x^2} + 2x - 7$$.
(a) Find the values of f(1) and f(–1).
(b) Hence, find the remainder of f(x) ÷ (x + 1).
• $$\begin{array}{1}a)\qquad f(1) \\= 2{(1)^3} + 6{(1)^2} + 2(1) - 7\\= 2 + 6 + 2 - 7\\= \underline{\underline 3} \end{array}$$

$$\begin{array}{1}f( - 1) \\ = 2{( - 1)^3} + 6{( - 1)^2} + 2( - 1) - 7\\ = - 2 + 6 - 2 - 7\\ = \underline{\underline { - 5}} \end{array}$$

(b) By the remainder theorem, we have
remainder = f(–1)
= –5

#### Understand the factor theorem

• Let $$h(x) = 6{x^3} - 4{x^2} - 9x + 6$$.
Determine whether each of the following is a factor of h(x) by using the factor theorem.
(a) $$x - 1$$
(b) $$2x + 3$$
• $$\begin{array}{1}a)\qquad h(1)\\ = 6{(1)^3} - 4{(1)^2} - 9(1) + 6\\ = - 1\\ \ne 0\end{array}$$
$$\therefore x -1$$ is not a factor of h(x).

$$\begin{array}{1}b)\qquad h( - \frac{3}{2}) \\= 6{( - \frac{3}{2})^3} - 4{( - \frac{3}{2})^2} - 9( - \frac{3}{2}) + 6\\ = - \frac{{39}}{4}\\ \ne 0\end{array}$$
$$\therefore 2x+ 3$$ is not a factor of h(x).