More about polynomials- (4.1-4.3)

Perform division of polynomials

  • Theory
  • Examples
    Find the quotient and the remainder of \((3x^2 – 6x + 2) ÷ (x – 1)\)
  • Solutions

    \(\begin{array}{c}\hspace{3cm}{\rm{ }}3x{\rm{ }} - 3\\x - 1\left){\vphantom{1\begin{array}{l}3{x^2} - 6x + 2\\\underline {3{x^2} - 3x{\rm{ }}} \\{\rm{ }} - 3x + 2\\{\rm{ }}\underline { - 3x + 3} \\{\rm{ }} - 1\end{array}}}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}3{x^2} - 6x + 2\\\underline {3{x^2} - 3x{\rm{ }}} \\{\rm{ }} - 3x + 2\\{\rm{ }}\underline { - 3x + 3} \\{\rm{ }} - 1\end{array}}}\end{array}\)

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

Understand the remainder theorem

  • Theory
  • Examples
    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).
  • Solutions
    \(\begin{array}{1}a)\qquad f(1) \\= 2{(1)^3} + 6{(1)^2} + 2(1) - 7\\= 2 + 6 + 2 - 7\\= \underline{\underline 3} \end{array}\)

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

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

Understand the factor theorem

  • Theory
  • Examples
    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) \(2x + 3\)
  • Solutions
    \(\begin{array}{1}a)\qquad h(1)\\ = 6{(1)^3} - 4{(1)^2} - 9(1) + 6\\ = - 1\\ \ne 0\end{array}\)
    \(\therefore x -1\) 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).