3FG2MoreExamples


 * Factorising by introducing i 2 **


 * Example 1 **

math \text{Factorise } \; z^2+10 \quad z\in C math


 * Solution:**


 * In the real field, this can't be factorised.
 * But introducing i 2 = –1 makes this a difference of two squares

math \\ . \qquad z^2+10 \\. \\ . \qquad = z^2-10i^2 \\. \\ . \qquad = \left( z+\sqrt{10}\,i \right) \left( z-\sqrt{10}\,i \right) math

Example 2

math \text{Factorise } \; z^2+6z+13 \quad z\in C math


 * Solution:**


 * Factorise by completing the square

math \\ . \qquad z^2+6z+13 \qquad \textit{halve the middle term then square to get the constant term to make a perfect square}\\ .\\ . \qquad = \underline{z^2+6z+9} \; -9+13 \qquad \textit{the first 3 terms form a perfect square}\\. \\ . \qquad = (z+3)^2+4 math


 * In the real field, this can't be factorised.
 * But introducing i 2 = –1 makes this a difference of two squares

math \\ . \qquad = (z+3)^2-4i^2 \\. \\ . \qquad = (z+3+2i)(z+3-2i) math

Example 3

math \text{Factorise }\; 2z^2+5z+5 \quad z\in C math


 * Solution:**

Take out 2 as a common factor, then complete the square

math \\ . \qquad 2z^2+5z+5 \\. \\ . \qquad =2 \left( z^2 + \frac{5}{2}z + \frac{5}{2} \right) \\. \\ . \qquad =2 \left( \underline{z^2 + \frac{5}{2}z + \frac{25}{16}} \; -\frac{25}{16}+\frac{5}{2} \right) \\. \\ . \qquad =2 \left( \left(z + \frac{5}{4} \right)^2 + \frac{15}{16} \right) math

Introduce i 2 = –1 to make this a difference of two squares

math \\ . \qquad =2 \left( \left(z + \frac{5}{4} \right)^2 - \frac{15}{16}i^2 \right) \\ .\\ . \qquad =2 \left( z + \dfrac{5}{4} + \dfrac{\sqrt{15}\,i}{4} \right) \left( z + \dfrac{5}{4} - \dfrac{\sqrt{15}\,i}{4} \right) math .