Question

How do you factor `c^ { 2} - 3c + 19`?

Answer 1

`c^2-3c+19 = (c-3/2-sqrt(67)/2i)(c-3/2+sqrt(67)/2i)`

Explanation:

We can factor this quadratic trinomial by completing the square, but only with non-real complex coefficients.

Ths difference of squares identity can be written:

`a^2-b^2 = (a-b)(a+b)`

We use this with `a=(c-3/2)` and `b=sqrt(67)/2i` as follows:

`c^2-3c+19 = c^2-2(3/2)c+9/4+67/4`

`color(white)(c^2-3c+19) = (c-3/2)^2+(sqrt(67)/2)^2`

`color(white)(c^2-3c+19) = (c-3/2)^2-(sqrt(67)/2i)^2`

`color(white)(c^2-3c+19) = ((c-3/2)-sqrt(67)/2i)((c-3/2)+sqrt(67)/2i)`

`color(white)(c^2-3c+19) = (c-3/2-sqrt(67)/2i)(c-3/2+sqrt(67)/2i)`

where `i` is the imaginary unit, satisfying `i^2=-1`