Question

`3x^2-8x+15`? (factoring trinomials without quadratic formula)

Answer 1

`3x^2-8x+15 = 3(x-4/3-sqrt(29)/3i)(x-4/3+sqrt(29)/3i)`

Explanation:

Given:

`3x^2-8x+15`

Note that this is in standard form:

`ax^2+bx+c`

with `a=3`, `b=-8` and `c=15`

This has discriminant `Delta` given by the formula:

`Delta = b^2-4ac = (color(blue)(-8))^2-4(color(blue)(3))(color(blue)(15)) = 64-180 = -116`

Since `Delta < 0` this quadatic has no real zeros and no linear factors with real coefficients.

We can factor it with complex coefficients by completing the square:

`3(3x^2-8x+15) = 9x^2-24x+45`

`color(white)(3(3x^2-8x+15)) = (3x)^2-2(3x)(4)+4^2+29`

`color(white)(3(3x^2-8x+15)) = (3x-4)^2+(sqrt(29))^2`

`color(white)(3(3x^2-8x+15)) = (3x-4)^2-(sqrt(29)i)^2`

`color(white)(3(3x^2-8x+15)) = ((3x-4)-sqrt(29)i)((3x-4)+sqrt(29)i)`

`color(white)(3(3x^2-8x+15)) = (3x-4-sqrt(29)i)(3x-4+sqrt(29)i)`

So:

`3x^2-8x+15 = 1/3(3x-4-sqrt(29)i)(3x-4+sqrt(29)i)`

`color(white)(3x^2-8x+15) = 3(x-4/3-sqrt(29)/3i)(x-4/3+sqrt(29)/3i)`