Question

How do you factor `21x^3 - 18x^2y + 24xy^2`?

Answer 1

`21x^3-18x^2y+24xy^2 = 3x(7x^2-6xy+8y^2)`

Explanation:

Note that all of the terms are divisible by `3x`, so we can separate that out as a factor first:

`21x^3-18x^2y+24xy^2 = 3x(7x^2-6xy+8y^2)`

The remaining quadratic is in the form `ax^2+bxy+cy^2` with `a=7`, `b=-6` and `c=8`.

This has discriminant `Delta` given by the formula:

`Delta = b^2-4ac = (-6)^2-4*7*8 = 36 - 224 = -188`

Since `Delta < 0` this quadratic has no linear factors with Real coefficients.

`color(white)()`
Complex coefficients

If we are determined to factor it, we can find some Complex coefficients using the quadratic formula to find zeros:

`x/y = (-b+-sqrt(b^2-4ac))/(2a)`

`color(white)(x/y) = (-b+-sqrt(Delta))/(2a)`

`color(white)(x/y) = (6+-sqrt(-188))/14`

`color(white)(x/y) = (6+-2sqrt(47)i)/14`

`color(white)(x/y) = (3+-sqrt(47)i)/7`

Hence:

`7x^2-6xy+8y^2 = 7(x - ((3+sqrt(47)i)/7)y)(x - ((3-sqrt(47)i)/7)y)`

`color(white)(7x^2-6xy+8y^2) = 1/7(7x - (3+sqrt(47)i)y)(7x - (3-sqrt(47)i)y)`