Hi Sam,

It would be impossible to completely factor -12a^{2}+20a-21 into the product of real numbers. The expression would be considered prime in that case. This is probably the answer you are looking for.

However, factoring would be possible if *complex* linear factors are allowed. It would require using the definition i = √(-1) and the quadratic formula:

a = (-B ± √(B^{2}-4AC))/2A

My first step would be to factor out -1/3, turning A into a positive square number:

-12a^{2}+20a-21 = -1/3(36a^{2}-60a+63)

From there, I would use the quadratic formula with A=36, B=-60 and C=63:

a = (-(-60) ± √((-60)^{2}-4(36)(63)))/(2(36))

= (60 ± √(3600-9072))/72 [Simplifying]

= (60 ± √(-5472))/72 [Subtracting]

= (60 ± 12i√(38))/72 [Simplifying the radical]

= 12(5 ± i√(38))/72 [Factoring the numerator]

= (5 ± i√(38))/6 [Reducing]

With a bit of rewriting this would give a factorization of:

-1/3(6a - 5 + i√(38))(6a - 5 - i√(38)) = -1/3(-6i a + √(38) + 5i)(6i a + √(38) - 5i)

Bonus:

It is also possible to complete the square using the expression -12a^{2}+20a-21:

-12a^{2}+20a-21 = -12(a^{2}-5/3a)-21 [Rewriting]

= -12(a^{2}-5/3a + 25/36)-21 + 25/3 [Completing the square]

= -12(a-5/6)^{2}-63/3+25/3 [Rewriting]

= -12(a-5/6)^{2}-38/3 [Simplifying]

This form is important, primarily because it shows that at a = 5/6 the expression reaches a maximum value of -38/3.