I can't seem to figure it out. Thanks!

## How do you factor the expression -12a^2+20a-21

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# 2 Answers

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))/2AMy 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.

This expression can not be factorized in the set of real numbers.

There is formula for factorization of quadratic expression: ax^2+bx+c=a(x-x,)(x-x,,) , where x, and x.. are roots of quadratic equation ax^2+bx+c=0 . To find the roots we can use the formula x = (-b ± √D)/2a , where D=b^2-4ac . There are 2 roots if D>0 , there
is 1root if D=0 , and zero roots in set of real numbers if D is negative.

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Let's use information above: -12 a^2 + 20a - 21 = 0

D = 20^2 -4(-12)(-21) < 0