^{2 }-9x-16,200

how do you factor x^{2 }-9x-16,200

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by calculating b2-4ac from quadratic formula. If b2-4ac doesn't become perfect square, it is prime.

Brian M. | MathmagicianMathmagician

x^2 – 9x

-16,200

First, the

coefficient of the first term (1) is prime, so we will not be able to factor a

constant out. Next, there is no variable in the last term (16,200 (there is no

"x")), so we cannot factor out a variable.

Next, we

must find factors of 16,200 that differ by 9, so that when added together we

can get the -9 value that we need in the middle term.

Since

122*131=15,982 (which is less than 16,200), and 123*132=16,236 (which is greater than 16,200), these would be the possible factors of 16,200 that might have combined to produce the -9 that we were after. Since neither pair are actually factors of 16,200, it would seem that 16,200 does not have factors that differ exactly by a value of 9. Therefore, the polynomial cannot be factored. It must be prime.

Since b^2 - 4ac =(-9)^2 - 4(1)(-16,200) = 64,881 is not a perfect square, x^2 - 9x - 16,200 is prime.However, you can factor it in irrational domain by completing the square or using quadratic formula.

Here is the way by completing the square:

x^2-9x+(9/2)^2 - (9/2)^2 - 16,200

= (x-4.5)^2 - 16,220.25

= [x-4.5+sqrt(16,220.25)][x-4.5-sqrt(16,220.25)]

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