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a^2 – 5a – 20 factor completely

Mathematics
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-5(4) right ?
No not this time I think you have copied your question incorrectly it should be a^2 - a - 20 right ?
no it has 5 in this one

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Other answers:

ok
i see it's prime so how do i show work
is the answer \[5+\sqrt{105}/2,5-\sqrt{105}/2\]
when it is prime, we just state "prime" and do not attempt to factor. Now if it said, solve then we solve using the quadratic formula
7.623,-2.623
yes or no?
no because it's prime
but it can be solved by discriminant method
it just said factor and show work so i don't think i have to use that .
correct, if the instructions were to factor write "prime" if the instructions were to solve then use the quadratic formula
It's prime! Applying quadratic formula, we get: a^2 – 5*a – 20 = 0 a = 1, b = - 5, c = - 20 b^2 - 4*a*c = 25 - 4*1*( - 20) = 105 a1 = ( - ( - 5) + sqrt(105))/2 = 5/2 + (1/2)*sqrt(105) a2 = 5/2 - (1/2)*sqrt(105)
ya??got it!

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