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a^2 – 5a – 20 factor completely

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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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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
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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