## AshhSmith a^2 – 5a – 20 factor completely one year ago one year ago

1. AshhSmith

-5(4) right ?

2. vishweshshrimali5

No not this time I think you have copied your question incorrectly it should be a^2 - a - 20 right ?

3. AshhSmith

no it has 5 in this one

4. vishweshshrimali5

ok

5. AshhSmith

i see it's prime so how do i show work

6. nitz

is the answer $5+\sqrt{105}/2,5-\sqrt{105}/2$

7. precal

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

8. nitz

7.623,-2.623

9. nitz

yes or no?

10. AshhSmith

no because it's prime

11. nitz

but it can be solved by discriminant method

12. AshhSmith

it just said factor and show work so i don't think i have to use that .

13. precal

correct, if the instructions were to factor write "prime" if the instructions were to solve then use the quadratic formula

14. nitz

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)

15. nitz

ya??got it!