## AshhSmith Group Title a^2 – 5a – 20 factor completely 2 years ago 2 years ago

1. AshhSmith Group Title

-5(4) right ?

2. vishweshshrimali5 Group Title

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

3. AshhSmith Group Title

no it has 5 in this one

4. vishweshshrimali5 Group Title

ok

5. AshhSmith Group Title

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

6. nitz Group Title

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

7. precal Group Title

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 Group Title

7.623,-2.623

9. nitz Group Title

yes or no?

10. AshhSmith Group Title

no because it's prime

11. nitz Group Title

but it can be solved by discriminant method

12. AshhSmith Group Title

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

13. precal Group Title

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

14. nitz Group Title

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 Group Title

ya??got it!