jchick one year ago solve and show all work for a2 – a – 20 and a2 – 5a – 20

1. myininaya

what are you solving?

2. jchick

3. myininaya

ok for the first one we have a^2-a-20 a^2-1a-20 so first step is to ask yourself what two integers multiply to by -20 and add up to be -1?

4. jchick

What do you mean?

5. myininaya

$ax^2+bx+c \\ \text{ when } a=1 \text{ and if the expression is "factorable" } \\ \text{ then you should be able to find } \\ \text{ two integers so that when you multiply them together you get } c \\ \text{ and when you add them together you get } b$ for example: $x^2+5x+6 \\ \text{ the question here first to answer is: } \\ \text{ what two integers multiply to be 6 } \\ \text{ and add up to be 5}$ well 2(3)=6 and 2+3=5 so the factored form of $x^2+5x+6 \text{ is } (x+2)(x+3)$

6. jchick

a2 – a – 20 (a )(a ) -4*5 10*-2 (a+4)(a-5) a2 -5a+4a-2 -a a2 – 5a – 20 (a )(a ) 10*-2 -5*4 (a-5)(a+4) a2 + 12ab + 27b2 (a )(a ) 9*3 (a+3b)(a+9b) a2 +9ab+3ab+27b a2 +12ab+27b

7. jchick

Right?

8. myininaya

yes to the first one: a^2-a-20=(a-5)(a+4) since -5*4=-20 and -5+4=-1 no to the second one: a^2-5a-20 this does not equal (a-5)(a+4) -5(4)=-20 but -5+4 isn't -5

9. myininaya

and the third one looks good

10. jchick

a2 – a – 20 (a )(a ) -4*5 10*-2 (a+4)(a-5)

11. myininaya

so you got 2/3 correct

12. myininaya

you just need to fix the second one

13. jchick

Ok

14. jchick

I will try

15. myininaya

a^2-5a-20 are there two integers that multiply to be -20 and add up to be -5?

16. myininaya

possible pairings that multiply to be -20: 1(-20) -1(20) 2(-10) -2(10) 4(-5) -4(5)

17. myininaya

do any of those pairs also add up to -5?

18. myininaya

which equation is true if any: 1+(-20)=-5 -1+20=-5 2+(-10)=-5 -2+10=-5 4+(-5)=-5 -4+5=-5

19. jchick

None

20. myininaya

you are right

21. jchick

They are all false

22. myininaya

so that means that this is a prime trinomial

23. myininaya

means it is not "factorable"

24. myininaya

I put factorable in quotation marks because it only really means it isn't factorable over the integers

25. jchick

Ok

26. jchick

So I am not sure on the second one

27. myininaya

it isn't factorable

28. myininaya

it is prime

29. jchick

Ok thanks

30. myininaya

a^2-5a-20 does not factor

31. myininaya

unless you can factor over the reals?

32. myininaya

which I don't think that is the instructions

33. jchick

No but how would I write this just say it is prime?

34. myininaya

you can either say not factorable or prime

35. myininaya

which ever vocabulary your teacher prefers really but either saying is fine in general

36. jchick

37. jchick

how does this look?

38. jchick

@myininaya