SOLUTION: see look i get this >> b + 7b - 30 ( b - 3 ) ( b + 10 ) u put " b " at the begining of both ( ) then u ask ur self wat times wat = -30? wich wud be -3

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: see look i get this >> b + 7b - 30 ( b - 3 ) ( b + 10 ) u put " b " at the begining of both ( ) then u ask ur self wat times wat = -30? wich wud be -3       Log On


   



Question 113336This question is from textbook Algebra structure and method book 1
: see look i get this >> b + 7b - 30
( b - 3 ) ( b + 10 )
u put " b " at the begining of both ( )
then u ask ur self wat times wat = -30?
wich wud be -3 times 10
Now u have to make sure that when u add -3 + 10 u get 7
So can u please tell me how to do this problem!
10e - 12e + 3
wat i dont get the only two #s u can x together to get 3 is 1 times 3
But when u add 1 + 3 it dont = to -12 !
Wat am i doin wrong?!?
By the way my name is Angie
This question is from textbook Algebra structure and method book 1

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)


Looking at the expression 10e%5E2-12e%2B3, we can see that the first coefficient is 10, the second coefficient is -12, and the last term is 3.



Now multiply the first coefficient 10 by the last term 3 to get %2810%29%283%29=30.



Now the question is: what two whole numbers multiply to 30 (the previous product) and add to the second coefficient -12?



To find these two numbers, we need to list all of the factors of 30 (the previous product).



Factors of 30:

1,2,3,5,6,10,15,30

-1,-2,-3,-5,-6,-10,-15,-30



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to 30.

1*30 = 30
2*15 = 30
3*10 = 30
5*6 = 30
(-1)*(-30) = 30
(-2)*(-15) = 30
(-3)*(-10) = 30
(-5)*(-6) = 30


Now let's add up each pair of factors to see if one pair adds to the middle coefficient -12:



First NumberSecond NumberSum
1301+30=31
2152+15=17
3103+10=13
565+6=11
-1-30-1+(-30)=-31
-2-15-2+(-15)=-17
-3-10-3+(-10)=-13
-5-6-5+(-6)=-11




From the table, we can see that there are no pairs of numbers which add to -12. So 10e%5E2-12e%2B3 cannot be factored.



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Answer:



So 10%2Ae%5E2-12%2Ae%2B3 doesn't factor at all (over the rational numbers).



So 10%2Ae%5E2-12%2Ae%2B3 is prime.