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

Algebra.Com's Answer #82474 by jim_thompson5910(35256) $\"\"$ $\"About$

You can put this solution on YOUR website!
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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 $\"-12\"$ :

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First Number	Second Number	Sum
1	30	1+30=31
2	15	2+15=17
3	10	3+10=13
5	6	5+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.

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