document.write( "Question 277983: factor trinomials 4x^2+9x+10= \n" ); document.write( "

Algebra.Com's Answer #202362 by richwmiller(17219) $\"\"$ $\"About$

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Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)

Looking at the expression $\"4x%5E2%2B9x%2B10\"$ , we can see that the first coefficient is $\"4\"$ , the second coefficient is $\"9\"$ , and the last term is $\"10\"$ .

Now multiply the first coefficient $\"4\"$ by the last term $\"10\"$ to get $\"%284%29%2810%29=40\"$ .

Now the question is: what two whole numbers multiply to $\"40\"$ (the previous product) and add to the second coefficient $\"9\"$ ?

To find these two numbers, we need to list all of the factors of $\"40\"$ (the previous product).

Factors of $\"40\"$ :

1,2,4,5,8,10,20,40

-1,-2,-4,-5,-8,-10,-20,-40

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

These factors pair up and multiply to $\"40\"$ .

1*40 = 40
2*20 = 40
4*10 = 40
5*8 = 40
(-1)*(-40) = 40
(-2)*(-20) = 40
(-4)*(-10) = 40
(-5)*(-8) = 40

Now let's add up each pair of factors to see if one pair adds to the middle coefficient $\"9\"$ :

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First Number	Second Number	Sum
1	40	1+40=41
2	20	2+20=22
4	10	4+10=14
5	8	5+8=13
-1	-40	-1+(-40)=-41
-2	-20	-2+(-20)=-22
-4	-10	-4+(-10)=-14
-5	-8	-5+(-8)=-13

From the table, we can see that there are no pairs of numbers which add to $\"9\"$ . So $\"4x%5E2%2B9x%2B10\"$ cannot be factored.

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Answer:

So $\"4%2Ax%5E2%2B9%2Ax%2B10\"$ doesn't factor at all (over the rational numbers).

So $\"4%2Ax%5E2%2B9%2Ax%2B10\"$ is prime.

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