document.write( "Question 449383: Hey, Im in much need of help to figure out how to factor/solve qudaratic equations that are not in X^2+bx+c=0 form. for example,how do you factor 4x^2=4x+5? \n" ); document.write( "

Algebra.Com's Answer #309129 by MathLover1(20849) $\"\"$ $\"About$

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\n" ); document.write( "\n" ); document.write( "factor $\"4x%5E2=4x%2B5\"$ ..first write it in $\"x%5E2%2Bbx%2Bc=0\"$ form\r
\n" ); document.write( "\n" ); document.write( " $\"4x%5E2-4x-5=0\"$ \r
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Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)

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

Now multiply the first coefficient $\"4\"$ by the last term $\"-5\"$ to get $\"%284%29%28-5%29=-20\"$ .

Now the question is: what two whole numbers multiply to $\"-20\"$ (the previous product) and add to the second coefficient $\"-4\"$ ?

To find these two numbers, we need to list all of the factors of $\"-20\"$ (the previous product).

Factors of $\"-20\"$ :

1,2,4,5,10,20

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

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

These factors pair up and multiply to $\"-20\"$ .

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

Now let's add up each pair of factors to see if one pair adds to the middle coefficient $\"-4\"$ :

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First Number	Second Number	Sum
1	-20	1+(-20)=-19
2	-10	2+(-10)=-8
4	-5	4+(-5)=-1
-1	20	-1+20=19
-2	10	-2+10=8
-4	5	-4+5=1

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

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Answer:

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

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

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