document.write( "Question 103799: Level 1, General Polynomials, Degree 2 \r
\n" ); document.write( "\n" ); document.write( "Write your answer in the form (ay + b) (cy +d) .... (ey +f)\r
\n" ); document.write( "\n" ); document.write( "New Problem, Factor: \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "- 4 y2 - y + 3
\n" ); document.write( " \n" ); document.write( "

Algebra.Com's Answer #75458 by jim_thompson5910(35256) $\"\"$ $\"About$

You can put this solution on YOUR website!
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

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

$\"-4%2Ay%5E2-y%2B3\"$ Start with the given expression.

$\"-%284y%5E2%2By-3%29\"$ Factor out the GCF $\"-1\"$ .

Now let's try to factor the inner expression $\"4y%5E2%2By-3\"$

---------------------------------------------------------------

Looking at the expression $\"4y%5E2%2By-3\"$ , we can see that the first coefficient is $\"4\"$ , the second coefficient is $\"1\"$ , and the last term is $\"-3\"$ .

Now multiply the first coefficient $\"4\"$ by the last term $\"-3\"$ to get $\"%284%29%28-3%29=-12\"$ .

Now the question is: what two whole numbers multiply to $\"-12\"$ (the previous product) and add to the second coefficient $\"1\"$ ?

To find these two numbers, we need to list all of the factors of $\"-12\"$ (the previous product).

Factors of $\"-12\"$ :

1,2,3,4,6,12

-1,-2,-3,-4,-6,-12

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

These factors pair up and multiply to $\"-12\"$ .

1*(-12) = -12
2*(-6) = -12
3*(-4) = -12
(-1)*(12) = -12
(-2)*(6) = -12
(-3)*(4) = -12

Now let's add up each pair of factors to see if one pair adds to the middle coefficient $\"1\"$ :

\n" ); document.write( "

First Number	Second Number	Sum
1	-12	1+(-12)=-11
2	-6	2+(-6)=-4
3	-4	3+(-4)=-1
-1	12	-1+12=11
-2	6	-2+6=4
-3	4	-3+4=1

From the table, we can see that the two numbers $\"-3\"$ and $\"4\"$ add to $\"1\"$ (the middle coefficient).

So the two numbers $\"-3\"$ and $\"4\"$ both multiply to $\"-12\"$ and add to $\"1\"$

Now replace the middle term $\"1y\"$ with $\"-3y%2B4y\"$ . Remember, $\"-3\"$ and $\"4\"$ add to $\"1\"$ . So this shows us that $\"-3y%2B4y=1y\"$ .

$\"4y%5E2%2Bhighlight%28-3y%2B4y%29-3\"$ Replace the second term $\"1y\"$ with $\"-3y%2B4y\"$ .

$\"%284y%5E2-3y%29%2B%284y-3%29\"$ Group the terms into two pairs.

$\"y%284y-3%29%2B%284y-3%29\"$ Factor out the GCF $\"y\"$ from the first group.

$\"y%284y-3%29%2B1%284y-3%29\"$ Factor out $\"1\"$ from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.

$\"%28y%2B1%29%284y-3%29\"$ Combine like terms. Or factor out the common term $\"4y-3\"$

--------------------------------------------------

So $\"-1%284y%5E2%2By-3%29\"$ then factors further to $\"-%28y%2B1%29%284y-3%29\"$

===============================================================

Answer:

So $\"-4%2Ay%5E2-y%2B3\"$ completely factors to $\"-%28y%2B1%29%284y-3%29\"$ .

In other words, $\"-4%2Ay%5E2-y%2B3=-%28y%2B1%29%284y-3%29\"$ .

Note: you can check the answer by expanding $\"-%28y%2B1%29%284y-3%29\"$ to get $\"-4%2Ay%5E2-y%2B3\"$ or by graphing the original expression and the answer (the two graphs should be identical).

\n" ); document.write( " \n" ); document.write( "

\n" );