document.write( "Question 260263: 2x^3+2x^2-4x over x^2-6x+9 simplify \n" ); document.write( "

Algebra.Com's Answer #191716 by richwmiller(17219) $\"\"$ $\"About$

You can put this solution on YOUR website!
(2x(x^2+x-2))/(x-3)^2\r
\n" ); document.write( "\n" ); document.write( " $\"2x%2A+%28x-1%29%2A%28x%2B2%29%2F%28x-3%29%5E2\"$
\n" ); document.write( "Below are instructions on how to factor x^2+x-2 and x^2-6x+9
\n" ); document.write( "Well worth learning. They are long but easy to follow.
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Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)

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

Now multiply the first coefficient $\"2\"$ by the last term $\"-2\"$ to get $\"%282%29%28-2%29=-4\"$ .

Now the question is: what two whole numbers multiply to $\"-4\"$ (the previous product) and add to the second coefficient $\"1\"$ ?

To find these two numbers, we need to list all of the factors of $\"-4\"$ (the previous product).

Factors of $\"-4\"$ :

1,2,4

-1,-2,-4

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

These factors pair up and multiply to $\"-4\"$ .

1*(-4) = -4
2*(-2) = -4
(-1)*(4) = -4
(-2)*(2) = -4

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

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First Number	Second Number	Sum
1	-4	1+(-4)=-3
2	-2	2+(-2)=0
-1	4	-1+4=3
-2	2	-2+2=0

From the table, we can see that there are no pairs of numbers which add to $\"1\"$ . So $\"2x%5E2%2Bx-2\"$ cannot be factored.

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Answer:

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

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

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

Looking at the expression $\"x%5E2-6x%2B9\"$ , we can see that the first coefficient is $\"1\"$ , the second coefficient is $\"-6\"$ , and the last term is $\"9\"$ .

Now multiply the first coefficient $\"1\"$ by the last term $\"9\"$ to get $\"%281%29%289%29=9\"$ .

Now the question is: what two whole numbers multiply to $\"9\"$ (the previous product) and add to the second coefficient $\"-6\"$ ?

To find these two numbers, we need to list all of the factors of $\"9\"$ (the previous product).

Factors of $\"9\"$ :

1,3,9

-1,-3,-9

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

These factors pair up and multiply to $\"9\"$ .

1*9 = 9
3*3 = 9
(-1)*(-9) = 9
(-3)*(-3) = 9

Now let's add up each pair of factors to see if one pair adds to the middle coefficient $\"-6\"$ :

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First Number	Second Number	Sum
1	9	1+9=10
3	3	3+3=6
-1	-9	-1+(-9)=-10
-3	-3	-3+(-3)=-6

From the table, we can see that the two numbers $\"-3\"$ and $\"-3\"$ add to $\"-6\"$ (the middle coefficient).

So the two numbers $\"-3\"$ and $\"-3\"$ both multiply to $\"9\"$ and add to $\"-6\"$

Now replace the middle term $\"-6x\"$ with $\"-3x-3x\"$ . Remember, $\"-3\"$ and $\"-3\"$ add to $\"-6\"$ . So this shows us that $\"-3x-3x=-6x\"$ .

$\"x%5E2%2Bhighlight%28-3x-3x%29%2B9\"$ Replace the second term $\"-6x\"$ with $\"-3x-3x\"$ .

$\"%28x%5E2-3x%29%2B%28-3x%2B9%29\"$ Group the terms into two pairs.

$\"x%28x-3%29%2B%28-3x%2B9%29\"$ Factor out the GCF $\"x\"$ from the first group.

$\"x%28x-3%29-3%28x-3%29\"$ Factor out $\"3\"$ 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.

$\"%28x-3%29%28x-3%29\"$ Combine like terms. Or factor out the common term $\"x-3\"$

$\"%28x-3%29%5E2\"$ Condense the terms.

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Answer:

So $\"x%5E2-6%2Ax%2B9\"$ factors to $\"%28x-3%29%5E2\"$ .

In other words, $\"x%5E2-6%2Ax%2B9=%28x-3%29%5E2\"$ .

Note: you can check the answer by expanding $\"%28x-3%29%5E2\"$ to get $\"x%5E2-6%2Ax%2B9\"$ or by graphing the original expression and the answer (the two graphs should be identical).

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