document.write( "Question 161761: 64.\r
\n" ); document.write( "\n" ); document.write( "Factor each polynomial.\r
\n" ); document.write( "\n" ); document.write( "h^2-9hs+9s^2\r
\n" ); document.write( "\n" ); document.write( "26.\r
\n" ); document.write( "\n" ); document.write( "Factor using AC method\r
\n" ); document.write( "\n" ); document.write( "21x^2+2x-3\r
\n" ); document.write( "\n" ); document.write( "72.Factor\r
\n" ); document.write( "\n" ); document.write( "3x^2-18x-48\r
\n" ); document.write( "\n" ); document.write( "18. Factor\r
\n" ); document.write( "\n" ); document.write( "2h^2-h-3
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Algebra.Com's Answer #119216 by MathLover1(20849) $\"\"$ $\"About$

You can put this solution on YOUR website!
Factor each polynomial.
\n" ); document.write( " $\"h%5E2-9hs%2B9s%5E2+\"$ \r
\n" ); document.write( "\n" ); document.write( "Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 9hs
\n" ); document.write( "First Number---- Second Number---- Sum
\n" ); document.write( "1ЕЕЕЕЕЕЕЕЕ...........9ЕЕЕЕЕЕЕЕЕЕ.........10
\n" ); document.write( "3ЕЕЕЕЕЕЕЕЕ...........3ЕЕЕЕЕЕЕЕЕЕ.........6
\n" ); document.write( "-1ЕЕЕЕЕЕЕЕ..........-9ЕЕЕЕЕЕЕЕЕ........Е-10
\n" ); document.write( "-3ЕЕЕЕЕЕЕЕ..........-3ЕЕЕЕЕЕЕЕЕ........Е-6\r
\n" ); document.write( "\n" ); document.write( "None of these factors add to 9hs.
\n" ); document.write( "polynomial $\"h%5E2-9hs%2B9s%5E2+\"$ cannot be factored\r
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\n" ); document.write( "\n" ); document.write( "26.
\n" ); document.write( "Factor using AC method \r
\n" ); document.write( "\n" ); document.write( " $\"21x%5E2%2B2x-3\"$ \r
\n" ); document.write( "\n" ); document.write( "\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

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

Looking at the expression $\"21x%5E2%2B2x-3\"$ , we can see that the first coefficient is $\"21\"$ , the second coefficient is $\"2\"$ , and the last term is $\"-3\"$ .

Now multiply the first coefficient $\"21\"$ by the last term $\"-3\"$ to get $\"%2821%29%28-3%29=-63\"$ .

Now the question is: what two whole numbers multiply to $\"-63\"$ (the previous product) and add to the second coefficient $\"2\"$ ?

To find these two numbers, we need to list all of the factors of $\"-63\"$ (the previous product).

Factors of $\"-63\"$ :

1,3,7,9,21,63

-1,-3,-7,-9,-21,-63

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

These factors pair up and multiply to $\"-63\"$ .

1*(-63) = -63
3*(-21) = -63
7*(-9) = -63
(-1)*(63) = -63
(-3)*(21) = -63
(-7)*(9) = -63

Now let's add up each pair of factors to see if one pair adds to the middle coefficient $\"2\"$ :

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First Number	Second Number	Sum
1	-63	1+(-63)=-62
3	-21	3+(-21)=-18
7	-9	7+(-9)=-2
-1	63	-1+63=62
-3	21	-3+21=18
-7	9	-7+9=2

From the table, we can see that the two numbers $\"-7\"$ and $\"9\"$ add to $\"2\"$ (the middle coefficient).

So the two numbers $\"-7\"$ and $\"9\"$ both multiply to $\"-63\"$ and add to $\"2\"$

Now replace the middle term $\"2x\"$ with $\"-7x%2B9x\"$ . Remember, $\"-7\"$ and $\"9\"$ add to $\"2\"$ . So this shows us that $\"-7x%2B9x=2x\"$ .

$\"21x%5E2%2Bhighlight%28-7x%2B9x%29-3\"$ Replace the second term $\"2x\"$ with $\"-7x%2B9x\"$ .

$\"%2821x%5E2-7x%29%2B%289x-3%29\"$ Group the terms into two pairs.

$\"7x%283x-1%29%2B%289x-3%29\"$ Factor out the GCF $\"7x\"$ from the first group.

$\"7x%283x-1%29%2B3%283x-1%29\"$ 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.

$\"%287x%2B3%29%283x-1%29\"$ Combine like terms. Or factor out the common term $\"3x-1\"$

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Answer:

So $\"21%2Ax%5E2%2B2%2Ax-3\"$ factors to $\"%287x%2B3%29%283x-1%29\"$ .

In other words, $\"21%2Ax%5E2%2B2%2Ax-3=%287x%2B3%29%283x-1%29\"$ .

Note: you can check the answer by expanding $\"%287x%2B3%29%283x-1%29\"$ to get $\"21%2Ax%5E2%2B2%2Ax-3\"$ or by graphing the original expression and the answer (the two graphs should be identical).

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\n" ); document.write( "\n" ); document.write( "72.Factor
\n" ); document.write( " $\"3x%5E2-18x-48+\"$ \r
\n" ); document.write( "\n" ); document.write( "\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

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

$\"3%2Ax%5E2-18%2Ax-48\"$ Start with the given expression.

$\"3%28x%5E2-6x-16%29\"$ Factor out the GCF $\"3\"$ .

Now let's try to factor the inner expression $\"x%5E2-6x-16\"$

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

Looking at the expression $\"x%5E2-6x-16\"$ , we can see that the first coefficient is $\"1\"$ , the second coefficient is $\"-6\"$ , and the last term is $\"-16\"$ .

Now multiply the first coefficient $\"1\"$ by the last term $\"-16\"$ to get $\"%281%29%28-16%29=-16\"$ .

Now the question is: what two whole numbers multiply to $\"-16\"$ (the previous product) and add to the second coefficient $\"-6\"$ ?

To find these two numbers, we need to list all of the factors of $\"-16\"$ (the previous product).

Factors of $\"-16\"$ :

1,2,4,8,16

-1,-2,-4,-8,-16

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

These factors pair up and multiply to $\"-16\"$ .

1*(-16) = -16
2*(-8) = -16
4*(-4) = -16
(-1)*(16) = -16
(-2)*(8) = -16
(-4)*(4) = -16

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	-16	1+(-16)=-15
2	-8	2+(-8)=-6
4	-4	4+(-4)=0
-1	16	-1+16=15
-2	8	-2+8=6
-4	4	-4+4=0

From the table, we can see that the two numbers $\"2\"$ and $\"-8\"$ add to $\"-6\"$ (the middle coefficient).

So the two numbers $\"2\"$ and $\"-8\"$ both multiply to $\"-16\"$ and add to $\"-6\"$

Now replace the middle term $\"-6x\"$ with $\"2x-8x\"$ . Remember, $\"2\"$ and $\"-8\"$ add to $\"-6\"$ . So this shows us that $\"2x-8x=-6x\"$ .

$\"x%5E2%2Bhighlight%282x-8x%29-16\"$ Replace the second term $\"-6x\"$ with $\"2x-8x\"$ .

$\"%28x%5E2%2B2x%29%2B%28-8x-16%29\"$ Group the terms into two pairs.

$\"x%28x%2B2%29%2B%28-8x-16%29\"$ Factor out the GCF $\"x\"$ from the first group.

$\"x%28x%2B2%29-8%28x%2B2%29\"$ Factor out $\"8\"$ 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.

$\"%28x-8%29%28x%2B2%29\"$ Combine like terms. Or factor out the common term $\"x%2B2\"$

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

So $\"3%28x%5E2-6x-16%29\"$ then factors further to $\"3%28x-8%29%28x%2B2%29\"$

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Answer:

So $\"3%2Ax%5E2-18%2Ax-48\"$ completely factors to $\"3%28x-8%29%28x%2B2%29\"$ .

In other words, $\"3%2Ax%5E2-18%2Ax-48=3%28x-8%29%28x%2B2%29\"$ .

Note: you can check the answer by expanding $\"3%28x-8%29%28x%2B2%29\"$ to get $\"3%2Ax%5E2-18%2Ax-48\"$ or by graphing the original expression and the answer (the two graphs should be identical).

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\n" ); document.write( "\n" ); document.write( "18. Factor
\n" ); document.write( " $\"2h%5E2-h-3\"$ \r
\n" ); document.write( "\n" ); document.write( "\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

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

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

Now multiply the first coefficient $\"2\"$ by the last term $\"-3\"$ to get $\"%282%29%28-3%29=-6\"$ .

Now the question is: what two whole numbers multiply to $\"-6\"$ (the previous product) and add to the second coefficient $\"-1\"$ ?

To find these two numbers, we need to list all of the factors of $\"-6\"$ (the previous product).

Factors of $\"-6\"$ :

1,2,3,6

-1,-2,-3,-6

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

These factors pair up and multiply to $\"-6\"$ .

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

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	-6	1+(-6)=-5
2	-3	2+(-3)=-1
-1	6	-1+6=5
-2	3	-2+3=1

From the table, we can see that the two numbers $\"2\"$ and $\"-3\"$ add to $\"-1\"$ (the middle coefficient).

So the two numbers $\"2\"$ and $\"-3\"$ both multiply to $\"-6\"$ and add to $\"-1\"$

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

$\"2x%5E2%2Bhighlight%282x-3x%29-3\"$ Replace the second term $\"-1x\"$ with $\"2x-3x\"$ .

$\"%282x%5E2%2B2x%29%2B%28-3x-3%29\"$ Group the terms into two pairs.

$\"2x%28x%2B1%29%2B%28-3x-3%29\"$ Factor out the GCF $\"2x\"$ from the first group.

$\"2x%28x%2B1%29-3%28x%2B1%29\"$ 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.

$\"%282x-3%29%28x%2B1%29\"$ Combine like terms. Or factor out the common term $\"x%2B1\"$

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

Answer:

So $\"2%2Ax%5E2-x-3\"$ factors to $\"%282x-3%29%28x%2B1%29\"$ .

In other words, $\"2%2Ax%5E2-x-3=%282x-3%29%28x%2B1%29\"$ .

Note: you can check the answer by expanding $\"%282x-3%29%28x%2B1%29\"$ to get $\"2%2Ax%5E2-x-3\"$ or by graphing the original expression and the answer (the two graphs should be identical).

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