document.write( "Question 454099: Please solve by grouping:\r
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Algebra.Com's Answer #311870 by rwm(914)\"\" \"About 
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Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)


Looking at the expression \"63q%5E2-52q-20\", we can see that the first coefficient is \"63\", the second coefficient is \"-52\", and the last term is \"-20\".



Now multiply the first coefficient \"63\" by the last term \"-20\" to get \"%2863%29%28-20%29=-1260\".



Now the question is: what two whole numbers multiply to \"-1260\" (the previous product) and add to the second coefficient \"-52\"?



To find these two numbers, we need to list all of the factors of \"-1260\" (the previous product).



Factors of \"-1260\":

1,2,3,4,5,6,7,9,10,12,14,15,18,20,21,28,30,35,36,42,45,60,63,70,84,90,105,126,140,180,210,252,315,420,630,1260

-1,-2,-3,-4,-5,-6,-7,-9,-10,-12,-14,-15,-18,-20,-21,-28,-30,-35,-36,-42,-45,-60,-63,-70,-84,-90,-105,-126,-140,-180,-210,-252,-315,-420,-630,-1260



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



These factors pair up and multiply to \"-1260\".

1*(-1260) = -1260
2*(-630) = -1260
3*(-420) = -1260
4*(-315) = -1260
5*(-252) = -1260
6*(-210) = -1260
7*(-180) = -1260
9*(-140) = -1260
10*(-126) = -1260
12*(-105) = -1260
14*(-90) = -1260
15*(-84) = -1260
18*(-70) = -1260
20*(-63) = -1260
21*(-60) = -1260
28*(-45) = -1260
30*(-42) = -1260
35*(-36) = -1260
(-1)*(1260) = -1260
(-2)*(630) = -1260
(-3)*(420) = -1260
(-4)*(315) = -1260
(-5)*(252) = -1260
(-6)*(210) = -1260
(-7)*(180) = -1260
(-9)*(140) = -1260
(-10)*(126) = -1260
(-12)*(105) = -1260
(-14)*(90) = -1260
(-15)*(84) = -1260
(-18)*(70) = -1260
(-20)*(63) = -1260
(-21)*(60) = -1260
(-28)*(45) = -1260
(-30)*(42) = -1260
(-35)*(36) = -1260


Now let's add up each pair of factors to see if one pair adds to the middle coefficient \"-52\":



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First NumberSecond NumberSum
1-12601+(-1260)=-1259
2-6302+(-630)=-628
3-4203+(-420)=-417
4-3154+(-315)=-311
5-2525+(-252)=-247
6-2106+(-210)=-204
7-1807+(-180)=-173
9-1409+(-140)=-131
10-12610+(-126)=-116
12-10512+(-105)=-93
14-9014+(-90)=-76
15-8415+(-84)=-69
18-7018+(-70)=-52
20-6320+(-63)=-43
21-6021+(-60)=-39
28-4528+(-45)=-17
30-4230+(-42)=-12
35-3635+(-36)=-1
-11260-1+1260=1259
-2630-2+630=628
-3420-3+420=417
-4315-4+315=311
-5252-5+252=247
-6210-6+210=204
-7180-7+180=173
-9140-9+140=131
-10126-10+126=116
-12105-12+105=93
-1490-14+90=76
-1584-15+84=69
-1870-18+70=52
-2063-20+63=43
-2160-21+60=39
-2845-28+45=17
-3042-30+42=12
-3536-35+36=1




From the table, we can see that the two numbers \"18\" and \"-70\" add to \"-52\" (the middle coefficient).



So the two numbers \"18\" and \"-70\" both multiply to \"-1260\" and add to \"-52\"



Now replace the middle term \"-52q\" with \"18q-70q\". Remember, \"18\" and \"-70\" add to \"-52\". So this shows us that \"18q-70q=-52q\".



\"63q%5E2%2Bhighlight%2818q-70q%29-20\" Replace the second term \"-52q\" with \"18q-70q\".



\"%2863q%5E2%2B18q%29%2B%28-70q-20%29\" Group the terms into two pairs.



\"9q%287q%2B2%29%2B%28-70q-20%29\" Factor out the GCF \"9q\" from the first group.



\"9q%287q%2B2%29-10%287q%2B2%29\" Factor out \"10\" 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.



\"%289q-10%29%287q%2B2%29\" Combine like terms. Or factor out the common term \"7q%2B2\"



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Answer:



So \"63%2Aq%5E2-52%2Aq-20\" factors to \"%289q-10%29%287q%2B2%29\".



In other words, \"63%2Aq%5E2-52%2Aq-20=%289q-10%29%287q%2B2%29\".



Note: you can check the answer by expanding \"%289q-10%29%287q%2B2%29\" to get \"63%2Aq%5E2-52%2Aq-20\" or by graphing the original expression and the answer (the two graphs should be identical).

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