Question 561239


Looking at the expression {{{7x^2+17x+6}}}, we can see that the first coefficient is {{{7}}}, the second coefficient is {{{17}}}, and the last term is {{{6}}}.



Now multiply the first coefficient {{{7}}} by the last term {{{6}}} to get {{{(7)(6)=42}}}.



Now the question is: what two whole numbers multiply to {{{42}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{17}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{42}}} (the previous product).



Factors of {{{42}}}:

1,2,3,6,7,14,21,42

-1,-2,-3,-6,-7,-14,-21,-42



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



These factors pair up and multiply to {{{42}}}.

1*42 = 42
2*21 = 42
3*14 = 42
6*7 = 42
(-1)*(-42) = 42
(-2)*(-21) = 42
(-3)*(-14) = 42
(-6)*(-7) = 42


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{17}}}:



<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td  align="center"><font color=black>1</font></td><td  align="center"><font color=black>42</font></td><td  align="center"><font color=black>1+42=43</font></td></tr><tr><td  align="center"><font color=black>2</font></td><td  align="center"><font color=black>21</font></td><td  align="center"><font color=black>2+21=23</font></td></tr><tr><td  align="center"><font color=red>3</font></td><td  align="center"><font color=red>14</font></td><td  align="center"><font color=red>3+14=17</font></td></tr><tr><td  align="center"><font color=black>6</font></td><td  align="center"><font color=black>7</font></td><td  align="center"><font color=black>6+7=13</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>-42</font></td><td  align="center"><font color=black>-1+(-42)=-43</font></td></tr><tr><td  align="center"><font color=black>-2</font></td><td  align="center"><font color=black>-21</font></td><td  align="center"><font color=black>-2+(-21)=-23</font></td></tr><tr><td  align="center"><font color=black>-3</font></td><td  align="center"><font color=black>-14</font></td><td  align="center"><font color=black>-3+(-14)=-17</font></td></tr><tr><td  align="center"><font color=black>-6</font></td><td  align="center"><font color=black>-7</font></td><td  align="center"><font color=black>-6+(-7)=-13</font></td></tr></table>



From the table, we can see that the two numbers {{{3}}} and {{{14}}} add to {{{17}}} (the middle coefficient).



So the two numbers {{{3}}} and {{{14}}} both multiply to {{{42}}} <font size=4><b>and</b></font> add to {{{17}}}



Now replace the middle term {{{17x}}} with {{{3x+14x}}}. Remember, {{{3}}} and {{{14}}} add to {{{17}}}. So this shows us that {{{3x+14x=17x}}}.



{{{7x^2+highlight(3x+14x)+6}}} Replace the second term {{{17x}}} with {{{3x+14x}}}.



{{{(7x^2+3x)+(14x+6)}}} Group the terms into two pairs.



{{{x(7x+3)+(14x+6)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(7x+3)+2(7x+3)}}} Factor out {{{2}}} 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.



{{{(x+2)(7x+3)}}} Combine like terms. Or factor out the common term {{{7x+3}}}



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Answer:



So {{{7x^2+17x+6}}} factors to {{{(x+2)(7x+3)}}}.



In other words, {{{7x^2+17x+6=(x+2)(7x+3)}}}.



Note: you can check the answer by expanding {{{(x+2)(7x+3)}}} to get {{{7x^2+17x+6}}} or by graphing the original expression and the answer (the two graphs should be identical).

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