Question 659398


Looking at the expression {{{3b^2+5b+2}}}, we can see that the first coefficient is {{{3}}}, the second coefficient is {{{5}}}, and the last term is {{{2}}}.



Now multiply the first coefficient {{{3}}} by the last term {{{2}}} to get {{{(3)(2)=6}}}.



Now the question is: what two whole numbers multiply to {{{6}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{5}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> 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 {{{5}}}:



<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>6</font></td><td  align="center"><font color=black>1+6=7</font></td></tr><tr><td  align="center"><font color=red>2</font></td><td  align="center"><font color=red>3</font></td><td  align="center"><font color=red>2+3=5</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>-6</font></td><td  align="center"><font color=black>-1+(-6)=-7</font></td></tr><tr><td  align="center"><font color=black>-2</font></td><td  align="center"><font color=black>-3</font></td><td  align="center"><font color=black>-2+(-3)=-5</font></td></tr></table>



From the table, we can see that the two numbers {{{2}}} and {{{3}}} add to {{{5}}} (the middle coefficient).



So the two numbers {{{2}}} and {{{3}}} both multiply to {{{6}}} <font size=4><b>and</b></font> add to {{{5}}}



Now replace the middle term {{{5b}}} with {{{2b+3b}}}. Remember, {{{2}}} and {{{3}}} add to {{{5}}}. So this shows us that {{{2b+3b=5b}}}.



{{{3b^2+highlight(2b+3b)+2}}} Replace the second term {{{5b}}} with {{{2b+3b}}}.



{{{(3b^2+2b)+(3b+2)}}} Group the terms into two pairs.



{{{b(3b+2)+(3b+2)}}} Factor out the GCF {{{b}}} from the first group.



{{{b(3b+2)+1(3b+2)}}} 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.



{{{(b+1)(3b+2)}}} Combine like terms. Or factor out the common term {{{3b+2}}}




So {{{3b^2+5b+2}}} factors to {{{(b+1)(3b+2)}}}.



In other words, {{{3b^2+5b+2=(b+1)(3b+2)}}}.


I'll leave the rest to you.


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