Question 530219


Looking at the expression {{{10x^2-29x-3}}}, we can see that the first coefficient is {{{10}}}, the second coefficient is {{{-29}}}, and the last term is {{{-3}}}.



Now multiply the first coefficient {{{10}}} by the last term {{{-3}}} to get {{{(10)(-3)=-30}}}.



Now the question is: what two whole numbers multiply to {{{-30}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{-29}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{-30}}} (the previous product).



Factors of {{{-30}}}:

1,2,3,5,6,10,15,30

-1,-2,-3,-5,-6,-10,-15,-30



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



These factors pair up and multiply to {{{-30}}}.

1*(-30) = -30
2*(-15) = -30
3*(-10) = -30
5*(-6) = -30
(-1)*(30) = -30
(-2)*(15) = -30
(-3)*(10) = -30
(-5)*(6) = -30


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{-29}}}:



<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td  align="center"><font color=red>1</font></td><td  align="center"><font color=red>-30</font></td><td  align="center"><font color=red>1+(-30)=-29</font></td></tr><tr><td  align="center"><font color=black>2</font></td><td  align="center"><font color=black>-15</font></td><td  align="center"><font color=black>2+(-15)=-13</font></td></tr><tr><td  align="center"><font color=black>3</font></td><td  align="center"><font color=black>-10</font></td><td  align="center"><font color=black>3+(-10)=-7</font></td></tr><tr><td  align="center"><font color=black>5</font></td><td  align="center"><font color=black>-6</font></td><td  align="center"><font color=black>5+(-6)=-1</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>30</font></td><td  align="center"><font color=black>-1+30=29</font></td></tr><tr><td  align="center"><font color=black>-2</font></td><td  align="center"><font color=black>15</font></td><td  align="center"><font color=black>-2+15=13</font></td></tr><tr><td  align="center"><font color=black>-3</font></td><td  align="center"><font color=black>10</font></td><td  align="center"><font color=black>-3+10=7</font></td></tr><tr><td  align="center"><font color=black>-5</font></td><td  align="center"><font color=black>6</font></td><td  align="center"><font color=black>-5+6=1</font></td></tr></table>



From the table, we can see that the two numbers {{{1}}} and {{{-30}}} add to {{{-29}}} (the middle coefficient).



So the two numbers {{{1}}} and {{{-30}}} both multiply to {{{-30}}} <font size=4><b>and</b></font> add to {{{-29}}}



Now replace the middle term {{{-29x}}} with {{{x-30x}}}. Remember, {{{1}}} and {{{-30}}} add to {{{-29}}}. So this shows us that {{{x-30x=-29x}}}.



{{{10x^2+highlight(x-30x)-3}}} Replace the second term {{{-29x}}} with {{{x-30x}}}.



{{{(10x^2+x)+(-30x-3)}}} Group the terms into two pairs.



{{{x(10x+1)+(-30x-3)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(10x+1)-3(10x+1)}}} 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.



{{{(x-3)(10x+1)}}} Combine like terms. Or factor out the common term {{{10x+1}}}



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



Answer:



So {{{10x^2-29x-3}}} factors to {{{(x-3)(10x+1)}}}.



In other words, {{{10x^2-29x-3=(x-3)(10x+1)}}}.



Note: you can check the answer by expanding {{{(x-3)(10x+1)}}} to get {{{10x^2-29x-3}}} or by graphing the original expression and the answer (the two graphs should be identical).



If you need more help, email me at <a href="mailto:jim_thompson5910@hotmail.com">jim_thompson5910@hotmail.com</a>


Also, please consider visiting my website: <a href="http://www.freewebs.com/jimthompson5910/home.html">http://www.freewebs.com/jimthompson5910/home.html</a> and making a donation. Thank you


Jim