Question 584928


Looking at the expression {{{4x^2+36x+35}}}, we can see that the first coefficient is {{{4}}}, the second coefficient is {{{36}}}, and the last term is {{{35}}}.



Now multiply the first coefficient {{{4}}} by the last term {{{35}}} to get {{{(4)(35)=140}}}.



Now the question is: what two whole numbers multiply to {{{140}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{36}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{140}}} (the previous product).



Factors of {{{140}}}:

1,2,4,5,7,10,14,20,28,35,70,140

-1,-2,-4,-5,-7,-10,-14,-20,-28,-35,-70,-140



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



These factors pair up and multiply to {{{140}}}.

1*140 = 140
2*70 = 140
4*35 = 140
5*28 = 140
7*20 = 140
10*14 = 140
(-1)*(-140) = 140
(-2)*(-70) = 140
(-4)*(-35) = 140
(-5)*(-28) = 140
(-7)*(-20) = 140
(-10)*(-14) = 140


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{36}}}:



<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>140</font></td><td  align="center"><font color=black>1+140=141</font></td></tr><tr><td  align="center"><font color=black>2</font></td><td  align="center"><font color=black>70</font></td><td  align="center"><font color=black>2+70=72</font></td></tr><tr><td  align="center"><font color=black>4</font></td><td  align="center"><font color=black>35</font></td><td  align="center"><font color=black>4+35=39</font></td></tr><tr><td  align="center"><font color=black>5</font></td><td  align="center"><font color=black>28</font></td><td  align="center"><font color=black>5+28=33</font></td></tr><tr><td  align="center"><font color=black>7</font></td><td  align="center"><font color=black>20</font></td><td  align="center"><font color=black>7+20=27</font></td></tr><tr><td  align="center"><font color=black>10</font></td><td  align="center"><font color=black>14</font></td><td  align="center"><font color=black>10+14=24</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>-140</font></td><td  align="center"><font color=black>-1+(-140)=-141</font></td></tr><tr><td  align="center"><font color=black>-2</font></td><td  align="center"><font color=black>-70</font></td><td  align="center"><font color=black>-2+(-70)=-72</font></td></tr><tr><td  align="center"><font color=black>-4</font></td><td  align="center"><font color=black>-35</font></td><td  align="center"><font color=black>-4+(-35)=-39</font></td></tr><tr><td  align="center"><font color=black>-5</font></td><td  align="center"><font color=black>-28</font></td><td  align="center"><font color=black>-5+(-28)=-33</font></td></tr><tr><td  align="center"><font color=black>-7</font></td><td  align="center"><font color=black>-20</font></td><td  align="center"><font color=black>-7+(-20)=-27</font></td></tr><tr><td  align="center"><font color=black>-10</font></td><td  align="center"><font color=black>-14</font></td><td  align="center"><font color=black>-10+(-14)=-24</font></td></tr></table>



From the table, we can see that there are no pairs of numbers which add to {{{36}}}. So {{{4x^2+36x+35}}} cannot be factored.



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Answer:



So {{{4x^2+36x+35}}} doesn't factor at all (over the rational numbers).



So {{{4x^2+36x+35}}} is prime.