Question 368669
I'm assuming that you want to factor this right? Please post the full instructions.




Looking at {{{5c^2-17c+14}}} we can see that the first term is {{{5c^2}}} and the last term is {{{14}}} where the coefficients are 5 and 14 respectively.


Now multiply the first coefficient 5 and the last coefficient 14 to get 70. Now what two numbers multiply to 70 and add to the  middle coefficient -17? Let's list all of the factors of 70:




Factors of 70:

1,2,5,7,10,14,35,70


-1,-2,-5,-7,-10,-14,-35,-70 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to 70

1*70

2*35

5*14

7*10

(-1)*(-70)

(-2)*(-35)

(-5)*(-14)

(-7)*(-10)


note: remember two negative numbers multiplied together make a positive number



Now which of these pairs add to -17? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to -17


<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center">1</td><td align="center">70</td><td>1+70=71</td></tr><tr><td align="center">2</td><td align="center">35</td><td>2+35=37</td></tr><tr><td align="center">5</td><td align="center">14</td><td>5+14=19</td></tr><tr><td align="center">7</td><td align="center">10</td><td>7+10=17</td></tr><tr><td align="center">-1</td><td align="center">-70</td><td>-1+(-70)=-71</td></tr><tr><td align="center">-2</td><td align="center">-35</td><td>-2+(-35)=-37</td></tr><tr><td align="center">-5</td><td align="center">-14</td><td>-5+(-14)=-19</td></tr><tr><td align="center">-7</td><td align="center">-10</td><td>-7+(-10)=-17</td></tr></table>



From this list we can see that -7 and -10 add up to -17 and multiply to 70



Now looking at the expression {{{5c^2-17c+14}}}, replace {{{-17c}}} with {{{-7c+-10c}}} (notice {{{-7c+-10c}}} adds up to {{{-17c}}}. So it is equivalent to {{{-17c}}})


{{{5c^2+highlight(-7c+-10c)+14}}}



Now let's factor {{{5c^2-7c-10c+14}}} by grouping:



{{{(5c^2-7c)+(-10c+14)}}} Group like terms



{{{c(5c-7)-2(5c-7)}}} Factor out the GCF of {{{c}}} out of the first group. Factor out the GCF of {{{-2}}} out of the second group



{{{(c-2)(5c-7)}}} Since we have a common term of {{{5c-7}}}, we can combine like terms


So {{{5c^2-7c-10c+14}}} factors to {{{(c-2)(5c-7)}}}



So this also means that {{{5c^2-17c+14}}} factors to {{{(c-2)(5c-7)}}} (since {{{5c^2-17c+14}}} is equivalent to {{{5c^2-7c-10c+14}}})




------------------------------------------------------------




     Answer:

So {{{5c^2-17c+14}}} factors to {{{(c-2)(5c-7)}}}