Question 118105


Looking at {{{3x^2+17x+10}}} we can see that the first term is {{{3x^2}}} and the last term is {{{10}}} where the coefficients are 3 and 10 respectively.


Now multiply the first coefficient 3 and the last coefficient 10 to get 30. Now what two numbers multiply to 30 and add to the  middle coefficient 17? Let's list all of the factors of 30:




Factors of 30:

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


-1,-2,-3,-5,-6,-10,-15,-30 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to 30

1*30

2*15

3*10

5*6

(-1)*(-30)

(-2)*(-15)

(-3)*(-10)

(-5)*(-6)


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">30</td><td>1+30=31</td></tr><tr><td align="center">2</td><td align="center">15</td><td>2+15=17</td></tr><tr><td align="center">3</td><td align="center">10</td><td>3+10=13</td></tr><tr><td align="center">5</td><td align="center">6</td><td>5+6=11</td></tr><tr><td align="center">-1</td><td align="center">-30</td><td>-1+(-30)=-31</td></tr><tr><td align="center">-2</td><td align="center">-15</td><td>-2+(-15)=-17</td></tr><tr><td align="center">-3</td><td align="center">-10</td><td>-3+(-10)=-13</td></tr><tr><td align="center">-5</td><td align="center">-6</td><td>-5+(-6)=-11</td></tr></table>



From this list we can see that 2 and 15 add up to 17 and multiply to 30



Now looking at the expression {{{3x^2+17x+10}}}, replace {{{17x}}} with {{{2x+15x}}} (notice {{{2x+15x}}} adds up to {{{17x}}}. So it is equivalent to {{{17x}}})


{{{3x^2+highlight(2x+15x)+10}}}



Now let's factor {{{3x^2+2x+15x+10}}} by grouping:



{{{(3x^2+2x)+(15x+10)}}} Group like terms



{{{x(3x+2)+5(3x+2)}}} Factor out the GCF of {{{x}}} out of the first group. Factor out the GCF of {{{5}}} out of the second group



{{{(x+5)(3x+2)}}} Since we have a common term of {{{3x+2}}}, we can combine like terms


So {{{3x^2+2x+15x+10}}} factors to {{{(x+5)(3x+2)}}}



So this also means that {{{3x^2+17x+10}}} factors to {{{(x+5)(3x+2)}}} (since {{{3x^2+17x+10}}} is equivalent to {{{3x^2+2x+15x+10}}})


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Answer:


So {{{3x^2+17x+10}}} factors to {{{(x+5)(3x+2)}}}