Question 117772


Looking at {{{2x^2+11x+5}}} we can see that the first term is {{{2x^2}}} and the last term is {{{5}}} where the coefficients are 2 and 5 respectively.


Now multiply the first coefficient 2 and the last coefficient 5 to get 10. Now what two numbers multiply to 10 and add to the  middle coefficient 11? Let's list all of the factors of 10:




Factors of 10:

1,2,5,10


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


These factors pair up and multiply to 10

1*10

2*5

(-1)*(-10)

(-2)*(-5)


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



Now which of these pairs add to 11? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 11


<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center">1</td><td align="center">10</td><td>1+10=11</td></tr><tr><td align="center">2</td><td align="center">5</td><td>2+5=7</td></tr><tr><td align="center">-1</td><td align="center">-10</td><td>-1+(-10)=-11</td></tr><tr><td align="center">-2</td><td align="center">-5</td><td>-2+(-5)=-7</td></tr></table>



From this list we can see that 1 and 10 add up to 11 and multiply to 10



Now looking at the expression {{{2x^2+11x+5}}}, replace {{{11x}}} with {{{1x+10x}}} (notice {{{1x+10x}}} adds up to {{{11x}}}. So it is equivalent to {{{11x}}})


{{{2x^2+highlight(1x+10x)+5}}}



Now let's factor {{{2x^2+1x+10x+5}}} by grouping:



{{{(2x^2+1x)+(10x+5)}}} Group like terms



{{{x(2x+1)+5(2x+1)}}} Factor out the GCF of {{{x}}} out of the first group. Factor out the GCF of {{{5}}} out of the second group



{{{(x+5)(2x+1)}}} Since we have a common term of {{{2x+1}}}, we can combine like terms


So {{{2x^2+1x+10x+5}}} factors to {{{(x+5)(2x+1)}}}



So this also means that {{{2x^2+11x+5}}} factors to {{{(x+5)(2x+1)}}} (since {{{2x^2+11x+5}}} is equivalent to {{{2x^2+1x+10x+5}}})


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Answer:


So {{{2x^2+11x+5}}} factors to {{{(x+5)(2x+1)}}}