Question 145612


Looking at {{{x^2+7x+10}}} we can see that the first term is {{{x^2}}} and the last term is {{{10}}} where the coefficients are 1 and 10 respectively.


Now multiply the first coefficient 1 and the last coefficient 10 to get 10. Now what two numbers multiply to 10 and add to the  middle coefficient 7? 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 7? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 7


<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 2 and 5 add up to 7 and multiply to 10



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


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



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



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



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


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



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




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     Answer:

So {{{x^2+7x+10}}} factors to {{{(x+5)(x+2)}}}