```Question 125883

Looking at {{{1x^2+5x+6}}} we can see that the first term is {{{1x^2}}} and the last term is {{{6}}} where the coefficients are 1 and 6 respectively.

Now multiply the first coefficient 1 and the last coefficient 6 to get 6. Now what two numbers multiply to 6 and add to the  middle coefficient 5? Let's list all of the factors of 6:

Factors of 6:

1,2,3,6

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

These factors pair up and multiply to 6

1*6

2*3

(-1)*(-6)

(-2)*(-3)

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

Now which of these pairs add to 5? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 5

<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center">1</td><td align="center">6</td><td>1+6=7</td></tr><tr><td align="center">2</td><td align="center">3</td><td>2+3=5</td></tr><tr><td align="center">-1</td><td align="center">-6</td><td>-1+(-6)=-7</td></tr><tr><td align="center">-2</td><td align="center">-3</td><td>-2+(-3)=-5</td></tr></table>

From this list we can see that 2 and 3 add up to 5 and multiply to 6

Now looking at the expression {{{1x^2+5x+6}}}, replace {{{5x}}} with {{{2x+3x}}} (notice {{{2x+3x}}} adds up to {{{5x}}}. So it is equivalent to {{{5x}}})

{{{1x^2+highlight(2x+3x)+6}}}

Now let's factor {{{1x^2+2x+3x+6}}} by grouping:

{{{(1x^2+2x)+(3x+6)}}} Group like terms

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

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

So {{{1x^2+2x+3x+6}}} factors to {{{(x+3)(x+2)}}}

So this also means that {{{1x^2+5x+6}}} factors to {{{(x+3)(x+2)}}} (since {{{1x^2+5x+6}}} is equivalent to {{{1x^2+2x+3x+6}}})

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