Question 131049


Looking at {{{1y^2+3y-40}}} we can see that the first term is {{{1y^2}}} and the last term is {{{-40}}} where the coefficients are 1 and -40 respectively.


Now multiply the first coefficient 1 and the last coefficient -40 to get -40. Now what two numbers multiply to -40 and add to the  middle coefficient 3? Let's list all of the factors of -40:




Factors of -40:

1,2,4,5,8,10,20,40


-1,-2,-4,-5,-8,-10,-20,-40 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to -40

(1)*(-40)

(2)*(-20)

(4)*(-10)

(5)*(-8)

(-1)*(40)

(-2)*(20)

(-4)*(10)

(-5)*(8)


note: remember, the product of a negative and a positive number is a negative number



Now which of these pairs add to 3? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 3


<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center">1</td><td align="center">-40</td><td>1+(-40)=-39</td></tr><tr><td align="center">2</td><td align="center">-20</td><td>2+(-20)=-18</td></tr><tr><td align="center">4</td><td align="center">-10</td><td>4+(-10)=-6</td></tr><tr><td align="center">5</td><td align="center">-8</td><td>5+(-8)=-3</td></tr><tr><td align="center">-1</td><td align="center">40</td><td>-1+40=39</td></tr><tr><td align="center">-2</td><td align="center">20</td><td>-2+20=18</td></tr><tr><td align="center">-4</td><td align="center">10</td><td>-4+10=6</td></tr><tr><td align="center">-5</td><td align="center">8</td><td>-5+8=3</td></tr></table>



From this list we can see that -5 and 8 add up to 3 and multiply to -40



Now looking at the expression {{{1y^2+3y-40}}}, replace {{{3y}}} with {{{-5y+8y}}} (notice {{{-5y+8y}}} adds up to {{{3y}}}. So it is equivalent to {{{3y}}})


{{{1y^2+highlight(-5y+8y)+-40}}}



Now let's factor {{{1y^2-5y+8y-40}}} by grouping:



{{{(1y^2-5y)+(8y-40)}}} Group like terms



{{{y(y-5)+8(y-5)}}} Factor out the GCF of {{{y}}} out of the first group. Factor out the GCF of {{{8}}} out of the second group



{{{(y+8)(y-5)}}} Since we have a common term of {{{y-5}}}, we can combine like terms


So {{{1y^2-5y+8y-40}}} factors to {{{(y+8)(y-5)}}}



So this also means that {{{1y^2+3y-40}}} factors to {{{(y+8)(y-5)}}} (since {{{1y^2+3y-40}}} is equivalent to {{{1y^2-5y+8y-40}}})




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     Answer:

So {{{y^2+3y-40}}} factors to {{{(y+8)(y-5)}}} which means that one of factors are either {{{y+8}}} or {{{y-5}}}