Question 139519


Looking at {{{20x^2+13x+2}}} we can see that the first term is {{{20x^2}}} and the last term is {{{2}}} where the coefficients are 20 and 2 respectively.


Now multiply the first coefficient 20 and the last coefficient 2 to get 40. Now what two numbers multiply to 40 and add to the  middle coefficient 13? 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 two negative numbers multiplied together make a positive number



Now which of these pairs add to 13? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 13


<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=41</td></tr><tr><td align="center">2</td><td align="center">20</td><td>2+20=22</td></tr><tr><td align="center">4</td><td align="center">10</td><td>4+10=14</td></tr><tr><td align="center">5</td><td align="center">8</td><td>5+8=13</td></tr><tr><td align="center">-1</td><td align="center">-40</td><td>-1+(-40)=-41</td></tr><tr><td align="center">-2</td><td align="center">-20</td><td>-2+(-20)=-22</td></tr><tr><td align="center">-4</td><td align="center">-10</td><td>-4+(-10)=-14</td></tr><tr><td align="center">-5</td><td align="center">-8</td><td>-5+(-8)=-13</td></tr></table>



From this list we can see that 5 and 8 add up to 13 and multiply to 40



Now looking at the expression {{{20x^2+13x+2}}}, replace {{{13x}}} with {{{5x+8x}}} (notice {{{5x+8x}}} adds up to {{{13x}}}. So it is equivalent to {{{13x}}})


{{{20x^2+highlight(5x+8x)+2}}}



Now let's factor {{{20x^2+5x+8x+2}}} by grouping:



{{{(20x^2+5x)+(8x+2)}}} Group like terms



{{{5x(4x+1)+2(4x+1)}}} Factor out the GCF of {{{5x}}} out of the first group. Factor out the GCF of {{{2}}} out of the second group



{{{(5x+2)(4x+1)}}} Since we have a common term of {{{4x+1}}}, we can combine like terms


So {{{20x^2+5x+8x+2}}} factors to {{{(5x+2)(4x+1)}}}



So this also means that {{{20x^2+13x+2}}} factors to {{{(5x+2)(4x+1)}}} (since {{{20x^2+13x+2}}} is equivalent to {{{20x^2+5x+8x+2}}})




{{{(5x+2)(4x+1)=0}}} Set the factorization equal to zero




Now set each factor equal to zero:

{{{5x+2=0}}} or  {{{4x+1=0}}} 


{{{x=-2/5}}} or  {{{x=-1/4}}}    Now solve for x in each case




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Answer:


So our solutions are


 {{{x=-2/5}}} or  {{{x=-1/4}}}