Question 585441
I'm assuming you want to factor this.




{{{-2x^2+3x+20}}} Start with the given expression.



{{{-(2x^2-3x-20)}}} Factor out the GCF {{{-1}}}.



Now let's try to factor the inner expression {{{2x^2-3x-20}}}



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Looking at the expression {{{2x^2-3x-20}}}, we can see that the first coefficient is {{{2}}}, the second coefficient is {{{-3}}}, and the last term is {{{-20}}}.



Now multiply the first coefficient {{{2}}} by the last term {{{-20}}} to get {{{(2)(-20)=-40}}}.



Now the question is: what two whole numbers multiply to {{{-40}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{-3}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{-40}}} (the previous product).



Factors of {{{-40}}}:

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

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



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{-40}}}.

1*(-40) = -40
2*(-20) = -40
4*(-10) = -40
5*(-8) = -40
(-1)*(40) = -40
(-2)*(20) = -40
(-4)*(10) = -40
(-5)*(8) = -40


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{-3}}}:



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



From the table, we can see that the two numbers {{{5}}} and {{{-8}}} add to {{{-3}}} (the middle coefficient).



So the two numbers {{{5}}} and {{{-8}}} both multiply to {{{-40}}} <font size=4><b>and</b></font> add to {{{-3}}}



Now replace the middle term {{{-3x}}} with {{{5x-8x}}}. Remember, {{{5}}} and {{{-8}}} add to {{{-3}}}. So this shows us that {{{5x-8x=-3x}}}.



{{{2x^2+highlight(5x-8x)-20}}} Replace the second term {{{-3x}}} with {{{5x-8x}}}.



{{{(2x^2+5x)+(-8x-20)}}} Group the terms into two pairs.



{{{x(2x+5)+(-8x-20)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(2x+5)-4(2x+5)}}} Factor out {{{4}}} from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.



{{{(x-4)(2x+5)}}} Combine like terms. Or factor out the common term {{{2x+5}}}



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So {{{-(2x^2-3x-20)}}} then factors further to {{{-(x-4)(2x+5)}}}



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Answer:



So {{{-2x^2+3x+20}}} completely factors to {{{-(x-4)(2x+5)}}}.



In other words, {{{-2x^2+3x+20=-(x-4)(2x+5)}}}.



Note: you can check the answer by expanding {{{-(x-4)(2x+5)}}} to get {{{-2x^2+3x+20}}} or by graphing the original expression and the answer (the two graphs should be identical).


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