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How to factor a trinomial with FOIL in Reverse.

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2 Answers

Genrally speaking:

  a trinomial  X^2 + bx + c

     Have to find 2 numbers whose sum is - b, and their product is c

     X^2 - 10X  +21 

    here -b = 10  c = 21

So the 2 numbers are 7, and 3 whose sum are 10 and product is 10

     then, X^2 -10x + 21 = (X-7)(X-3)

           if we work back ward

         (x -3)(X-7) = ( X-3)X -7( x-3)

                X^2 -3X -7x +21 = X^2-10X +21

     That was for coefficient of X^2 equals 1

       for  2X^2 -23X +38 to factor  we have to find 2 numbers such that their Sum is 23 and their product is 2(38)=76

    To find that we write 76= 2(2)(19)

    we see that factors of 4,19 can be added to 23

  2X^2-4x -19x +38= 2X (x -2) -19 (x -2)

 then factoring by grouping 2X^2 -23 x  + 38 = ( 2X -19 )(X- 2 )

                                                

Here's an example of factoring a trinomial using reverse FOIL:  

Suppose you want to factor x2 + 3x - 10

Create your two binomial factors with x at the beginning of each factor:

(x   ) (x   )

If the last term in the original trinomial is positive then you will use positive/positive or negative/negative signs.

If the last term in the original trinomial is negative then you will use positive/negative signs.  This gives us . . .

(x +  ) (x -  )

Finally choose factors of the last term that sum to the coefficient of the middle term.

In this case 5 * -2  = -10 and 5 + (-2) = 3

So your final answer is . . .

(x + 5) (x - 2)

You can always verify by using FOIL to see if you get the original trinomial.

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