Lesson Advanced factoring

Advanced factoring

Problem 1

 =  =  =  = .

Problem 2

 =  =  =  = .

Problem 3

According to the solution to the  Problem 2, 


 = 


is the decomposition of the number    into the product of two factors    and  .

Problem 4

 =  =  =  = .

Problem 5

 =  =  =  = .


You can factor polynomial   further similar to that it was done in the solution to Problem 4:


     =  =  =  =    and  


So,   = ..

Problem 6

 =  =  =  =  = .

Problem 7

 =  -  =  =  = .

Problem 8

One solution is OBVIOUS: it is k= 18.

Then  x^2 -18x + 81 = (x-9)*(x-9) = .      (1)



But there is another solution, too:  k= -18, which makes THIS factoring

      x^2 + 18x + 81 = (x+9)*(x+9) = .     (2)


Any student, familiar with standard binomial identities 

     (a+b)^2 = a^2 + 2ab + b^2,   (a-b)^2 = a^2 - 2ab + b^2,      (3)

should see it MOMENTARILY (!).



Even more interesting fact is that there are MANY OTHER SOLUTIONS over integer numbers.

They come from decomposition of the number 81 into the product of integer factors

    81 = 1*81 = 3*27 = (-1)*(-81) = (-3)*(-27)          (4)

(I just do not include the cases 81 = 9*9 = (-9)*(-9), which I considered above).



These decompositions produce the following factoring over integer numbers

    81 = 1*81        -->  k = 1 + 81 = 82       -->  x^2 - k + 81 = x^2 - 82 + 81 = (x-1)*(x-81)

    81 = 3*27        -->  k = 3 + 27 = 30       -->  x^2 - k + 81 = x^2 - 30 + 81 = (x-3)*(x-27)

    81 = (-1)*(-81)  -->  k = -1 + (-81) = -82  -->  x^2 - k + 81 = x^2 + 82 + 81 = (x+1)*(x+81)

    81 = (-3)*(-27)  -->  k = -3 + (-27) = -30  -->  x^2 - k + 81 = x^2 + 30 + 81 = (x+3)*(x+27)



So, over integer numbers we have 6 (six, SIX) solutions for k:  18, - 18, 82, -82, 30 and - 30, that provide 

factoring of  x^2 - k + 81 into the product of two binomials.