SOLUTION: If one prime linear Factor of f(x) = x³ + 2x² - 51x + 108 is x + 9, find the other two prime linear factors.

Algebra ->  Square-cubic-other-roots -> SOLUTION: If one prime linear Factor of f(x) = x³ + 2x² - 51x + 108 is x + 9, find the other two prime linear factors.      Log On


   



Question 183103This question is from textbook Algebra 2
: If one prime linear Factor of f(x) = x³ + 2x² - 51x + 108 is x + 9,
find the other two prime linear factors.
This question is from textbook Algebra 2

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!

f(x) = x³ + 2x² - 51x + 108 

Divide x³ + 2x² - 51x + 108 by x + 9, 
either by long division:


             x² -  7x +  12 
x + 9)x³ +  2x² - 51x + 108 
      x³ +  9x²  
           -7x² - 51x  
           -7x² - 63x
                  12x + 108 
                  12x + 108
                          0
                
or by, what amounts to the same thing,
synthetic division, if you've studied
that:

 -9|1   2  -51  108
   |   -9   63 -108  
    1  -7   12    0

Either way, thus far you have factored the
original cubic polynomial f(x) as:

(x + 9)(x² -  7x +  12)

Now we can factor the quadratic
polynomial in the second parentheses
as (x - 3)(x - 4) and the complete
factorization of f(x) into prime
linear factors is:

(x + 9)(x - 3)(x - 4)

So the other two prime linear factors of
f(x) are (x - 3) and (x - 4).

Edwin