SOLUTION: Find all zeros using synthetic division. Show your work. Q(x)= x^3 + (k+3)x^2 - k(2k-3)x - 6k^2 where k is a real number

Algebra ->  Functions -> SOLUTION: Find all zeros using synthetic division. Show your work. Q(x)= x^3 + (k+3)x^2 - k(2k-3)x - 6k^2 where k is a real number      Log On


   

Question 1123636: Find all zeros using synthetic division. Show your work.
Q(x)= x^3 + (k+3)x^2 - k(2k-3)x - 6k^2 where k is a real number
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
Q(x)= x³ + (k+3)x² - k(2k-3)x - 6k²

Multiply out the coefficient -k(2k-3)

Q(x)= x³ + (k+3)x² + (-2k²+3k)x - 6k²


The potential zeros are ± the  factors of -6k², which are

±1, ±2, ±3, ±6, ±k, ±2k, ±3k, ±6k, ±k², ±2k², ±3k², ±6k²

That's a huge bunch to try! (lol), but I have a hunch that one
of the zeros is k.  Let's see if I'm lucky:

 k | 1    k+3   -2k²+3k   -6k   
   |      k      2k²+3k    6k    
    ------------------------- 
     1   2k+3        6k     0

The remainder is 0, so I was lucky! lol

So k is a zero of Q(x), thus x-k is a factor
of Q(x) and we have partially factored Q(x) as

Q(x) = (x-k)[x² + (2k+3)x + 6k]

So we set the quadratic factor equal zero and solve:

x² + (2k+3)x + 6k = 0

x+=+%28-b+%2B-+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29+

x+=+%28-%282k%2B3%29+%2B-+sqrt%28+%282k%2B3%29%5E2-4%281%29%286k%29+%29%29%2F%282%281%29%29+

x+=+%28-2k-3+%2B-+sqrt%284k%5E2%2B12k%2B9-24k%29%29%2F2+

x+=+%28-2k-3+%2B-+sqrt%284k%5E2-12k%2B9%29+%29%2F2+

x+=+%28-2k-3+%2B-+sqrt%28%282k-3%29%5E2%29%29%2F2+

x+=+%28-2k-3+%2B-+%282k-3%29%29%2F2+

Using the + of ±

x+=+%28-2k-3+%2B+%282k-3%29%29%2F2+

x+=+%28-2k-3+%2B+2k-3%29%2F2+

x+=+%28-6%29%2F2+

x+=+-3, so -3 is a zero of Q(x)

Using the - of ±

x+=+%28-2k-3+-+%282k-3%29%29%2F2+

x+=+%28-2k-3+-+2k%2B3%29%2F2+

x+=+%28-4k%29%2F2+

x+=+-2k, so -2k is a zero of Q(x)

So the three zeros of Q(x) are k, -3 and -2k

Edwin