SOLUTION: Find all the real zeros of the function. h(x)= {{{x^4-4x^3-9x^2+16x+20}}}

Algebra ->  Test -> SOLUTION: Find all the real zeros of the function. h(x)= {{{x^4-4x^3-9x^2+16x+20}}}      Log On


   

Question 825381: Find all the real zeros of the function.
h(x)= x%5E4-4x%5E3-9x%5E2%2B16x%2B20
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
As per the rational zero theorem,
the possible rational zeros of h%28x%29 are integers factors of 20 , meaning
-20, -5, -4, -2, -1, 1, 2, 4, 5, and 20.

We can try them all, starting by the easiest ones.
Trying 1 and -1, we find x=-1 is one zero:
h%281%29=1-4-9%2B16%2B20=-3-9%2B16%2B20=-12%2B16%2B20=4%2B20=24
h%28-1%29=1%2B4-9-16%2B20=5-9-16%2B20=-4-16%2B20=-20%2B20=0

Since x=-1 is a zero of h%28x%29 , x-%28-1%29=x%2B1 must be a factor of h%28x%29 .
That means that h%28x%29 divides evenly by x%2B1 , with no remainder.
Dividing h%28x%29 by x%2B1 we get
Q%28x%29=x%5E3-5x%5E2-4x%2B20 , so
h%28x%29=%28x%2B1%29Q%28x%29 .
(The division can be done by "synthetic division", or whatever method you have been taught in class).

As per the rational zero theorem,
the possible rational zeros of Q%28x%29 are integers factors of 20 , meaning
-20, -5, -4, -2, -1, 1, 2, 4, 5, and 20.
We know that Q%281%29%3C%3E0 because otherwise
We can find that Q%282%29=0 either by substituting 2 for x,
or by dividing Q%28x%29 by x-2 .
Q%282%29=2%5E3-5%2A2%5E2-4%2A2%2B20=8-5%2A4-8%2B20=8-20-8%2B20=0 ,
and dividing we find that
Q%28x%29=%28x-2%29%28x%5E2-3x-10%29 .

So,
h%28x%29=%28x%2B1%29Q%28x%29=%28x%2B1%29%28x-2%29%28x%5E2-3x-10%29 .

Quadratic polynomial x%5E2-3x-10 is easy to factor as
x%5E2-3x-10=%28x%2B2%29%28x-5%29 , so
.
highlight%28h%28x%29=%28x%2B1%29%28x-2%29%28x%2B2%29%28x-5%29%29
and h%28x%29=0 for
highlight%28x=-2%29 , highlight%28x=-1%29 , highlight%28x=2%29 , and highlight%28x=5%29 .