Polynomial Function

Question

Find a bound on the real zeros of the polynomial function.
f(x) = x4 - 7x2 - 18

 

step by step instructions

Answer 1

f(x) =x⁴-7x²-18

 

2 methods:

x⁴-7x²-18 = 0

 

(x² - 9) (x² + 2) = 0

 

x² - 9 = 0      x² + 2 = 0

x² = 9           x² = -2  

x = +/- 3       x = +/- sqrt(-2)  -> no real solutions or +/- i(sqrt(2)

 

So x = 3, -3.

 

 

Notice that if you are more comfortable with quadratics,

 

you can substitute u for x^2 which gives you the following:

 

u² - 7u - 18 = 0

 

(u - 9) (u + 2) = 0

 

u = 9     u = -2

 

Now replace u with x^2.

 

x² = 9    x² = -2

 

The rest is the same as above.

 

 

 

Hope this helps.
Jason

 

Answer 2

On the assumption that Bella meant to write

 

f(x) = x^4 - (7)(x^2) - 18

 

It is so easy to show that the real solutions of this polynomial are -3 and 3 that the "proof" hardly needs mentioning.

 

I'm not exactly sure what constitutes a "bound" in situations like this.  (-3.5 and 3.5)? (-4 and 4)?

Answer 3

f(x) = x4 - 7x2 - 18
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