Find a bound on the real zeros of the polynomial function.

Question

x⁴-7x²-18

 

please step by step instructions.

 

Thanks ;-)

Answer 1

ƒ(x) = x⁴ - 7x² - 18  ...... (1)

To find the zeros of this polynomial function we need to set f(x) equal to 0 and solve it for "x" .

Assume that x² = y , then

y² - 7y - 18 = 0  ....... (2)

Let's factor the equation (2)

- 18 = (- 9) * 2
- 7   = - 9 + 2

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

y₁ = 9  ---->  x² = 9  ----->  x₁₂ = ±√9 = ± 3 

y₂ = - 2  ----->  x² = - 2 < 0 this equation does not have real roots.

Thus, given polynomial function has two real zeros ±3

Answer 2

You approach this problem very similarly to a quadratic equation.

 

First determine what factors can add to -7 and multiply to -18

 

The factors are +2,-9

 

This leads to the equation:

 

x⁴-7x²-18=(x²-9)(x²+2)

 

(x²+2)  cannot be broken down further

 

But:

 

(x²-9)  can be broken into:

 

(x-3)(x+3)

 

Leading to the conclusion that:

 

x⁴-7x²-18=(x-3)(x+3)(x²+2)

 

Set the equation equal to 0

 

(x-3)(x+3)(x²+2) = 0

 

You find that x = -3, 3, and +√(-2) (which is an imaginary root)

 

Leading to the conclusion that the real zeros are when

 

x = -3 or +3