Question

Find the real zeros for:
F(x)=x^5-x^2+6

Answer 1

x=-1.334

Explanation:

Let first use descrates rule of signs to find possible zeroes.

First, find the number of signs changes to find possible positive zeroes.

Here,

we have

`x {}^(5) - {x + 6}`

We have 2 signs changes so we have either

2 positive zeroes, or none.

To find. negative subsitue -x for x.

`( - x) {}^(5) - ( - x) + 6`

`- x {}^(5) + x + 6`

There is one sign change so we have 1 negative zzeroes.

The fundamental theorem algebra tells us that for a function which a leading degree n has exactly n roots at most.

The leading degree has 5 roots so we have either

2 positive zeroes 1 positive zero, or 2 complex zeroes.

or

4 complex zeroes, 1 positive zero.

So let see first.

`x {}^(5) - {x}^(2) + 6`

Rational Roots Theroem isn't applicable, because we have no zeroes. that are rational.

We must use a table of values to find a zero.

If you plug in -2 and -1, the sings oscillate so we have a zero between

-2 and-1.

`f( - 2) = { - 2}^(5) - ( - 2) {}^(2) + 6 = - 30`

`f( - 1) = - 1 {}^(5) - ( - 1) {}^(2) + 6 = 4`

So using the Theorem, the zero has to between -30 and 4.

In fact, if we plug the function in a graphing calculator, we get this.

So we have one real zero x=-1.334