Question

According to Descartes's rule of sign, how many possible positive and negative roots are there for the equation 0=-4x^6 - 3x^5+2x² - 4x +1? Drag the choices to the boxes to correctly complete the table.

Answer 1

$-4x^6 - 3x^5+2x^2 - 4x +1 = 0$  has 1 or 3 Positive roots and one only Negative roots.

Descartes' rule of sign.

Function has the same number of positive real zeroes as there are number of changes in the sign of the coefficients, or an even number less than the found number.

Now, here the given expression is : p(x) = $-4x^6 - 3x^5+2x^2 - 4x +1$

Now here, the number of sign changes are:

First : + to - in $- 3x^5 => + 2x^2$

Second : - to + in  $+2x² => - 4x$

Third: + to - in  $- 4x => +1$

So, the total sign changes = 3

⇒ The total positive roots of p(x)  = 3 or a EVEN NUMBER LESS THAN 3 = 1.

So, p(x) has either 3 or 1 POSITIVE ROOTS.

⇒ The number of negative roots p(x) has is either 4 or 6.

Hence  has 1 or 3 Positive roots.

Now I look at f (−x). That is, having changed the sign on x, I'm now doing the negative-root case:

p(-x) = $-4(-x)^6 - 3(-x)^5+2(-x)^2 - 4(-x) +1$

       = $-4x^6 + 3x^5+2x^2 + 4x + 1$

the number of sign changes are:

First : + to - in $-4x^6 => +3x^5$

So, the total sign changes = 1.

so there is exactly one negative root. (In this case, I don't try to count down by two's, because the first subtraction would give me a negative number.)

Hence has 1 or 3 Positive roots and one only Negative roots.

For more deatils about Descartes's rule of sign visit link below

brainly.com/question/14387230

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