Question

Find the number of positive and negative roots of the polynomial
P(x) = x3 − 3x2 + 4x − 5. Find P(2) and P′(2) using synthetic division method.

Answer 1

1. Find the number of positive and negative roots of the polynomial

$P(x) = x^{3} - 3x^{2} + 4x - 5$ Find $\mathsf{P}(2)$ and $\mathsf{P}'(2)$ using synthetic division method.

Answer:

As per Descartes's rule of signs, the number of positive roots is equal to change in sign of $P(x)$ , or is less than that by an even number.

Given polynomial is $P(x) = x^3 - 3x^2 + 4x - 5$

The signs of coefficients are in the form $+ - + -$

Since there are 3 changes of sign, there are 3 or 1 positive real roots

Similarly, for the number of negative real roots we have to find $P(-x)$ and we have to count again

$P(-x) = -x^{3} - 3x^{2} - 4x - 5$

(It is same as reversing the signs on terms of odd degree)

The signs of coefficients are in the form $- - - -$

Since there is no change of sign, there are zero negative real roots.

As the total number of zeros of $P(x)$ is 3, that means it has 0 or 2 non-real Complex zeroes.

To evaluate $P(2)$

Using factor-remainder theorem we know that if a polynomial $P(x)$ is divided by $x - a$ , the remainder is $P(a)$ .

So the value of $P(2)$ by synthetic division as below

Hence,

$P(2) = -1$

To evaluate $P^{\prime}(2)$

$P^{\prime}(x) = 3x^{2} - 6x + 4$

Hence,

$P^{\prime}(2) = 4$