Find the negative of remainder when $p (x) = 4 x^{3} - 3 x^{2} + 2 x - 4$ is divided by $x - 1$

Open in App

Solution

Given $p (x) = 4 x^{3} - 3 x^{2} + 2 x - 4$ .
Now by remainder theorem we've the remainder when $p (x)$ is divided by $x - 1$ is $p (1)$ .
Now
$p (1) = 4 - 3 + 2 - 4 = - 1$ .
So the negative of the remainder is $1$ .

Suggest Corrections

Similar questions

p (x) = x^{3} - 2 x^{2} - 4 x - 1, g (x) = x + 1

p (x) = x^{3} - 3 x^{2} + 4 x + 50, g (x) = x - 3

p (x) = 4 x^{3} - 12 x^{2} + 14 x - 3, g (x) = 2 x - 1

p (x) = x^{3} - 6 x^{2} + 2 x - 4, g (x) = 1 - \frac{3}{2} x

4 x^{3} - 3 x^{2} + 2 x - 4

x + 3

p (x) = x^{3} - 2 x^{2} - 4 x - 1, g (x) = x + 1

p (x) = x^{3} - 3 x^{2} + 4 x + 50, g (x) = x - 3

p (x) = 4 x^{3} - 12 x^{2} + 14 x - 3, g (x) = 2 x - 1

p (x) = x^{3} - 6 x^{2} + 2 x - 4, g (x) = 1 - \frac{3}{2} x

p (x) = x^{4} - 3 x^{2} + 2 x + 5

p (x)

(x - 1)

Statement1: The polynomial $P (x) = 4 x^{3} - 3 x^{2} + 5 x - 6$ when divided by $x - 1$ gives zero as the remainder.

Statement2: $(x - 1)$ is a factor of the polynomial $P (x) = 4 x^{3} - 3 x^{2} + 5 x - 6$ .

Join BYJU'S Learning Program