Question

Any one answer the question in picture please

Answer 1

Answer:

Step-by-step explanation:

just divide (xcube+3x square+3x+1)with ( x+1) ...dividing will give u a quotient which u will have to factorise

Answer 2

According to remainder theorem when a polynomial $f(x)$ is divided by $ax + b$ then the remainder obtained is $f(\frac{-b}{a})$.

Hence remainder when $f(x) = x^{3} + 3x^{2} + 3x + 1$ is divided by $x + 1$ is $f(-1) = (-1)^{3} + 3(-1)^{2} + 3(-1) + 1 = -1 + 3 - 3 + 1 = 0$.

Remainder when $f(x) = x^{3} + 3x^{2} + 3x + 1$ is divided by $x - \frac{1}{2}$ is $f(\frac{1}{2}) = (\frac{1}{2})^{3} + 3(\frac{1}{2})^{2} + 3(\frac{1}{2}) + 1 = \frac{1}{8} + \frac{3}{4} + \frac{3}{2} + 1 = \frac{27}{8}$.

Remainder when $f(x) = x^{3} + 3x^{2} + 3x + 1$ is divided by $x$ is $f(0) = (0)^{3} + 3(0)^{2} + 3(0) + 1 = 0 + 0 + 0 + 1 = 1$.

Remainder when $f(x) = x^{3} + 3x^{2} + 3x + 1$ is divided by $x + \pi$ is $f(-\pi) = (-\pi)^{3} + 3(-\pi)^{2} + 3(-\pi) + 1 = -\pi^{3} + 3\pi^{2} - 3\pi + 1$.

Remainder when $f(x) = x^{3} + 3x^{2} + 3x + 1$ is divided by $5 + 2x$ is $f(\frac{-5}{2}) = (\frac{-5}{2})^{3} + 3(\frac{-5}{2})^{2} + 3(\frac{-5}{2}) + 1 = -\frac{125}{8} + \frac{75}{4} - \frac{15}{2} + 1 = -\frac{27}{8}$