Find, in each case, the remainder when: $x^{4} - 3 x^{2} + 2 x + 1$ is divided by $x - 1$ .

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Solution

From remainder theorem, we know that when a polynomial $f (x)$ is divided by $(x - a)$ , then the remainder is $f (a) .$
Given, $f (x) = x^{4} - 3 x^{2} + 2 x + 1$ is divided by $x - 1$
So, remainder $= f (1) = (1)^{4} - 3 (1)^{2} + 2 (1) + 1 = 1 - 3 + 2 + 1 = 1$

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(iii) $x^{4} + 1 i s d i v i d e d b y x + 1.$

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The polynomial x⁴- 2x³+ 3x²- ax + b when divided by x +1 and x -1 give remainder 19 and 5 respectively . Find the remainder when the polynomial is divided by x -3

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Find, in each case, the remainder when: x4−3x2+2x+1 is divided by x−1.

From remainder theorem, we know that when a polynomial f(x) is divided by (x–a), then the remainder is f(a).Given, f(x)=x4–3x2+2x+1 is divided by x–1So, remainder =f(1)=(1)4–3(1)2+2(1)+1=1–3+2+1=1

Find, in each case, the remainder when: $x^{4} - 3 x^{2} + 2 x + 1$ is divided by $x - 1$ .

From remainder theorem, we know that when a polynomial $f (x)$ is divided by $(x - a)$ , then the remainder is $f (a) .$
Given, $f (x) = x^{4} - 3 x^{2} + 2 x + 1$ is divided by $x - 1$
So, remainder $= f (1) = (1)^{4} - 3 (1)^{2} + 2 (1) + 1 = 1 - 3 + 2 + 1 = 1$