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Zeroes of a Polynomial

When the poly...

Question

When the polynomial $x^{4} + x^{2} + 1$ is divided by $(x + 1) (x^{2} - x + 1)$ then the remainder is $a x + b$ , then $a + b$ is equal to

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Solution

The correct option is A $1$
$\frac{x^{4} + x^{2} + 1}{(x + 1) (x^{2} - x + 1)}$

$= \frac{x^{4} + 2 x^{2} + 1 - x^{2}}{(x + 1) (x^{2} - x + 1)}$

$= \frac{{(x^{2} + 1)}^{2} - x^{2}}{(x + 1) (x^{2} - x + 1)}$

$= \frac{(x^{2} - x + 1) (x^{2} + x + 1)}{(x + 1) (x^{2} - x + 1)}$

$= \frac{(x^{2} + x + 1)}{(x + 1)}$

$= \frac{x (x + 1) + 1}{(x + 1)}$

$= x + \frac{1}{x + 1}$

Remainder $= 1$ is of the form $a x + b$

$\Rightarrow a = 0, b = 1$

$∴ a + b = 0 + 1 = 1$

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When the polynomial x4+x2+1 is divided by (x+1)(x2−x+1) then the remainder is ax+b , then a+b is equal to

The correct option is A 1x4+x2+1(x+1)(x2−x+1)=x4+2x2+1−x2(x+1)(x2−x+1)=(x2+1)2−x2(x+1)(x2−x+1)=(x2−x+1)(x2+x+1)(x+1)(x2−x+1)=(x2+x+1)(x+1)=x(x+1)+1(x+1)=x+1x+1Remainder=1 is of the form ax+b⇒a=0,b=1∴a+b=0+1=1

When the polynomial $x^{4} + x^{2} + 1$ is divided by $(x + 1) (x^{2} - x + 1)$ then the remainder is $a x + b$ , then $a + b$ is equal to