
If ${x^3} + 1,x + 1,{x^2} - x + 1$ and 0 are dividend, divisor, quotient and remainder respectively. Then verify the division algorithm.
Answer
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Hint: It is given in the question that If ${x^3} + 1,x + 1,{x^2} - x + 1$ and 0 are dividend, divisor, quotient and remainder respectively.
Then, we have to verify the division algorithm.
Thus, to verify the division algorithm we use the formula $p\left( x \right) = q\left( x \right) \times d\left( x \right) + r\left( x \right)$ and substitute the appropriate values in the formula.
Complete step-by-step answer:
It is given in the question that If ${x^3} + 1,x + 1,{x^2} - x + 1$ and 0 are dividend, divisor, quotient and remainder respectively.
Then, we have to verify the division algorithm.
Since, we have given that,
Dividend $p\left( x \right) = {x^3} + 1$
Divisor $q\left( x \right) = x + 1$
Quotient $d\left( x \right) = {x^2} - x + 1$
Remainder $r\left( x \right) = 0$
The division algorithm states that for any polynomial \[p\left( x \right)\] and \[q\left( x \right)\] , there exists unique polynomial \[d\left( x \right)\] and \[r\left( x \right)\] such that
$p\left( x \right) = q\left( x \right) \times d\left( x \right) + r\left( x \right)$
First, we will right the L.H.s of the division algorithm i.e.
$p\left( x \right) = {x^3} + 1$
Now, we are going to find the R.H.S of the division algorithm i.e.
$q\left( x \right) \times d\left( x \right) + r\left( x \right)$
$\therefore \left( {x + 1} \right) \times \left( {{x^2} - x + 1} \right) + 0$
$\therefore {x^3} - {x^2} + x + {x^2} - x + 1$
After cancellation the equal terms of opposite signs, we get ${x^3} + 1$ which is same as $p\left( x \right)$
Hence,
$L.H.S = R.H.S$
Therefore, the division algorithm is verified.
Note: Division algorithm: A division algorithm is an algorithm which, given two integers N and D, compute their quotient and/or remainder, the result of division.
$p\left( x \right) = q\left( x \right) \times d\left( x \right) + r\left( x \right)$ where,
$p\left( x \right)$ is dividend, $q\left( x \right)$ is divisor, $d\left( x \right)$ is quotient and $r\left( x \right)$ is remainder.
Then, we have to verify the division algorithm.
Thus, to verify the division algorithm we use the formula $p\left( x \right) = q\left( x \right) \times d\left( x \right) + r\left( x \right)$ and substitute the appropriate values in the formula.
Complete step-by-step answer:
It is given in the question that If ${x^3} + 1,x + 1,{x^2} - x + 1$ and 0 are dividend, divisor, quotient and remainder respectively.
Then, we have to verify the division algorithm.
Since, we have given that,
Dividend $p\left( x \right) = {x^3} + 1$
Divisor $q\left( x \right) = x + 1$
Quotient $d\left( x \right) = {x^2} - x + 1$
Remainder $r\left( x \right) = 0$
The division algorithm states that for any polynomial \[p\left( x \right)\] and \[q\left( x \right)\] , there exists unique polynomial \[d\left( x \right)\] and \[r\left( x \right)\] such that
$p\left( x \right) = q\left( x \right) \times d\left( x \right) + r\left( x \right)$
First, we will right the L.H.s of the division algorithm i.e.
$p\left( x \right) = {x^3} + 1$
Now, we are going to find the R.H.S of the division algorithm i.e.
$q\left( x \right) \times d\left( x \right) + r\left( x \right)$
$\therefore \left( {x + 1} \right) \times \left( {{x^2} - x + 1} \right) + 0$
$\therefore {x^3} - {x^2} + x + {x^2} - x + 1$
After cancellation the equal terms of opposite signs, we get ${x^3} + 1$ which is same as $p\left( x \right)$
Hence,
$L.H.S = R.H.S$
Therefore, the division algorithm is verified.
Note: Division algorithm: A division algorithm is an algorithm which, given two integers N and D, compute their quotient and/or remainder, the result of division.
$p\left( x \right) = q\left( x \right) \times d\left( x \right) + r\left( x \right)$ where,
$p\left( x \right)$ is dividend, $q\left( x \right)$ is divisor, $d\left( x \right)$ is quotient and $r\left( x \right)$ is remainder.
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