Remainder Calculation for \(x^6\) Divided by \(x^2+1\)
This problem asks us to find the remainder when the polynomial \(x^6\) is divided by the polynomial \(x^2 + 1\). We can solve this using algebraic manipulation, specifically by leveraging the properties of polynomial division and substitution.
Understanding Polynomial Division and Remainders
When we divide a polynomial \(P(x)\) by another polynomial \(D(x)\), we get a quotient \(Q(x)\) and a remainder \(R(x)\) such that:
\(P(x) = D(x) \cdot Q(x) + R(x)\)The degree of the remainder \(R(x)\) must be strictly less than the degree of the divisor \(D(x)\). In our case:
- \(P(x) = x^6\) (the dividend)
- \(D(x) = x^2 + 1\) (the divisor)
Since the degree of the divisor \(D(x)\) is 2 (the highest power of \(x\) is \(x^2\)), the remainder \(R(x)\) must have a degree less than 2. This means the remainder will be of the form \(ax + b\), where \(a\) and \(b\) are constants.
Method: Using Substitution
A quick way to find the remainder is to use the relationship derived from the divisor.
- Set the divisor to zero to find the values of \(x\) that make it zero: \(x^2 + 1 = 0\)
- Solve for \(x^2\): \(x^2 = -1\)
- Now, we rewrite the dividend \(x^6\) in terms of \(x^2\): \(x^6 = (x^2)^3\)
- Substitute the value \(x^2 = -1\) into the expression for \(x^6\): \(x^6 = (-1)^3\)
- Calculate the result: \((-1)^3 = -1 \times -1 \times -1 = -1\)
This value, -1, represents the remainder when \(x^6\) is divided by \(x^2+1\). Because this result is a constant, it fits the requirement for the remainder \(R(x)\) (degree less than 2).
Step-by-Step Algebraic Approach
Alternatively, we can use algebraic steps:
- Start with the dividend \(x^6\).
- We know \(x^2 = -1\) implies \(x^2+1=0\). We want to express \(x^6\) using terms of \(x^2+1\).
- Rewrite \(x^6\) as \((x^2)^3\).
- Substitute \(x^2 = -1 + (x^2+1)\): \(x^6 = (x^2)^3 = (-1 + (x^2+1))^3\)
- Expanding this using the binomial theorem would be complex. Instead, let's directly use \(x^2 \equiv -1 \pmod{x^2+1}\).
- \(x^6 = (x^2)^3 \equiv (-1)^3 \pmod{x^2+1}\)
- \(x^6 \equiv -1 \pmod{x^2+1}\)
This shows that \(x^6\) leaves a remainder of \(-1\) when divided by \(x^2+1\).
Conclusion
Both methods confirm that the remainder when \(x^6\) is divided by \(x^2 + 1\) is \(-1\).
Similar Questions
- Let \(p(x)\) be a polynomial. When \(p(x)\) is divided by \((x-1)\), it leaves 2 as the remainder. When \(p(x)\) is divided by \((x-2)\), it leaves 1 as the remainder. What is the remainder when \(p(x)\) is divided by \((x - 1)(x-2)\)?
- \((x+2)\) is a factor of which one of the following?
What is the HCF of the polynomials x⁸ + x⁴ + 1 and x⁴ + x² + 1?
- If 2 is a zero of the polynomial \(p(x) = x^3 + 3x^2 - 6x - a\), then what is the sum of the squares of the other zeros of the polynomial?
- Suppose \(p(x) = x^4 + a_3x^3 + a_2x^2 + a_1x + a_0\) and \(q(x) = x^4 + b_3x^3 + b_2x^2 + b_1x + b_0\) are the polynomials. If \(\alpha, \beta, \gamma, \delta\) are zeros of \(p(x)\) and \(\alpha, \beta, \gamma, \lambda\) are zeros of \(q(x)\), then what is \(\frac{p(x) - q(x)}{(x - \alpha) (x - \beta) (x - \gamma)}\) equal to ?
- Consider the following in respect of the polynomial \(x^{4k} + x^{4k+2} + x^{4k+4} + x^{4k+6}\):
1. The remainder is zero when the polynomial is divided by \(x^2 + 1\).
2. The remainder is zero when the polynomial is divided by \(x^4 + 1\).
Which of the statements given above is/are correct? - If \(x^3 + px^2 + qx + r\) is an integer for all integral values of \(x\), then consider the following statements :
I. \(p\) must be an integer
II. \(q\) must be an integer
III. \(r\) must be an integer
Which of the statements given above is/are correct? - Which of the following expressions can divide both the polynomials \(x^3+2x^2-5x+2\) and \(x^3+4x^2+x-6\) exactly ?
I. \(x-1\)
II. \(x+1\)
III. \(x+2\)
Select the correct answer using the code given below : If x² - 5x + 4 is a factor of x⁴ - px² + q, then what are the values of p and q respectively?
Important Questions from Polynomials
If y 2= y + 7, then what is the value of y 3?
Factorize x 2- y 2- 9z 2+ 6yz
If one of the zeros of the polynomial x 3+ ax 2+ bx + c is - 1, then the product of other two zeros is equal to :
If a(a + b + c) 2 = 1792; b(a + b + c) 2 = 1536; c(a + b + c) 2 = 768, then what will be the value of b?
If x = 3 so, what is the value of x 2 + 2x + 5 ?
