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Polynomial

Unit: 2
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Class 10: Optional Math

Introduction to Polynomial, Remainder Theorem with proof, Division Relation, Factor Theorem with proof

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    Introduction to Polynomial

    An algebraic expression in which the exponent of each variable is a whole number is called a polynomial. 

    Example: \(2x^{\frac{1}{2}}\) (not polynomial)  \(2x^{2}\) (polynomial)

    The general form of a n degree Polynomial can be written as.

    \(f(x) = a_nx^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + …… ….. ….. +  a_2x^2 + a_1x^1 +a_0 \)

    Where, 

    1. \(a_n, a_{n-1}, a_{n-2}, ….. ….. ….. a_2, a_1, a_0 \) are are real number coefficients.
    2. n  is a whole number (the degree of the polynomial)
    3. The exponent of the variable x  must be either zero or a positive whole number only
    Key Properties
    • Since a polynomial is a function, it is used not only in mathematics but also in practical applications such as demand, supply, velocity, acceleration, cost, production, and other fields.
    • Polynomials can be classified by degree: Linear (degree 1), Quadratic (degree 2), Cubic (degree 3), Quartic (degree 4), etc.

    Examples: \(f(x) = x^4 – 3x^3 + 5x^2 + 4\) (4 degree polynomial), \(x^3 + 2x^2 + 3x - 6\) (3 degree polynomial

     

    Remainder Theorem

    Statement:

    When a polynomial $f(x)$ of degree $n$ is divided by $(x - a)$, the remainder is $f(a)$ and the degree of the quotient is $(n-1)$.

    Proof

    When the polynomial $f(x)$ is divided by the divisor $(x - a)$, there exist the quotient $q(x)$ and the remainder $R$. Then, according to the ordinary division method, the relation among $f(x)$, $(x - a)$, $q(x)$, and $R$ is:

    $f(x) = q(x) \times (x - a) + R \quad \ldots (i)$

    Substituting $x = a$ in equation $(i)$:

    $f(a) = q(a) \times (a - a) + R$

    $f(a) = q(a) \times 0 + R$

    $\boxed{f(a) = R}$

    Therefore, $R = f(a)$. Hence, proved. $\blacksquare$

     

    Division Relation

    If $f(x)$ is divided by $d(x)$ with quotient $q(x)$ and remainder $R$:

    $f(x) = d(x) \times q(x) + R$

    Or equivalently:

    $\frac{f(x)}{d(x)} = q(x) + \frac{R}{d(x)}$

    If $R = 0$, then $\frac{f(x)}{d(x)} = q(x)$, meaning $f(x)$ is exactly divisible by $d(x)$.

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