Remainder Theorem - Practice Questions & MCQ

Equation of higher degree

is known as the polynomial equation of degree n which has exactly n roots (i.e., number of real roots + number of imaginary roots = n)

Relation between its coefficients and roots

$
\text { sum of all roots }=\sum \alpha_1=\alpha_1+\alpha_2+\ldots+\alpha_{n-1}+\alpha_n=(-1) \frac{\mathrm{a}_1}{\mathrm{a}_0}
$
sum of products taken two at a time
$
\sum \alpha_1 \alpha_2=\alpha_1 \alpha_2+\alpha_1 \alpha_3+\ldots+\alpha_1 \alpha_{\mathrm{n}}+\alpha_2 \alpha_3+\ldots+\alpha_2 \alpha_{\mathrm{n}}+\ldots+\alpha_{\mathrm{n}-1} \alpha_{\mathrm{n}}=(-1)^2 \frac{a_2}{a_0}
$
sum of products taken three at a time
$
\sum \alpha_1 \alpha_2 \alpha_3=(-1)^3 \frac{\mathrm{a}_3}{\mathrm{a}_0}
$
product of all roots $=\alpha_1 \alpha_2 \ldots \alpha_n=(-1)^n \frac{a_n}{a_0}$
For example,
Suppose $\mathrm{n}=3$ and $a x^3+b x^2+c x+d=0$ is polynomial equation with $\mathrm{a} \neq 0$ and ?,? and $?$ are the roots of the equation then :
$
\begin{aligned}
& \alpha+\beta+\gamma=-\frac{b}{a} \\
& \sum \alpha \beta=\alpha \beta+\beta \gamma+\gamma \alpha=(-1)^2 \frac{\mathrm{c}}{\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{a}} \\
& \alpha \beta \gamma=(-1)^3 \frac{d}{\mathrm{a}}=-\frac{d}{\mathrm{a}}
\end{aligned}
$

Transformation of roots

For transformation of roots, we can use the same procedure we used in case of quadratic equations.

Remainder theorem

The remainder theorem states that if a polynomial f(x) is divided by a linear function (x - k), then the remainder is f(k).

In Division,

        Dividend = Divisor x Quotient + Remainder

For polynomials also we can use this theorem

        f(x) = d(x).q(x) + r(x)

where f(x) is the divisor, d(x) is the divisor, q(x) is the quotient and r(x) is the remainder. And these 4 are polynomials

Degree of remainder r(x) is always less than degree of divisor d(x)

Now, if divisor d(x) is a linear polynomial (x-k). Let q(x) be the quotient, remainder r(x) will be a constant value equal to R:

                f(x) = (x - k)q(x) + R

Now if we put x = k

i.e.          f(k) = (k - k)q(x) + R = 0 + R

               f(k) = R 

So, remainder is f(k), when f(x) is divided by a linear polynomial (x-k)

Eg, To find remainder when f(x) = 2x³ - 3x - 4 is divided by (x-3),

Here k = 3, So remainder will be f(k) = f(3) = 2.(3)³ - 3(3) - 4 = 54 - 9 - 4 = 41

Factor Theorem

Now if f(k) = 0, then this means that remainder when f(x) is divided by (x-k) is 0. 

As remainder is 0, so (x-k) is a factor of f(x)

So, factor theorem states that if f(k) - 0, then (x-k) is a factor of f(x).

Eg, f(x) = x³ + 3x - 4

Now we can observe by hit and trial that f(1) = 1 + 3 - 4 = 0, so (x-1) is a factor of f(x).