
What is the remainder theorem?
Answer
465.6k+ views
Hint: Theoretically, an approach of Euclidean division of polynomials is known as the Remainder Theorem. This Theorem states that if we are given with a polynomial $p\left( x \right)$ which is divided by a factor $\left( {x - c} \right)$. Then the remainder obtained would be $p\left( x \right)$ at $x = c$.
Complete step-by-step answer:
According to the basic definition of Remainder Theorem, if there is a polynomial $p\left( x \right)$ whose variable is $x$, and that polynomial is divided by a linear factor $\left( {x - a} \right)$, where $a$ is some constant. And the result obtained would be some polynomial quotient $q\left( x \right)$ and a remainder $r$. And also, the remainder $r\left( x \right)$ obtained is actually $p\left( x \right)$ at $x = a$ or $p\left( a \right)$;
Proof of Remainder theorem:
Since, we know that dividend$ = $divisor$ \times $quotient$ + $remainder.
From the above statement we considered $p\left( x \right)$ as dividend, $\left( {x - a} \right)$ as divisor, $q\left( x \right)$ as quotient and $r$ as remainder.
Substituting these values above, we get:
$p\left( x \right) = \left( {x - a} \right)q\left( x \right) + r$
Substituting $x = a$:
$
p\left( a \right) = \left( {a - a} \right)q\left( a \right) + r \\
p\left( a \right) = 0.q\left( a \right) + r \\
p\left( a \right) = 0 + r \\
p\left( a \right) = r \\
$
Which states that $p\left( a \right)$, is the remainder for the polynomial $p\left( x \right)$ at $x = a$.
Let’s check by taking a polynomial;
$p\left( x \right) = {x^3} - 4{x^2} - 7x + 10$, which is divided by a linear equation $\left( {x - 2} \right)$.
So, according to the remainder theorem, the remainder can be determined by $p\left( 2 \right)$;
So, substituting $2$ in place of $x$ in $p\left( x \right)$, and we get:
$
p\left( x \right) = {x^3} - 4{x^2} - 7x + 10 \\
p\left( 2 \right) = {2^3} - 4{\left( 2 \right)^2} - 7 \times 2 + 10 \\
$
On, further solving the equation, we get:
$
p\left( 2 \right) = {2^3} - 4{\left( 2 \right)^2} - 7 \times 2 + 10 \\
p\left( 2 \right) = 8 - 16 - 14 + 10 \\
p\left( 2 \right) = 18 - 30 \\
p\left( 2 \right) = - 12 \\
$
which is the remainder for the polynomial $p\left( x \right) = {x^3} - 4{x^2} - 7x + 10$, when divided by $\left( {x - 2} \right)$.
Note: We can also check the remainder is correct or not by using the formula; dividend$ = $divisor$ \times $quotient$ + $remainder. Just substitute the values and solve.
In some case if the remainder is zero, then both the divisor and quotient are the factors of the dividend.
Complete step-by-step answer:
According to the basic definition of Remainder Theorem, if there is a polynomial $p\left( x \right)$ whose variable is $x$, and that polynomial is divided by a linear factor $\left( {x - a} \right)$, where $a$ is some constant. And the result obtained would be some polynomial quotient $q\left( x \right)$ and a remainder $r$. And also, the remainder $r\left( x \right)$ obtained is actually $p\left( x \right)$ at $x = a$ or $p\left( a \right)$;
Proof of Remainder theorem:
Since, we know that dividend$ = $divisor$ \times $quotient$ + $remainder.
From the above statement we considered $p\left( x \right)$ as dividend, $\left( {x - a} \right)$ as divisor, $q\left( x \right)$ as quotient and $r$ as remainder.
Substituting these values above, we get:
$p\left( x \right) = \left( {x - a} \right)q\left( x \right) + r$
Substituting $x = a$:
$
p\left( a \right) = \left( {a - a} \right)q\left( a \right) + r \\
p\left( a \right) = 0.q\left( a \right) + r \\
p\left( a \right) = 0 + r \\
p\left( a \right) = r \\
$
Which states that $p\left( a \right)$, is the remainder for the polynomial $p\left( x \right)$ at $x = a$.
Let’s check by taking a polynomial;
$p\left( x \right) = {x^3} - 4{x^2} - 7x + 10$, which is divided by a linear equation $\left( {x - 2} \right)$.
So, according to the remainder theorem, the remainder can be determined by $p\left( 2 \right)$;
So, substituting $2$ in place of $x$ in $p\left( x \right)$, and we get:
$
p\left( x \right) = {x^3} - 4{x^2} - 7x + 10 \\
p\left( 2 \right) = {2^3} - 4{\left( 2 \right)^2} - 7 \times 2 + 10 \\
$
On, further solving the equation, we get:
$
p\left( 2 \right) = {2^3} - 4{\left( 2 \right)^2} - 7 \times 2 + 10 \\
p\left( 2 \right) = 8 - 16 - 14 + 10 \\
p\left( 2 \right) = 18 - 30 \\
p\left( 2 \right) = - 12 \\
$
which is the remainder for the polynomial $p\left( x \right) = {x^3} - 4{x^2} - 7x + 10$, when divided by $\left( {x - 2} \right)$.
Note: We can also check the remainder is correct or not by using the formula; dividend$ = $divisor$ \times $quotient$ + $remainder. Just substitute the values and solve.
In some case if the remainder is zero, then both the divisor and quotient are the factors of the dividend.
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