When we divide a polynomial $P(x)$ by another polynomial of degree $1$ of the form $(x-a)$, what is the remainder?
There is a very simple way of finding this, without actually performing the polynomial division.
We need to evaluate the polynomial $P(x)$ at $x = a$, and that should give us the remainder.
Let us find out why.
When we divide a polynomial $P(x)$ by another polynomial $G(x)$, the quotient and the remainder both are polynomials, and let us call them $Q(x)$ and $R(x)$.
So, from the basic rule of division:
$dividend = quotient \times divisor + remainder$
So, when we divide $P(x)$ by $G(x)$ and get $Q(x)$ and $R(x)$ as quotient and remainder respectively, we can say:
$P(x) = G(x) \times Q(x) + R(x)$
We know that $G(x)$ is a linear polynomial of the form $(x-a)$, whose degree is $1$. We have also seen that the degree of the remainder is necessarily less than the degree of the divisor. Since here the degree of the divisor is $1$, the degree of the remainder has to be $0$, which means that the remainder is a constant value, independent of $x$, let's say $r$.
$\texttip{\therefore}{therefore} P(x) = (x - a) \times Q(x) + r$
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Substituting $x = a$ in the above equation, we get:
$P(a) = (a - a) \times Q(a) + r$
$P(a) = 0 \times Q(a) + r$
$\texttip{\therefore}{therefore} r = P(a)$, which is nothing but the polynomial $P(x)$ evaluated at $x = a$.
Now let us try the following problem:
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