Question

Divide the polynomial $2x^4-4x^3-3x-1$ by $(x-1)$ and verify the remainder with zero of the divisor.

Answer 1

We need divide $f\left(x\right)=2x^4-4x^3-3x-1$ by $(x-1)$.

Procedure:
Step 1:

$\frac{2x^4}{x}=2x^3$
Now multiply $(x-1)\left(2x^3\right)=2x^4-2x^3$ and then subtract.

Step 2:

$\frac{-2x^3}{x}=-2x^2$

Now multiply $(x-1)(-2x^2)=-2x^3+2x^2$ and then subtract.

Step 3:

$\frac{-2x^2}{x}=-2x$

Now multiply $(x-1)(-2x)=-2x^2+2x$ and then subtract.

Step 4:

$\frac{-5x}{x}=-5$

Now multiply $(x-1)(-5)=-5x+5$ and then subtract.

Here the quotient is $2x^3-2x^2-2x-5$ and the remainder is $-6$.
Now, the zero of the polynomial $(x-1)$ is $1$.
Put $x=1$ in $f\left(x\right),f\left(x\right)=2x^4-4x^3-3x-1$
$f\left(1\right)=2(1{)}^4-4(1{)}^3-3\left(1\right)-1$
$=2\left(1\right)-4\left(1\right)-3\left(1\right)-1$
$=2-4-3-1$
$=-6$
Is the remainder same as the value of the polynomial $f\left(x\right)$ at zero of $(x-1)$?
From the above examples we shall now state the fact in the form of the following theorem.
It gives a remainder without actual division of a polynomial by a linear polynomial in one variable.

Given polynomial is $f\left(x\right)=2x^4-4x^3-3x-1$ and divided by $(x-1)$

Put $x=1$ in the given polynomial, we get

$f\left(x\right)=2x^4-4x^3-3x-1$

$\Rightarrow f\left(x\right)=2(1{)}^4-4(1{)}^3-3\left(1\right)-1$

$\Rightarrow f\left(x\right)=2-4-3-1$

$\Rightarrow f\left(x\right)=-6$

Then $2x^4-4x^3-3x-1$ divided by $(x-2)$ the reminder is $-6$.

Then polynomial $2x^4-4x^3-3x-1$ divided by $(x-2)$ the reminder is not zero.