Check if polynomial ($x^{3} \ +\ 3x^{2} \ +\ 3x\ +\ 1$) is divided by ($x\ +\ 1$).

Given: $x^{3} \ +\ 3x^{2} \ +\ 3x\ +\ 1$

To check: Here we have to check if polynomial ($x^{3} \ +\ 3x^{2} \ +\ 3x\ +\ 1$) is divided by ($x\ +\ 1$).

Solution:

If $x\ +\ 1$ is a factor, then $x\ =\ -1$ should be a zero of the polynomial $x^{3} \ +\ 3x^{2} \ +\ 3x\ +\ 1$. 

Putting $x\ =\ -1$ in $x^{3} \ +\ 3x^{2} \ +\ 3x\ +\ 1$:

$x^{3} \ +\ 3x^{2} \ +\ 3x\ +\ 1$

$=\ (-1)^{3} \ +\ 3(-1)^{2} \ +\ 3(-1)\ +\ 1$

$=\ -1\ +\ 3(1) \ -\ 3\ +\ 1$

$=\ -1\ +\ 3 \ -\ 2$

$=\ 0$

It is clear that $x\ +\ 1$ is a factor of $x^{3} \ +\ 3x^{2} \ +\ 3x\ +\ 1$.

So, polynomial ($x^{3} \ +\ 3x^{2} \ +\ 3x\ +\ 1$) is divisible by ($x\ +\ 1$).

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Updated on: 10-Oct-2022

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