Question

Given, \[x\ + \ 1\ \] is a factor of the polynomial
a).\[ x^{3} + x^{2}- x + 1\]
b).\[x^{3} + x^{2} + x + 1\]
c).\[x^{4} + x^{3} + x^{2} + 1\]
d).\[x^{4} + 3x^{3} + \ 3x^{2} + x + 1\]

Answer 1

Hint: In this question, we need to find that \[x + 1\] is a factor of which polynomial. By using the factor theorem, we can find the polynomial. Factor theorem is nothing but it links the factors and zeros of the polynomial. If \[(k-1)\] is the factor of the polynomial \[f(k)\] then \[f(1)\] is equal to \[0\]. This is the factor theorem. In order to find the correct polynomial, we need to check the given polynomials separately.

Complete answer:
Given, the factor \[x + 1\]
Now we need to find the polynomial for the factor \[x + 1\]
Let us consider the polynomial as \[f(x)\]
By factor theorem, if \[x + 1\] is the factor of \[f(x)\] , then \[f( - 1)\ = 0\]
a).\[\ x^{3} + \ x^{2}- x\ + 1\]
Let \[f\left( x \right) = \ x^{3} + \ x^{2}\ x\ + 1\]
Now we need to find \[f( - 1)\] that is we need to substitute \[- 1\] in the place of \[x\]
\[f( - 1) = ( - 1)\ + ( - 1)\ - ( - 1)\ + 1\]
On simplifying,
We get,
\[f( - 1)\ = 2\] which is not equal to 0.
So, \[x + 1\] is not the factor of \[x^{3} + \ x^{2}- x\ + 1\]
b).\[x^{3} + \ x^{2} + \ x + 1\]
Let \[f\left( x \right) = x^{3} + \ x^{2} + \ x + 1\]
Now we need to find \[f( - 1)\]
\[f\left( - 1 \right) = \left( - 1 \right)^{3} + \left( - 1 \right)^{2} + \left( - 1 \right) + 1\]
On simplifying,
We get,
\[f\left( - 1 \right) = 0\]
Since \[f\left( x \right)\] is equal to \[0\] , \[x + 1\] is the factor of \[x^{3} + \ x^{2} + \ x + 1\]
We can also check the other two polynomials also.
c).\[x^{4} + \ x^{3} + \ x^{2} + 1\]
Let \[f\left( x \right) = x^{4} + \ x^{3} + \ x^{2} + 1\]
Now we need to find \[f( - 1)\]
\[f\left( - 1 \right) = \left( - 1 \right)^{4} + \left( - 1 \right)^{3} + \left( - 1 \right)^{2} + 1\]
On simplifying,
We get,
\[f( - 1)\ = 2\] which is not equal to \[0\]
So \[x + 1\] is not the factor of \[x^{4} + \ x^{3} + \ x^{2} + 1\]
d).\[x^{4} + \ 3x^{3} + \ 3x^{2} + \ x\ + 1\]
Let \[f\left( x \right) = x^{4} + \ 3x^{3} + \ 3x^{2} + \ x\ + 1\]
Now need to find \[f( - 1)\],
\[f( - 1)\ = ( - 1)\ 4 + 3( - 1)\ + 3( - 1)\ + ( - 1)\ + 1\]
On simplifying,
We get,
\[f\left( - 1 \right) = 1\] which is not equal to \[0\]
Since \[x + 1\] is not the factor of \[x^{4} + \ 3x^{3} + \ 3x^{2} + \ x\ + 1\]
Hence \[\left( x + 1 \right)\] is the factor of \[x^{3} + \ x^{2} + \ x + 1\]
Final answer :
\[(x + 1)\] is the factor of \[x^{3} + \ x^{2}+ \ x + 1\]
Option b) \[x^{3} + \ x^{2}+ \ x + 1\] is correct.

Note:
The concept used in this question to find the polynomial factor is factor theorem. The simple example for the factor theorem is if \[(x-a)\] is the factor of the polynomial \[p(x)\] then \[p(a) =0\]. It is commonly used to find the roots of the polynomial and also for factoring a polynomial.