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Determine which of the following polynomials has $ (x+1) $ a factor:
(i) $ x^{3}+x^{2}+x+1 $
(ii) $ x^{4}+x^{3}+x^{2}+x+1 $
(iii) $ x^{4}+3 x^{3}+3 x^{2}+x+1 $
(iv) $ x^{3}-x^{2}-(2+\sqrt{2}) x+\sqrt{2} $
Given :
The given term is $(x + 1)$.
The given polynomials are
(i) \( x^{3}+x^{2}+x+1 \)
(ii) \( x^{4}+x^{3}+x^{2}+x+1 \)
(iii) \( x^{4}+3 x^{3}+3 x^{2}+x+1 \)
(iv) \( x^{3}-x^{2}-(2+\sqrt{2}) x+\sqrt{2} \)
To find :
We have to check whether the given polynomials have $(x + 1)$ as a factor.
Solution :
According to the factor theorem,
If $(x+1)$ is a factor of given polynomial $P(x)$, then at $x= -1$, $p(x)=0$.
(i) $x^{3}+x^{2}+x+1$
Let $p(x)= x^{3}+x^{2}+x+1$
Substituting $x= -1$
$p(−1)=(−1)^3+(−1)^2+(−1)+1 =−1+1−1+1=0$
Hence, by factor theorem, $x+1$ is a factor of $x^{3}+x^{2}+x+1$.
(ii) $x^{4}+x^{3}+x^{2}+x+1 $
Let $p(x)=x^{4}+x^{3}+x^{2}+x+1 $
Substituting $x= -1$
$p(−1)=(−1)^4+(-1)^3+(−1)^2+(−1)+1 =1−1+1−1+1=1$
Hence, by factor theorem, $x+1$ is not a factor of $x^{4}+x^{3}+x^{2}+x+1$.
(iii) $x^{4}+3x^{3}+3x^{2}+x+1 $
Let $p(x)=x^{4}+3x^{3}+3x^{2}+x+1 $
Substituting $x= -1$
$p(−1)=(−1)^4+3(-1)^3+3(−1)^2+(−1)+1 =1−3+3−1+1=1$
Hence, by factor theorem, $x+1$ is not a factor of $x^{4}+3x^{3}+3x^{2}+x+1$.
(iv) $x^{3}-x^{2}-(2+\sqrt{2}) x+\sqrt{2}$
Let $p(x)=x^{3}-x^{2}-(2+\sqrt{2}) x+\sqrt{2}$
Substituting $x= -1$
$p(−1)=(−1)^3-(-1)^2-(2+\sqrt{2}) (-1)+\sqrt{2}=-1-1+(2+\sqrt{2})+\sqrt{2}$
$=-2+2+\sqrt{2}+\sqrt{2}$
$=2\sqrt{2}$
Hence, by factor theorem, $x+1$ is not a factor of $x^{3}-x^{2}-(2+\sqrt{2}) x+\sqrt{2}$.
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