Question

Determine which of the following polynomials has $(x + 1)$ as 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 $

Answer 1

Hint: First, we have to define what the terms we need to solve the problem are.
Since the factor is given as in the form of linear equation with degree one and also called the straight line, we need to find the value of $x$ and then solving further to find which are equations are polynomial to that factor $(x + 1)$

Complete step-by-step answer:
let every polynomial can be written us in the form of equation and vise versa (by Cayley Hamilton theorem), using this we can rewrite the polynomial factor as $(x + 1) = 0 \Rightarrow x = - 1$ (which is also called the remainder of the value $x$, we obtained this by adding $ - 1$ on both sides). Thus, we have the value of $(x + 1) = 0 \Rightarrow x = - 1$; now we are going to apply this in every option to check which are the polynomials having the facto r$(x + 1)$.

$(i){x^3} + {x^2} + x + 1$ substituting the value $x = - 1$
we get, ${x^3} + {x^2} + x + 1 = {( - 1)^3} + {( - 1)^2} + - 1 + 1$(minus one power odd values are again the minus one, but even powers get positive one).
Thus, we get ${( - 1)^3} + {( - 1)^2} + - 1 + 1 = - 1 + 1 - 1 + 1 \Rightarrow 0$ (equals to zero gives the polynomial a factor)
Hence $(i){x^3} + {x^2} + x + 1$ has the factor $(x + 1)$

$(ii){x^4} + {x^3} + {x^2} + x + 1$ substituting the value $x = - 1$ we get,
${x^4} + {x^3} + {x^2} + x + 1 = {( - 1)^4} + {( - 1)^3} + {( - 1)^2} + - 1 + 1$ (minus one power odd values are again the minus one, but even powers get positive one).
Thus, we get ${( - 1)^4} + {( - 1)^3} + {( - 1)^2} + - 1 + 1 = 1 - 1 + 1 - 1 + 1 \Rightarrow 1$(equals to one does not give the polynomial a factor)
Hence $(ii){x^4} + {x^3} + {x^2} + x + 1$ is not the factor of $(x + 1)$

$(iii){x^4} + 3{x^3} + 3{x^2} + x + 1$ substituting the value $x = - 1$
we get, $(iii){x^4} + 3{x^3} + 3{x^2} + x + 1 = {( - 1)^4} + 3{( - 1)^3} + 3{( - 1)^2} + - 1 + 1$ (minus one power odd values are again the minus one, but even powers get positive one). Thus, we get ${( - 1)^4} + 3{( - 1)^3} + 3{( - 1)^2} + - 1 + 1 = 1 - 3 + 3 - 1 + 1 \Rightarrow 1$ (equals to one does not give the polynomial a factor)
Hence $(iii){x^4} + 3{x^3} + 3{x^2} + x + 1$ is not the factor of $(x + 1)$

$(iv){x^3} - {x^2} - (2 + \sqrt 2 )x + \sqrt 2 $ substituting the value $x = - 1$ we get, ${x^3} - {x^2} - (2 + \sqrt 2 )x + \sqrt 2 = {( - 1)^3} - {( - 1)^2} - (2 + \sqrt 2 )( - 1) + \sqrt 2 $ (minus one power odd values are again the minus one, but even powers get positive one). Thus, we get ${( - 1)^3} - {( - 1)^2} - (2 + \sqrt 2 )( - 1) + \sqrt 2 = - 1 + 1 + 2 - \sqrt 2 + \sqrt 2 \Rightarrow 0$ (equals to zero gives the polynomial a factor)
Hence $(iv){x^3} - {x^2} - (2 + \sqrt 2 )x + \sqrt 2 $ has the factor $(x + 1)$

Thus, the options are $(i){x^3} + {x^2} + x + 1$ and $(iv){x^3} - {x^2} - (2 + \sqrt 2 )x + \sqrt 2 $ are correct.

Note: Since to find the factors of the polynomial we first need to find the value of the unknowns and then substituting into the given polynomials if there exist zero it is the factor. The values of the x may be above two integers too; like $x = 1, - 1$ also possible but not in linear (one degree).