Determine whether (x + 1) is a factor of the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) According to the Factor Theorem,$(x + 1)$ is a factor of $p(x)$ if and only if $p(-1) = 0$.Given $p(x) = x^{4} + x^{3} + x^{2} + x + 1$.Substituti

Vedclass · Accepted Answer

No

Vedclass · Answer

(B) According to the Factor Theorem,$(x + 1)$ is a factor of $p(x)$ if and only if $p(-1) = 0$.Given $p(x) = x^{4} + x^{3} + x^{2} + x + 1$.Substituting $x = -1$ into the polynomial:$p(-1) = (-1)^{4} + (-1)^{3} + (-1)^{2} + (-1) + 1$Calculating each term:$p(-1) = 1 - 1 + 1 - 1 + 1$Simplifying the expression:$p(-1) = 1$Since $p(-1) 
eq 0$,by the Factor Theorem,$(x + 1)$ is not a factor of the given polynomial.

Determine whether $(x + 1)$ is a factor of the polynomial $p(x) = x^{4} + x^{3} + x^{2} + x + 1$.

Similar Questions

Find the value of $k$,if $x - 1$ is a factor of $p(x)$ in this case: $p(x) = x^{2} + x + k$.

Factorise: $2y^{3} + y^{2} - 2y - 1$

Factorise the following: $64 a^{3}-27 b^{3}-144 a^{2} b+108 a b^{2}$

Classify the following as linear,quadratic,and cubic polynomials:
$(i)$ $x^{2}+x$
$(ii)$ $x-x^{3}$
$(iii)$ $y+y^{2}+4$

Find the zero of the polynomial: $p(x) = x - 5$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module