Determine which of the following polynomials has $(x + 1)$ a factor : $x^{4}+x^{3}+x^{2}+x+1$.
For $x+1=0,$ we have $x=-1$.
$\therefore $ The zero of $x+1$ is $-1$.
$\because$ $p ( x ) = x ^{4}+ x ^{3}+ x ^{2}+ x +1 $
$\therefore$ $p (-1) =(-1)^{4}+(-1)^{3}+(-1)^{2}+(-1)+1$
$=1-1+1-1+1=3-2=1$
$\because$ $f (-1) \neq 0 $
$\therefore$ $p ( x )$ is not divisible by $x +1$.
Factorise : $2 y^{3}+y^{2}-2 y-1$
Evaluate the following using suitable identities : $(102)^{3}$
Find the value of the polynomial $5x -4x^2+ 3$ at $x = 0$.
Without actually calculating the cubes, find the value of each of the following : $(28)^{3}+(-15)^{3}+(-13)^{3}$
Verify whether the following are zeroes of the polynomial, indicated against them.
$p(x)=2 x+1, \,\,x=\frac{1}{2}$