Maths M.C.Q

Let $p(x)=2 x^2+k x$
Since, $(x+1)$ is a factor of $p(x)$, then
$p(-1)=0$
$2(-1)^2+k(-1)=0$
$\Rightarrow 2-k=0$
$\Rightarrow k=2$

Let $p(x)=2 x^2+7 x-4$
$=2 x^2+8 x-x-4[\text { by splitting middle term }]$
$=2 x(x+4)-1(x+4)$
$=(2 x-1)(x+4)$
For zeroes of $p(x)$, put $p(x)=0 \Rightarrow(2 x-1)(x+4)=0$
$\Rightarrow 2 x-1=0 \text { and } x+4=0$
$\Rightarrow \mathrm{x}=\frac{1}{2}$ and $\mathrm{x}=-4$
Hence, one of the zeroes of the polynomial $p(x)$ is $\frac{1}{2}$.

Now, $a^3+b^3+c^3=(a+b+c)\left(a^2+b^2+c^2-a b-b e-c a\right)+3 a b c$
$\left[\right.$ Using identity, $\left.a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b e-c a\right)\right]=0+3 a b c$
$\therefore a+b+c=0, \text { given }$
$a^3+b^3+c^3=3 a b c$

Let assume $(x+1)$ is a factor of $x^3+x^2+x+1$
So, $x=-1$ is zero of $x^3+x^2+x+1$
$(-1)^3+(-1)^2+(-1)+1=0$
$\Rightarrow-1+1-1+1=0$
$\Rightarrow 0=0$
Hence, our assumption is true.

$(x+y)^3-\left(x^3+y^3\right)=x^3+y^3+3 x y(x+y)-x^3-y^3$
${\left[(a+b)^3=a^3+b^3+3 a b(a+b)\right]}$
$=3 x y(x+y)$
So, $3 x y$ is a factor of $(x+y)^3-\left(x^3+y^3\right)$.