Question

If g(x) and h(x) are two polynomials such that $p\left( x\right) =g\left( x^{3}\right) +xh\left( x^{3}\right)$ and p(x) is divisible by $x^{2}+x+1$, then
A. g(1) + h(1)= -1
B. g(1) + h(1) =0
C. g(1) - h(1) $\neq$0
D. g(1)= h(1)= -1

Answer 1

Hint: In this question it is given that If g(x) and h(x) are two polynomials such that $p\left( x\right) =g\left( x^{3}\right) +xh\left( x^{3}\right)$ and p(x) is divisible by $x^{2}+x+1$, then we have to find the relation between g(1) and h(1). So to find the solution we need to know that since $x^{2}+x+1$ is the factor of p(x) then the zeros of $x^{2}+x+1$ is must be the zeros of p(x), so after finding the zeros we will put it in the given equation $p\left( x\right) =g\left( x^{3}\right) +xh\left( x^{3}\right)$, from where we will get the required solution.

Complete step-by-step answer:
First we are going to find the zeros of the polynomial $x^{2}+x+1$,
So, $x^{2}+x+1=0$.....(1)
Now as we know that if any equation is in the form of $ax^{2}+bx+c=0$ then the quadratic formula is,
$x=\dfrac{-b\pm \sqrt{b^{2}-4ac} }{2a}$
Now by comparing (1) with the above equation we can write,
a=1, b=1 and c=1
So by quadratic formula we can write,
$x=\dfrac{-1\pm \sqrt{1^{2}-4\cdot 1\cdot 1} }{2\cdot 1}$
$x=\dfrac{-1\pm \sqrt{1-4} }{2}$
$x=\dfrac{-1\pm \sqrt{-3} }{2}$
$x=\dfrac{-1\pm \sqrt{3} i}{2}$ [since,i is the square root of -1 therefore, $i=\sqrt{-1}$]
$\therefore x=\dfrac{-1-\sqrt{3} i}{2} ,\ x=\dfrac{-1+\sqrt{3} i}{2}$
Now as we know that if $\omega$ is the cube root of unity, then
$\therefore \omega =\dfrac{-1-\sqrt{3} i}{2} ,\ w^{2}=\dfrac{-1+\sqrt{3} i}{2}$
and $\omega^{3} =1$
So ultimately we get,
$x=\omega ,\ x=w^{2}$
So if we put the values of x in p(x) then the polynomial p(x) becomes zero. i.e, $p\left( \omega \right) =0,\ p\left( \omega^{2} \right) =0$
Here the given equation,
$p\left( x\right) =g\left( x^{3}\right) +xh\left( x^{3}\right)$.......(2)
Now when $x=\omega$, we get from equation (2),
$p\left( \omega \right) =g\left( \omega^{3} \right) +\omega h\left( \omega^{3} \right)$
$\Rightarrow 0=g\left( 1\right) +\omega h\left( 1\right)$
$\Rightarrow g\left( 1\right) +\omega h\left( 1\right) =0$.......(3)
Again when $x=w^{2}$, from equation (2) we get,
p\left( \omega^{2} \right) =g\left\\{ \left( \omega^{2} \right)^{3} \right\\} +\omega^{2} h\left\\{ \left( \omega^{2} \right)^{3} \right\\}
$\Rightarrow 0=g\left( \omega^{6} \right) +\omega^{2} h\left( \omega^{6} \right)$
\Rightarrow 0=g\left\\{ \left( \omega^{3} \right)^{2} \right\\} +\omega^{2} h\left\\{ \left( \omega^{3} \right)^{2} \right\\}
\Rightarrow 0=g\left\\{ \left( 1\right)^{2} \right\\} +\omega^{2} h\left\\{ \left( 1\right)^{2} \right\\} [since, $\omega^{3} =1$]
$\Rightarrow 0=g\left( {}1\right) +\omega^{2} h\left( 1\right)$
$\Rightarrow g\left( {}1\right) +\omega^{2} h\left( 1\right) =0$
$\Rightarrow g\left( {}1\right) =-\omega^{2} h\left( 1\right)$......(4)
Now putting the value of g(1) in equation (3), we get,
$-\omega^{2} h\left( 1\right) +\omega h\left( 1\right) =0$
\Rightarrow h\left( 1\right) \left\\{ \omega -\omega^{{}2} \right\\} =0
Iff, $h\left( 1\right) =0$
And by putting the value of h(1) in equation (4) we get,
$g\left( 1\right) =-\omega^{2} \times 0$
$g\left( 1\right) =0$
Therefore, g(1)+h(1)=0+0=0
Hence the correct option is option B.

Note: While solving this type of question you need to know that the zeros of a polynomial are some values of the variable for which the polynomial will be zero, which is also called the root of a polynomial.