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.