Question

If $f\left( x \right) = g\left( {{x^3}} \right) + xh\left( {{x^3}} \right)$ is divisible by ${x^2} - x + 1$ then which of the following is not true
A. Both $g\left( x \right)$ and $h\left( x \right)$ are divisible by $x + 1$
B. $g\left( x \right)$ is divisible by $x + 1$
C. $h\left( x \right)$ is divisible by $x + 1$
D. The statement (A) is not true

Answer 1

First, we shall analyze the given information. It is given that the given equation $f\left( x \right) = g\left( {{x^3}} \right) + xh\left( {{x^3}} \right)$is divisible by ${x^2} - x + 1$.
Here, we need to calculate the roots ${x^2} - x + 1 = 0$.
And we shall apply the roots in place of $x$ in the given equation.
Formula to be used:
The quadratic formula of the form $a{x^2} + bx + c = 0$ is as follows.
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step by step answer:
The given equation is $f\left( x \right) = g\left( {{x^3}} \right) + xh\left( {{x^3}} \right)$.
It is given that the given equation $f\left( x \right) = g\left( {{x^3}} \right) + xh\left( {{x^3}} \right)$ is divisible by ${x^2} - x + 1$.
We shall calculate the roots of ${x^2} - x + 1 = 0$
Here,
$a = 1$,
$b = - 1$,
$c = 1$
Applying these values in the quadratic formula, we have
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
$x = \dfrac{{ - \left( { - 1} \right) \pm \sqrt {{{\left( { - 1} \right)}^2} - 4 \times 1 \times 1} }}{{2 \times 1}}$
$\Rightarrow x = \dfrac{{1 \pm \sqrt {1 - 4} }}{2}$
$\Rightarrow x = \dfrac{{1 \pm \sqrt { - 3} }}{2}$
$\Rightarrow x = \dfrac{{1 \pm i\sqrt 3 }}{2}$
Where $i$ is known as imaginary number.
We know that the value of the roots $\omega$ and ${\omega ^2}$ are$\dfrac{{ - 1 \pm i\sqrt 3 }}{2}$ .
Here, we got the roots as $x = \dfrac{{1 \pm i\sqrt 3 }}{2}$ which can also be written as $- \omega$ and $- {\omega ^2}$
Hence, the roots of ${x^2} - x + 1 = 0$ are $- \omega$ and $- {\omega ^2}$.
It is given that the given equation $f\left( x \right) = g\left( {{x^3}} \right) + xh\left( {{x^3}} \right)$ is divisible by ${x^2} - x + 1$.
Therefore, $- \omega$ and $- {\omega ^2}$are the factors of $f(x)$
So, we need to replace $x$ by $- \omega$ and $- {\omega ^2}$ in the given equation.
$f\left( { - \omega } \right) = g\left( { - {\omega ^3}} \right) - \omega h\left( { - {\omega ^3}} \right)$
$\Rightarrow f\left( { - \omega } \right) = g\left( { - {\omega ^3}} \right) - \omega h\left( { - {\omega ^3}} \right) = 0$
We know that the value of ${\omega ^3} = 1$
$\Rightarrow g\left( { - 1} \right) - \omega h\left( { - 1} \right) = 0$ ……….$\left( 1 \right)$
Similarly,
$f\left( { - {\omega ^2}} \right) = g\left( {{{\left( { - {\omega ^2}} \right)}^3}} \right) - {\omega ^2}h\left( {{{\left( { - {\omega ^2}} \right)}^3}} \right)$
$\Rightarrow f\left( { - {\omega ^2}} \right) = g\left( { - {\omega ^6}} \right) - {\omega ^2}h\left( { - {\omega ^6}} \right) = 0$
We know that the value of ${\omega ^6} = 1$
$\Rightarrow g\left( { - 1} \right) - {\omega ^2}h\left( { - 1} \right) = 0$ ……………..$\left( 2 \right)$
Now, we shall add $\left( 1 \right)$and $\left( 2 \right)$.
$\Rightarrow g\left( { - 1} \right) - \omega h\left( { - 1} \right) + g\left( { - 1} \right) - {\omega ^2}h\left( { - 1} \right) = 0$
$\Rightarrow 2g\left( { - 1} \right) - \left( {\omega + {\omega ^2}} \right)h\left( { - 1} \right) = 0$
We know that $\omega + {\omega ^2} = - 1$ $(1 + \omega + {\omega ^2} = 0)$
$\Rightarrow 2g\left( { - 1} \right) - \left( { - 1} \right)h\left( { - 1} \right) = 0$
$\Rightarrow 2g\left( { - 1} \right) + h\left( { - 1} \right) = 0$
$\Rightarrow h\left( { - 1} \right) = - 2g\left( { - 1} \right)$ …………..$\left( 3 \right)$
We need to substitute the above result in $\left( 1 \right)$

$$g\left( { - 1} \right) - \omega h\left( { - 1} \right) = 0 \\\ \Rightarrow g\left( { - 1} \right) - \omega \times - 2g\left( { - 1} \right) = 0 \\\$$

$\Rightarrow g\left( { - 1} \right) + 2\omega g\left( { - 1} \right) = 0$
$\Rightarrow \left( {1 + 2\omega } \right)g\left( { - 1} \right) = 0$
$\Rightarrow g\left( { - 1} \right) = 0$
Therefore, $\left( {x + 1} \right)$ is a factor of $g\left( x \right)$ . Hence, option B is correct.
Now, we shall substitute $g\left( { - 1} \right) = 0$ in $\left( 3 \right)$

$$h\left( { - 1} \right) = - 2g\left( { - 1} \right) \\\ \Rightarrow h\left( { - 1} \right) = - 2 \times 0 \\\$$

$\Rightarrow h\left( { - 1} \right) = 0$
Therefore, $\left( {x + 1} \right)$ is a factor of $h\left( x \right)$ . Hence, option C is true.
Also, option A is true.

So, the correct answer is “Option D”.

Note: We know that the value of the roots $\omega$ and ${\omega ^2}$ are $\dfrac{{ - 1 \pm i\sqrt 3 }}{2}$ . The sum of the cube root of unity is zero (i.e.) $1 + \omega + {\omega ^2} = 0$. Also, it is known that ${\omega ^3} = 1$. Generally, we express the square root of a negative number in terms of an imaginary number. So, here we replaced $i = - 1$