Question

if

$f(x) = g( {x}^{3}) + x.h( {x}^{3}) \: is \: divisible \: by \: {x}^{2} + x + 1$


then show that g(x) and h(x) are divisible by x - 1.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm \: f(x) = g( {x}^{3}) + x.h( {x}^{3}) \: is \: divisible \: by \: {x}^{2} + x + 1$

Now, Consider

$\rm \: {x}^{2} + x + 1 = 0$

Its a quadratic equation, so to find the solution, we use Quadratic Formula.

Thus,

$\rm \: x = \dfrac{ - 1 \: \pm \: \sqrt{ {1}^{2} - 4 \times 1 \times 1} }{2 \times 1}$

$\rm \: x = \dfrac{ - 1 \: \pm \: \sqrt{ 1 - 4} }{2}$

$\rm \: x = \dfrac{ - 1 \: \pm \: \sqrt{- 3} }{2}$

$\rm \: x = \dfrac{ - 1 \: \pm \: i \sqrt{3} }{2}$

$\rm\implies \:\rm \: x = \dfrac{ - 1 \: + \: i \sqrt{3} }{2} \: \: or \: \: \dfrac{ - 1 \: - \: i \sqrt{3} }{2}$

$\rm\implies \:\rm \: x = \omega \: \: or \: \: {\omega}^{2}$

where,

$\omega = \dfrac{ - 1 \: + \: i \sqrt{3} }{2}$

and

$\rm \: {\omega}^{2} = \dfrac{ - 1 \: - \: i \sqrt{3} }{2}$

So, it means

$\rm \: f(x) = g( {x}^{3}) + x.h( {x}^{3}) \: is \: divisible \: by \: (x - \omega) \: and \: {(x - \omega}^{2})$

Now, We know

Factor Theorem :- Factor theorem states that if a polynomial f(x) is divisible by x - a, then f(a) = 0.

So, using this Factor theorem, we have

$\rm \: f(\omega) = 0 \: \: and \: \: f( {\omega}^{2}) = 0$

So,

$\rm\implies \:g( {\omega}^{3}) + \omega.h( {\omega}^{3}) = 0$

We know

$\boxed{\tt{ \: \: {\omega}^{3} \: = \: 1 \: \: }}$

So, using this, we get

$\rm\implies \:g(1) + \omega.h(1) = 0 - - - (1)$

Also,

$\rm \: f( {\omega}^{2}) = 0$

$\rm\implies \:g( {\omega}^{6}) + {\omega}^{2} .h( {\omega}^{6}) = 0$

We know,

$\boxed{\tt{ \: \: {\omega}^{6} \: = {( {\omega}^{3}) }^{2} = \: 1 \: \: }}$

So, using this, we get

$\rm\implies \:g(1) + {\omega}^{2} .h(1) = 0 - - - - (2)$

On Subtracting equation (2) from equation (1), we get

$\rm\implies \:\rm \: h(1)[\omega - {\omega}^{2}] = 0$

As

$\rm \: \omega - {\omega}^{2} \: \ne \: 0$

$\rm\implies \:h(1) = 0 - - - (3)$

On substituting h(1) = 0 in equation (1), we get

$\rm\implies \:g(1) = 0 - - - (4)$

$\rm\implies \:g(1) = h(1) = 0$

So, By Factor Theorem,

$\rm\implies \:g(x) \: and \: h(x) \: both \: are \: divisible \: by \: x - 1$

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ADDITIONAL INFORMATION

SHORT CUT TRICK TO FIND ARGUMENT OF COMPLEX NUMBER

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Complex \: number & \bf arg(z) \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf x + iy & \sf  {tan}^{ - 1}\bigg |\dfrac{y}{x} \bigg|   \\ \\ \sf  - x + iy & \sf \pi - {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \\ \\ \sf  - x - iy & \sf  - \pi + {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \\ \\ \sf x - iy & \sf  - {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \end{array}} \\ \end{gathered}$