If ω is the cube root of unity and x=ω−ω2−2 then the value of x4+3x3+2x2-11x-6 is equal to ____.

Question

If ω is the cube root of unity and x=ω−ω2−2 then the value of x4+3x3+2x2-11x-6 is equal to ____.

Accepted Answer

We are given that ωis the cube root of unity
We are also given that the value of x as x=ω−

ω^{2}

−2
Now, let us try to form an equation of xby squaring the terms in the above equation by rearranging the terms then we get
⇒x+2= ω−

ω^{2}

⇒

{(x + 2)}^{2}

=

{(ω - ω^{2})}^{2}

We know that the formula of algebra that is

{(a + b)}^{2}

=

a^{2}

+2ab+

b^{2}

{(a - b)}^{2}

=

a^{2}

-2ab+

b^{2}

By using this formula in the above equation then we get
⇒

x^{2} + 4 x + 4 = ω^{2} - 2 ω^{3} + ω^{4}

We know that the standard results for the cube root of unity as

ω^{3} = 1

1+ω+

ω^{2}

=0
By using these standard results in the above equation then we get
⇒

x^{2} + 4 x + 4 = ω^{2} - 2 + ω

⇒

x^{2} + 4 x + 4 = - 2 + 1

⇒

x^{2} + 4 x + 7

............equation(i)
We are asked to find the value of

x^{4} + 3 x^{3} + 2 x^{2} - 11 x - 6

Let us assume that the given expression as
⇒f(x)=

x^{4} + 3 x^{3} + 2 x^{2} - 11 x - 6

Now, let us divide the polynomial f(x)with g(x)=

x^{2} + 4 x + 7

by using the long division method

x^{2} + 4 x + 7

)

\underset{̲}{x^{4} + 3 x^{3} + 2 x^{2} - 11 x - 6}

We know that in the long division method we eliminate the first term of the dividend by multiplying the certain term for divisor so that the multiplied term will be the quotient
By using the above process let us multiply the divisor with

x^{2}

then we get

x^{2}

\underset{̲}{{x^{2} + 4 x + 7) x}^{4} + 3 x^{3} + 2 x^{2} - 11 x - 6 - (x^{4} + 4 x^{3} + 7 x^{2}) = (- x^{3} - 5 x^{2} - 11 x - 6)}

Here, we can see that we can continue the division further.
By using the same process of long division that is multiplying the divisor with (−x) then we get

x^{2}

-

x

\underset{̲}{{x^{2} + 4 x + 7) x}^{4} + 3 x^{3} + 2 x^{2} - 11 x - 6 - (x^{4} + 4 x^{3} + 7 x^{2}) = (- x^{3} - 5 x^{2} - 11 x - 6) = (- x^{3} - 4 x^{2} - 7 x) = (- x^{2} - 4 x - 6)}

Here, we can see that we can continue the division further.
By using the same process of long division that is multiplying the divisor with (−1) then we get

x^{2}

-

x

-1

\underset{̲}{{x^{2} + 4 x + 7) x}^{4} + 3 x^{3} + 2 x^{2} - 11 x - 6 - (x^{4} + 4 x^{3} + 7 x^{2}) = (- x^{3} - 5 x^{2} - 11 x - 6) = (- x^{3} - 4 x^{2} - 7 x) = (- x^{2} - 4 x - 6) = 1}

We know that the formula of division that is
Dividend=Divisor×Quotient+RemainderDividend=Divisor×Quotient+Remair
By using this formula to above division then we get
⇒