Binomial expansion help

[Hrishi98]

. [7 marks]
Consider the polynomial

x^4+ax^3+bx^2+cx+d

 

, where a, b, c, d are real numbers
Given that 1 + i and 1 − 2i are zeros of the above polynomial, find the values of a, b, c and d.

 

I'm having tough time as this is very long in solving. Could anyone help out in doing this?

[Ossih]

. [7 marks]

Consider the polynomial

x^4+ax^3+bx^2+cx+d

 

, where a, b, c, d are real numbers

Given that 1 + i and 1 − 2i are zeros of the above polynomial, find the values of a, b, c and d.

 

I'm having tough time as this is very long in solving. Could anyone help out in doing this?

 

So if 1+i is a zero, that means 1-i is a zero as well right? so (x-1-i) and (x-1+i) are factors

Now, 1-2i is a zero so 1+2i is a zero as well, so (x-1+2i) and (x-1-2i) are factors

 

So now, the polynomial is (x-1-i)(x-1+i)(x-1+2i)(x-1-2i)

One way to expand them is to group them like this --> [(x-1)-i][(x-1)+i] [(x-1)+2i][(x-1)-2i]

Now you can expand the first two and last two using (a-b)(a+b) = a^2 - b^2

so this is

[(x-1)^2 -i^2] [(x-1)^2 - 4i^2]

= (x^2 - 2x + 1 + 1)(x^2 - 2x + 1 + 4)

= (x^2 - 2x + 2)(x^2 - 2x + 5)

now expand this out

= x^4 - 2x^3 + 5x^2 - 2x^3 +4x^2 - 10x + 2x^2 -4x + 10

= x^4 - 4x^3 + 11x^2 - 4x + 10

 

Hope this helped

[Hrishi98]

 

. [7 marks]

Consider the polynomial

x^4+ax^3+bx^2+cx+d

 

, where a, b, c, d are real numbers

Given that 1 + i and 1 − 2i are zeros of the above polynomial, find the values of a, b, c and d.

 

I'm having tough time as this is very long in solving. Could anyone help out in doing this?

 

So if 1+i is a zero, that means 1-i is a zero as well right? so (x-1-i) and (x-1+i) are factors

Now, 1-2i is a zero so 1+2i is a zero as well, so (x-1+2i) and (x-1-2i) are factors

 

So now, the polynomial is (x-1-i)(x-1+i)(x-1+2i)(x-1-2i)

One way to expand them is to group them like this --> [(x-1)-i][(x-1)+i] [(x-1)+2i][(x-1)-2i]

Now you can expand the first two and last two using (a-b)(a+b) = a^2 - b^2

so this is

[(x-1)^2 -i^2] [(x-1)^2 - 4i^2]

= (x^2 - 2x + 1 + 1)(x^2 - 2x + 1 + 4)

= (x^2 - 2x + 2)(x^2 - 2x + 5)

now expand this out

= x^4 - 2x^3 + 5x^2 - 2x^3 +4x^2 - 10x + 2x^2 -4x + 10

= x^4 - 4x^3 + 11x^2 - 4x + 10

 

Hope this helped

 

Thanks!!! I made a small mistake in the factors thing.