Find the polynomial HELP can anyone solve these with work?!?

A) P(x)=(x-4i)(x+4i)(x-1)^2

=>

P(x)=(x^2+16)(x-1)^2

=>

P(x)=(x^2+16)(x^2-2x+1)

=>

p(x)=x^4-2x^3+17x^2-32x+16

B) P(x)=(x+1)(x+2i)(x-2i)

=>

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

=>

P(x)=x^3+x^2+4x+4

{note that i=sqr(-1) & i^2= -1}

OK, here you go sweetheart :)

1) We are given the roots x = 4i and x = 1 (multiplicity 2). Another root must be x = - 4i as roots always occur in conjugate pairs (by the fundamental theorem of algebra)!

So the roots are :

x₁ = - 4i

x₂ = 4i

x₃ = 1

x₄ = 1

This can be factored as :

f(x) = (x - 1)(x - 1)(x - 4i)(x + 4i)

Expanding this will you the required polynomial of degree 4, (a quartic!) :

f(x)

= (x - 1)²[(x² + 4ix - 4ix - 16i²)]

= (x - 1)²(x² + 16)

= (x² - 2x + 1)(x² + 16)

= x⁴ - 2x³ + 17x² - 32x + 16

Hence the required polynomial is :

f(x) = x⁴ - 2x³ + 17x² - 32x + 16

2) We are given the roots :

x₁ = - 1

x₂ = - 2i

x₃ = 2i

The polynomial can be factored as :

f(x) = (x +1)(x + 2i)(x - 2i)

Expanding this will give you the required polynomial of degree 3 (a cubic) :

f(x)

= (x + 1)[x² - 2ix + 2ix - 4i²]

= (x + 1)(x² + 4)

= x³ + x² + 4x + 4

Hence the required polynomial is :

f(x) = x³ + x² + 4x + 4

Hope this helps !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

P.S. (Don't forget to vote me best answer as being the first to thoroughly answer your question!) ☺

Assuming your polynomial has real coefficients, then any imaginary solutions will occur as complex conjugate pairs. If the coefficients are complex, than the solutions are not necessarily conjugate pairs. In this case, the coefficients are said to be integers so they are real, and the 4i zero is paired with a -4i zero.

The polynomial is:

P(x) = (x+4i)(x-4i)(x-1)²

= (x² + 16)(x² - 2x + 1)

= x⁴ + (-2 )x³ + (1+16 )x² +(-32 )x + 16

= x⁴ - 2x³ + 17x² - 32x + 16

A) P(x) = (x-1)^2(x^2+16)

B) f(x) = (x+1)(x^2+4)

If x = 4i is a zero, then x = -4i will also be a zero.

So (x^2 + 16) will be a factor.

Also (x-1)^2 will be a factor.

That gives you P(x) = (x^2 - 2x + 1)(x^2 + 16)

= x^4 - 2x^3 + 17x^2 - 32x + 16.

As the coefficient of x^4 is already 1, we're finished.