Find a polynomial (lowest degree) with 0, i and -5+7 as its roots (zeros). Answer in expanded form.

Question

asked Jul 10, 2024 204k views

1 Answer

Umar Abdullah · Answer 1 · 2024-07-15T18:15:41+0000

Answer:

P(x)=x^5+10x^4+75x^3+10x^2+74x

Explanation:

Given information:

For any complex number
z=a+bi , the complex conjugate of the number is defined as
z^*=a-bi .

If f(z) is a polynomial with real coefficients, and z₁ is a root of f(z)=0, then its complex conjugate z₁* is also a root of f(z)=0.

Therefore, if P(x) is a polynomial, and i is a root of f(x)=0, then its complex conjugate -i is also a root of P(x)=0.

Similarly, if P(x) is a polynomial, and -5+7i is a root of f(x)=0, then its complex conjugate -5-7i is also a root of P(x)=0.

Therefore, the polynomial in factored form is:

P(x)=x(x-i)(x-(-i))(x-(-5+7i))(x-(-5-7i))

P(x)=x(x-i)(x+i)(x+5-7i)(x+5+7i)

Expand the polynomial:

P(x)=x(x^2-i^2)(x^2+10x+25-49i^2)

P(x)=x(x^2-(-1))(x^2+10x+25-49(-1))

P(x)=x(x^2+1)(x^2+10x+74)

P(x)=x(x^4+10x^3+75x^2+10x+74)

P(x)=x^5+10x^4+75x^3+10x^2+74x