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

1 Answer

4 votes

Answer:


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

Explanation:

Given information:

  • Polynomial function with real coefficients.
  • Roots: 0, i and -5+7i

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

User Umar Abdullah
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