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Find a degree 3 polynomial with real coefficients having zeros 3 and 4 i and a lead coefficient of 1. Write P in expanded form. Be sure to write the full equation, including P ( x ) = .

User Dprice
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1 Answer

3 votes

Answer:


P(x)=x^3-3x^2+16x-48

Explanation:

Given information:

  • Polynomial function with real coefficients.
  • Zeros: 3 and 4i
  • Lead coefficient of 1.

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 with real coefficients, and 4i is a root of f(x)=0, then its complex conjugate -4i is also a root of P(x)=0.

Therefore, the polynomial in factored form is:


P(x)=1(x-3)(x-4i)(x-(-4i))


P(x)=(x-3)(x-4i)(x+4i)

Expand the polynomial:


P(x)=(x-3)(x^2+4ix-4ix-16i^2)


P(x)=(x-3)(x^2-16(-1))


P(x)=(x-3)(x^2+16)


P(x)=x^3+16x-3x^2-48


P(x)=x^3-3x^2+16x-48

User Kaadzia
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