Question

Find a degree 3 polynomial with real coefficients having zeros 5 and 3i and a lead coefficient of 1. Write P in expanded form. Be sure to write the full equation, including P(x).

Answer 1

If a polynomial with real coefficients has a complex root $z$, then its conjugate $\bar{z}$ must be a root as well.

So, the three roots of the polynomial are $5$, $3i$ and its conjugate $-3i$

This leads to the factorization

$P(x) = (x-5)(x-3i)(x+3i) = (x-5)(x^2+9) = x^3+9x-5x^2-45$

So, your polynomial is

$P(x) = x^3-5x^2+9x-45$

Answer 2

The polynomial equation with real coefficients having zeros 5 and 3i and a lead coefficient of 1 is $P(x)=x^3-5x^2+9x-45$. This can be obtained by the factors method.

Polynomial equation:

• Polynomial equations are the one that forms with variables, coefficients, and exponents.
• The order of the polynomial is the highest degree of the variable in the equation.
• They are linear equations (degree 1), quadratic (degree 2), cubic (degree 3), and so on.
• we have standard form as $P(x)=a(x-x_1)(x-x_2)(x-x_3)....$

Finding the polynomial equation:

It is given that, the equation has zeros 5 and 3i

3i is an imaginary number its conjugate also be a zero for the equation

Here zeros are nothing but factors and the lead coefficient is 1

So,

$P(x)=1(x-5)(x-3i)(x+3i)$

From $(a-b)(a+b)=a^2+b^2$

$P(x)=1(x-5)(x^2-(3i)^2)$

$P(x)=1(x-5)(x^2+9i^2)$

Since $i^2=-1$

$P(x)=1(x-5)(x^2-9)$

$P(x)=x^3+9x-5x^2-45$

$P(x)=x^3-5x^2+9x-45$

Therefore, the required polynomial is $P(x)=x^3-5x^2+9x-45$ for the given values.

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