Question

Write the equation for a 3rd degree polynomial function that has roots at x = 1–3i and x = 2. The y‑intercept is (0,10). Write an equation for this function in factored form with real values.

Answer 1

Answer:

$\displaystyle P(x)=-\frac{1}{2}(x^2+2x+10)(x-2)$

Step-by-step explanation:

Factored Form Of Polynomials

If we know the roots of a polynomial as

$\alpha _1,\alpha _2,\alpha _3$

the polynomial can be expressed in factored form as

$a(x-\alpha _1)(x-\alpha _2)(x-\alpha _3)$

We are given two of the three roots of the polynomial:

$\alpha _1=1-3i$

$\alpha _2=2$

The other root must be the conjugate of the complex root:

$\alpha _3=1+3i$

Recall the product of two complex conjugate numbers is

$(1-3i)(1+3i)=1^2-(3i)^2=1+9=10$

The required polynomial is

$P(x)=a\left[ x-(1-3i)\right ]\left[ x-(1+3i)\right ](x-2)$

$P(x)=a(x^2+2x+10)(x-2)$

This is the factored form of the polynomial where only real numbers appear

We need to find the value of a, such as

$P(0)=10$

$P(0)=a(0^2+2(0)+10)(0-2)=10$

$-20a=10$

Thus the value of a is

$\displaystyle a=-\frac{1}{2}$

The expression of the required polynomial is

$\boxed{P(x)=-\frac{1}{2}(x^2+2x+10)(x-2)}$