Question

find a degree 3 polynomial with real coefficients having zeros 3 and $2-3i$ and a lead coefficient of 1. write $p$ in expanded form.
$p(x) = \square$
question help: video read written example

Explanation:

Step1: Identify all zeros

For real coefficients, complex zeros come in conjugate pairs. If $2-3i$ is a zero, then $2+3i$ is also a zero. The zeros are $3$, $2-3i$, $2+3i$.

Step2: Write factored form

Using the factor theorem, the polynomial is:
$P(x)=(x-3)(x-(2-3i))(x-(2+3i))$

Step3: Simplify complex factors first

Group the complex factors:
$(x-(2-3i))(x-(2+3i)) = [(x-2)+3i][(x-2)-3i]$
Use the difference of squares: $(a+b)(a-b)=a^2-b^2$, where $a=x-2$, $b=3i$:
$(x-2)^2-(3i)^2 = x^2-4x+4 - 9i^2$
Since $i^2=-1$, substitute:
$x^2-4x+4 + 9 = x^2-4x+13$

Step4: Multiply by remaining factor

Multiply $(x-3)$ with $x^2-4x+13$:
$P(x)=(x-3)(x^2-4x+13)$
$=x(x^2-4x+13)-3(x^2-4x+13)$
$=x^3-4x^2+13x-3x^2+12x-39$

Step5: Combine like terms

Combine $x^2$ terms: $-4x^2-3x^2=-7x^2$
Combine $x$ terms: $13x+12x=25x$

Answer:

$x^3 - 7x^2 + 25x - 39$