Question

question 20
find all complex solutions (real and non - real). answer exactly
$x^3 - x^2 + x + 39 = 0$
real: $x = $
non - real: $x = $

question 21
construct a polynomial function with real coefficients and of least degree possible using the given information.
real zeros: $2, - 3, 3$ and contains the point $(-2, 4)$.
$f(x) = $

question 22
solve the quadratic equation by using the quadratic formula. separate your answers with commas, if necessary.
$4x^2 + 20x + 41 = 0$

Explanation:

Question 20

Step1: Find real root via Rational Root Theorem

Test possible rational roots $\pm1,\pm3,\pm13,\pm39$. Substitute $x=-3$:
$$(-3)^3 - (-3)^2 + (-3) + 39 = -27 -9 -3 +39 = 0$$
So $x=-3$ is a real root, and $(x+3)$ is a factor.

Step2: Factor polynomial via polynomial division

Divide $x^3 - x^2 + x + 39$ by $(x+3)$:
$$x^3 - x^2 + x + 39 = (x+3)(x^2 -4x +13)$$

Step3: Solve quadratic for non-real roots

Use quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ for $x^2-4x+13=0$, where $a=1,b=-4,c=13$:
$$x=\frac{4\pm\sqrt{(-4)^2-4(1)(13)}}{2(1)}=\frac{4\pm\sqrt{16-52}}{2}=\frac{4\pm\sqrt{-36}}{2}=\frac{4\pm6i}{2}=2\pm3i$$

Step1: Write polynomial from zeros

For real zeros $2,-3,3$, the polynomial is:
$$f(x)=a(x-2)(x+3)(x-3)$$
Simplify $(x+3)(x-3)=x^2-9$, so:
$$f(x)=a(x-2)(x^2-9)$$

Step2: Solve for leading coefficient $a$

Substitute point $(-2,4)$:
$$4=a(-2-2)((-2)^2-9)$$
$$4=a(-4)(4-9)$$
$$4=a(-4)(-5)$$
$$4=20a$$
$$a=\frac{4}{20}=\frac{1}{5}$$

Step3: Expand to final form

$$f(x)=\frac{1}{5}(x-2)(x^2-9)=\frac{1}{5}(x^3-2x^2-9x+18)=\frac{1}{5}x^3-\frac{2}{5}x^2-\frac{9}{5}x+\frac{18}{5}$$

Step1: Identify quadratic coefficients

For $4x^2+20x+41=0$, $a=4,b=20,c=41$

Step2: Apply quadratic formula

Use $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$:
$$x=\frac{-20\pm\sqrt{20^2-4(4)(41)}}{2(4)}$$

Step3: Simplify discriminant and result

$$x=\frac{-20\pm\sqrt{400-656}}{8}=\frac{-20\pm\sqrt{-256}}{8}=\frac{-20\pm16i}{8}=-\frac{5}{2}\pm2i$$

Answer:

Real: $x = -3$
Non-real: $x = 2 \pm 3i$

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Question 21