Question

find all real solutions of the polynomial equation. (enter your answers as a comma-separated list.)
$x^5 - x^4 - 3x^3 + 5x^2 - 2x = 0$
$x = $

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1. -/1 points

find a polynomial function with real coefficients that has the given zeros. (there are many correct answers.)
$6, 4 - 9i$
$f(x) = $

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1. -/1 points

find a polynomial function with real coefficients that has the given zeros. (there are many correct answers.)
$-5, -5, 1 + \sqrt{2}i$
$f(x) = $

Explanation:

(First Problem):

Step1: Factor out common $x$

$x(x^4 - x^3 - 3x^2 + 5x - 2) = 0$

Step2: Rational root test for quartic

Test $x=1$: $1 - 1 - 3 + 5 - 2 = 0$, so $(x-1)$ is a factor. Use polynomial division or synthetic division to get:
$x(x-1)(x^3 - 3x + 2) = 0$

Step3: Factor the cubic

Test $x=1$ again: $1 - 3 + 2 = 0$, so $(x-1)$ is a factor. Divide to get:
$x(x-1)^2(x^2 + x - 2) = 0$

Step4: Factor quadratic

$x(x-1)^2(x+2)(x-1) = x(x-1)^3(x+2) = 0$

Step5: Solve for real roots

Set each factor equal to 0:
$x=0$, $x-1=0 \implies x=1$, $x+2=0 \implies x=-2$

(Second Problem):

Step1: Use conjugate root theorem

Since $4-9i$ is a zero, $4+9i$ is also a zero.

Step2: Write linear factors

$(x-6)(x-(4-9i))(x-(4+9i))$

Step3: Multiply complex factors

First compute $(x-(4-9i))(x-(4+9i)) = (x-4)^2 - (9i)^2 = x^2 -8x +16 +81 = x^2 -8x +97$

Step4: Multiply with linear factor

$(x-6)(x^2 -8x +97) = x^3 -8x^2 +97x -6x^2 +48x -582$

Step5: Simplify polynomial

$x^3 -14x^2 +145x -582$

(Third Problem):

Step1: Use conjugate root theorem

Since $1+\sqrt{2}i$ is a zero, $1-\sqrt{2}i$ is also a zero.

Step2: Write linear factors

$(x+5)^2(x-(1+\sqrt{2}i))(x-(1-\sqrt{2}i))$

Step3: Multiply complex factors

$(x-(1+\sqrt{2}i))(x-(1-\sqrt{2}i))=(x-1)^2 - (\sqrt{2}i)^2 = x^2 -2x +1 +2 = x^2 -2x +3$

Step4: Expand $(x+5)^2$

$(x+5)^2 = x^2 +10x +25$

Step5: Multiply quadratics

$(x^2 +10x +25)(x^2 -2x +3) = x^4 -2x^3 +3x^2 +10x^3 -20x^2 +30x +25x^2 -50x +75$

Step6: Simplify polynomial

$x^4 +8x^2 +8x^3 -20x +75 = x^4 +8x^3 +8x^2 -20x +75$

Answer:

(First Problem):
$0, 1, -2$

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