Question

directions: solve each polynomial.

1. $x^3 - 5x^2 + 14x - 24 = 0$, zero at $x = 3$
2. $2x^3 - 8x^2 - x + 9 = 0$, zero at $x = -1$
3. $x^3 + x^2 - 7x - 15 = 0$, zero at $x = 3$
4. $x^3 - x + 60 = 0$, zero at $x = -4$
5. $x^3 + 8 = 6x - 2x^2$, zero at $x = -4$
6. $x^4 + 2x^3 = 8x + 16$, zero at $x = -2$

Explanation:

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Problem 1: $x^3 - 5x^2 + 14x - 24 = 0$, zero at $x=3$

Step1: Use polynomial division

Divide $x^3 - 5x^2 + 14x - 24$ by $(x-3)$:
$\frac{x^3 - 5x^2 + 14x - 24}{x-3} = x^2 - 2x + 8$

Step2: Solve quadratic equation

Set $x^2 - 2x + 8 = 0$. Use quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where $a=1,b=-2,c=8$:
$x=\frac{2\pm\sqrt{4-32}}{2}=\frac{2\pm\sqrt{-28}}{2}=1\pm i\sqrt{7}$

Answer:

(1):
$x=3$, $x=1+i\sqrt{7}$, $x=1-i\sqrt{7}$

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Problem 2: $2x^3 - 8x^2 - x + 9 = 0$, zero at $x=-1$

Step1: Use polynomial division

Divide $2x^3 - 8x^2 - x + 9$ by $(x+1)$:
$\frac{2x^3 - 8x^2 - x + 9}{x+1} = 2x^2 - 10x + 9$

Step2: Solve quadratic equation

Set $2x^2 - 10x + 9 = 0$. Use quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where $a=2,b=-10,c=9$:
$x=\frac{10\pm\sqrt{100-72}}{4}=\frac{10\pm\sqrt{28}}{4}=\frac{5\pm\sqrt{7}}{2}$