Question

1. $x^4 - 81 = 0$

$(x^2 + 3)(x^2 - 3x + 9)$

1. $10x^3 + 12x^2 + 2x = 0$

1. $8x^3 - 125 = 0$

Explanation:

Problem 8: $x^4 - 81 = 0$

Step1: Rewrite as difference of squares

$x^4 - 81 = (x^2)^2 - 9^2 = (x^2 - 9)(x^2 + 9)$

Step2: Factor $x^2-9$ further

$x^2 - 9 = (x-3)(x+3)$, so:
$(x-3)(x+3)(x^2 + 9) = 0$

Step3: Solve each factor = 0

1. $x-3=0 \implies x=3$
2. $x+3=0 \implies x=-3$
3. $x^2+9=0 \implies x^2=-9 \implies x=\pm 3i$

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Problem 10: $10x^3 + 12x^2 + 2x = 0$

Step1: Factor out GCF $2x$

$2x(5x^2 + 6x + 1) = 0$

Step2: Factor quadratic trinomial

$5x^2 + 6x + 1 = (5x+1)(x+1)$, so:
$2x(5x+1)(x+1) = 0$

Step3: Solve each factor = 0

1. $2x=0 \implies x=0$
2. $5x+1=0 \implies x=-\frac{1}{5}$
3. $x+1=0 \implies x=-1$

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Problem 12: $8x^3 - 125 = 0$

Step1: Rewrite as difference of cubes

$8x^3 - 125 = (2x)^3 - 5^3$

Step2: Apply difference of cubes formula

$a^3 - b^3=(a-b)(a^2+ab+b^2)$, so:
$(2x-5)(4x^2 + 10x + 25) = 0$

Step3: Solve each factor = 0

1. $2x-5=0 \implies x=\frac{5}{2}$
2. For $4x^2+10x+25=0$, use quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$:

$x=\frac{-10\pm\sqrt{100-400}}{8}=\frac{-10\pm\sqrt{-300}}{8}=\frac{-10\pm10i\sqrt{3}}{8}=\frac{-5\pm5i\sqrt{3}}{4}$

Answer:

Problem 8: $x=3$, $x=-3$, $x=3i$, $x=-3i$
Problem 10: $x=0$, $x=-1$, $x=-\frac{1}{5}$
Problem 12: $x=\frac{5}{2}$, $x=\frac{-5+5i\sqrt{3}}{4}$, $x=\frac{-5-5i\sqrt{3}}{4}$