Question

this is a 2-page document
directions: solve each equation by factoring. simplify all irrational and complex solutions.

1. $12x^3 - 3x^2 = 0$

$3x^2(4x - 1)= 0$
$3x^2=0 \
ightarrow x^2=0 \
ightarrow x=0$
$4x - 1=0 \
ightarrow 4x=1 \
ightarrow x=\frac{1}{4}$

1. $x^4 - 64 = 0$
2. $2x^4 = 24x^2$
3. $x^3 + 512 = 0$
4. $8x^3 - 125 = 0$
5. $81x^4 = 3x$

Explanation:

Problem 2: $x^4 - 64 = 0$

Step1: Factor as difference of squares

$x^4 - 64 = (x^2 - 8)(x^2 + 8) = 0$

Step2: Factor first quadratic further

$x^2 - 8 = (x - 2\sqrt{2})(x + 2\sqrt{2}) = 0$

Step3: Solve each factor for $x$

$x - 2\sqrt{2}=0 \Rightarrow x=2\sqrt{2}$; $x + 2\sqrt{2}=0 \Rightarrow x=-2\sqrt{2}$; $x^2 + 8=0 \Rightarrow x^2=-8 \Rightarrow x=\pm2i\sqrt{2}$

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Problem 3: $2x^4 = 24x^2$

Step1: Rearrange to standard form

$2x^4 - 24x^2 = 0$

Step2: Factor out GCF

$2x^2(x^2 - 12) = 0$

Step3: Factor quadratic term

$x^2 - 12 = (x - 2\sqrt{3})(x + 2\sqrt{3}) = 0$

Step4: Solve each factor for $x$

$2x^2=0 \Rightarrow x=0$; $x - 2\sqrt{3}=0 \Rightarrow x=2\sqrt{3}$; $x + 2\sqrt{3}=0 \Rightarrow x=-2\sqrt{3}$

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Problem 4: $x^3 + 512 = 0$

Step1: Factor sum of cubes

$x^3 + 8^3 = (x + 8)(x^2 - 8x + 64) = 0$

Step2: Solve linear factor

$x + 8 = 0 \Rightarrow x=-8$

Step3: Solve quadratic via quadratic formula

For $x^2 - 8x + 64=0$, $x=\frac{8\pm\sqrt{64-256}}{2}=\frac{8\pm\sqrt{-192}}{2}=4\pm4i\sqrt{3}$

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Problem 5: $8x^3 - 125 = 0$

Step1: Factor difference of cubes

$(2x)^3 - 5^3 = (2x - 5)(4x^2 + 10x + 25) = 0$

Step2: Solve linear factor

$2x - 5 = 0 \Rightarrow x=\frac{5}{2}$

Step3: Solve quadratic via quadratic formula

For $4x^2 + 10x + 25=0$, $x=\frac{-10\pm\sqrt{100-400}}{8}=\frac{-10\pm\sqrt{-300}}{8}=-\frac{5}{4}\pm\frac{5i\sqrt{3}}{4}$

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Problem 6: $81x^4 = 3x$

Step1: Rearrange to standard form

$81x^4 - 3x = 0$

Step2: Factor out GCF

$3x(27x^3 - 1) = 0$

Step3: Factor difference of cubes

$27x^3 - 1 = (3x - 1)(9x^2 + 3x + 1) = 0$

Step4: Solve linear factors

$3x=0 \Rightarrow x=0$; $3x - 1=0 \Rightarrow x=\frac{1}{3}$

Step5: Solve quadratic via quadratic formula

For $9x^2 + 3x + 1=0$, $x=\frac{-3\pm\sqrt{9-36}}{18}=\frac{-3\pm\sqrt{-27}}{18}=-\frac{1}{6}\pm\frac{i\sqrt{3}}{6}$

Answer:

1. $x=0$, $x=\frac{1}{4}$
2. $x=2\sqrt{2}$, $x=-2\sqrt{2}$, $x=2i\sqrt{2}$, $x=-2i\sqrt{2}$
3. $x=0$, $x=2\sqrt{3}$, $x=-2\sqrt{3}$
4. $x=-8$, $x=4+4i\sqrt{3}$, $x=4-4i\sqrt{3}$
5. $x=\frac{5}{2}$, $x=-\frac{5}{4}+\frac{5i\sqrt{3}}{4}$, $x=-\frac{5}{4}-\frac{5i\sqrt{3}}{4}$
6. $x=0$, $x=\frac{1}{3}$, $x=-\frac{1}{6}+\frac{i\sqrt{3}}{6}$, $x=-\frac{1}{6}-\frac{i\sqrt{3}}{6}$