Question

1. $3x^5 + 54x = 33x^3$ 4. $x^3 - 4x^3 = x - 4$

Explanation:

Problem 3: $3x^5 + 54x = 33x^3$

Step1: Rearrange all terms to left

$3x^5 - 33x^3 + 54x = 0$

Step2: Factor out common term $3x$

$3x(x^4 - 11x^2 + 18) = 0$

Step3: Factor quartic as quadratic

Let $u=x^2$, so $u^2 -11u +18=0$. Factor: $(u-2)(u-9)=0$. Substitute back $u=x^2$: $3x(x^2 - 2)(x^2 - 9) = 0$

Step4: Factor difference of squares

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

Step5: Solve for $x$

Set each factor to 0:
$3x=0 \implies x=0$
$x-\sqrt{2}=0 \implies x=\sqrt{2}$
$x+\sqrt{2}=0 \implies x=-\sqrt{2}$
$x-3=0 \implies x=3$
$x+3=0 \implies x=-3$

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Problem 4: $x^3 - 4x^3 = x - 4$

Step1: Combine like terms on left

$-3x^3 = x - 4$

Step2: Rearrange all terms to left

$-3x^3 - x + 4 = 0$

Step3: Multiply by -1 to simplify

$3x^3 + x - 4 = 0$

Step4: Factor by rational root test

Test $x=1$: $3(1)^3 +1 -4=0$, so $(x-1)$ is a factor. Use polynomial division or synthetic division to get: $(x-1)(3x^2 + 3x + 4) = 0$

Step5: Solve linear and quadratic factors

$x-1=0 \implies x=1$
For $3x^2 +3x +4=0$, use quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where $a=3, b=3, c=4$:
$x=\frac{-3\pm\sqrt{9-48}}{6}=\frac{-3\pm\sqrt{-39}}{6}=\frac{-3\pm i\sqrt{39}}{6}$

Answer:

Problem 3 solutions: $\boldsymbol{x=-3,\ x=-\sqrt{2},\ x=0,\ x=\sqrt{2},\ x=3}$
Problem 4 solutions: $\boldsymbol{x=1,\ x=\frac{-3+i\sqrt{39}}{6},\ x=\frac{-3-i\sqrt{39}}{6}}$