DC CAPE Algebra II Polynomial, Rational, and Radical Relationships. Practice it free.

Cross-multiply to clear the fraction: $3(5x - 2y) = 2(x + 3y)$, so $15x - 6y = 2x + 6y$. Collect like terms to get $13x = 12y$. Divide both sides by $y$, then by $13$, to obtain $\dfrac{x}{y} = \dfrac{12}{13}\textsf{.}$

For $x \neq -2$, the factor $x+2$ cancels, giving $x-8 = 5$, so $x = 13\textsf{.}$

$C(4,1) = \dfrac{4!}{1! \cdot 3!} = \dfrac{24}{1 \cdot 6} = 4$.

Find two numbers with product $6$ and sum $-5$: $-2$ and $-3$. So $p(x)=(x-2)(x-3)$. After factoring, set each linear factor equal to zero: $(x-2)=0 \Rightarrow x=2$; $(x-3)=0 \Rightarrow x=3$. The zeros are $2$ and $3$.

$x^{3} + 125 = x^{3} + 5^{3} = (x + 5)(x^{2} - 5x + 25)\textsf{.}$

Factor: $\dfrac{(x+2)(x+3)}{(x+1)(x+2)}=\dfrac{x+3}{x+1}$ for $x \neq -1$ and $x \neq -2\textsf{.}$

$\sqrt{18}=3\sqrt{2}$, $\sqrt{50}=5\sqrt{2}$, and $\sqrt{32}=4\sqrt{2}$. Adding gives $3\sqrt{2}+5\sqrt{2}+4\sqrt{2}=12\sqrt{2}$.

Divide each exponent by $\Large 4$: $\Large a^{\tfrac{8}{4}} b^{\tfrac{12}{4}} = a^2 b^3$.