SOL Algebra II Expressions and Operations. Practice it free.

$\sqrt{27}=3\sqrt{3}$, $\sqrt{48}=4\sqrt{3}$, and $\sqrt{75}=5\sqrt{3}$. Adding gives $3\sqrt{3}+4\sqrt{3}+5\sqrt{3}=12\sqrt{3}$.

Let $u=x^2$. Then $u^2-10u+9=(u-1)(u-9)$, so $p(x)=(x^2-1)(x^2-9)=(x-1)(x+1)(x-3)(x+3)$. Set each linear factor equal to zero: $(x-1)=0 \Rightarrow x=1$; $(x+1)=0 \Rightarrow x=-1$; $(x-3)=0 \Rightarrow x=3$; $(x+3)=0 \Rightarrow x=-3$. The zeros are $-3$, $-1$, $1$, and $3$.

Multiply the coefficients: $3 \cdot 4 = 12\textsf{.}$ Add the exponents on $x$: $x^{2+1} = x^{3}\textsf{.}$ So the product is $12x^{3}\textsf{.}$

Distribute the minus sign: $(5x^2 + 3) - x^2 + 2x\textsf{.}$ Combine $x^2$ terms: $5x^2 - x^2 = 4x^2\textsf{.}$ The $x$ term is $2x\textsf{.}$ The constant stays $3\textsf{.}$ Result: $4x^2 + 2x + 3\textsf{.}$

Expand: $(3x - 1)(x + 2) = 3x^2 + 6x - x - 2 = 3x^2 + 5x - 2$.

Divide each exponent by $\Large 3$: $\Large u^{\tfrac{15}{3}} v^{\tfrac{9}{3}} w^{\tfrac{6}{3}} = u^5 v^3 w^2$.

The product is $75 \cdot 8 = 600$. Factor $600 = 100 \cdot 6$, where $100$ is the largest perfect-square factor. So $\sqrt{600} = \sqrt{100 \cdot 6} = 10\sqrt{6}$, and $6 = 2 \cdot 3$ is square-free.

$27^{-1/3} = \dfrac{1}{27^{1/3}} = \dfrac{1}{3}$.