How to multiply the binomials $(2x^3 - x)\left(\sqrt{x} + \frac{2}{x}\right)$ I am sorry if the numbers are not formatted, I have searched but found nothing on how. I am trying to multiply $$(2x^3 - x)\left(\sqrt{x} + \frac {2}{x}\right)$$ together and I arrive at a different answer that does not match the one given. The approach I take is to assign the numbers powers then multiply them by the lattice method. So $$(2x^3) (X^{\frac {1}{2}}) + (2x^3) (2x^{-1}) + (x^{\frac {1}{2}}) (-x)) + (2x^{-1}) (-x)$$. I am not asking for a solution but rather an explanation on how to go about multiplying the terms with negative and fractional exponents.
 A: We distribute just as we would distribute $$(a + b)(c + d) = a(c+d) + b(c + d) = ac + ad + bc + bd.$$

$$(2x^3 - x)(\color{blue}{\sqrt x + 2x^{-1}}) = 2x^3(\color{blue}{\sqrt x + 2x^{-1}}) - x(\color{blue}{\sqrt x + 2x^{-1}})\\ =2x^3\sqrt x + 4x^3x^{-1} - x\sqrt x - 2xx^{-1}\\ = 2x^{6/2}x^{1/2} + \frac {4x^3}{x} - x^{2/2}x^{1/2} - \frac {2x}{x}\\
= 2x^{6/2 + 1/2} + 4x^2 - x^{2/2 + 1/2} - 2 \\
= 2x^{7/2} + 4x^2 - x^{3/2} - 2$$
We are using the fact that $$a^ba^c = a^{b+c}$$
This holds for negative exponents as well. So when we approach this simply by looking at products of terms with the same base, each raised to an exponent, we can see that $4x^3x^{-1} = 4x^{3+ (-1)} = 4x^2$.
Note that this is consistent with what I wrote, e.g., $$4x^3x^{-1} = \frac{4x^3}{x}.$$ We can see that we can cancel the common factor of $x$ in the numerator and denominator: $\dfrac{4x^3}x = 4x^2$ and in doing so, we are, essentially, subtracting the exponent in the denominator from the exponent in the numerator. We can generalize this into a handy "rule":
$$\frac {a^b}{a^c} = a^{b-c}$$
A: Terms with negative and fractional exponents are dealt with in very much the same way as natural exponents.
you have


*

*$x^a*x^b=x^{a*b}$

*$x^{-a}=\frac{1}{x^a}$


Sometimes fractional exponents are written as roots. For example the exponent $\frac{1}{2}$ is more commonly known as the square root: $x^{\frac{1}{2}}=\sqrt{x}$ But this works equally for any fraction as exponent, so $x^{\frac{1}{3}}$ would be the third root, i.e. the (positive) solution (in $y$) to the equation $y^3=x$ and so on.
A: Try this:
Put $u^2=x$,   (i.e $u=x^{1/2})$:
$$\begin{align}
(2x^3-x)(\sqrt{x}+\frac 2x)&=(2u^6-u^2)(u+\frac2{u^2})\\
&=2u^7+4u^4-u^3-2\\
&=2x^{7/2}+4x^2-x^{3/2}-2\qquad \blacksquare\end{align}$$
