Use the distributive property to obtain each term:

$\displaystyle 4x^{2}y\left ( 2xy + 6xy^{2} + 3y^{{3}}\right ) =$ 

$\displaystyle 4x^{2}y(2xy) + 4x^{2}y(6xy^{2}) + 4x^{2}y(3y^{3}) =$

$\displaystyle 8x^{3}y^{2} + 24x^{3}y^{3} + 12x^{2}y^{4}$

\(\displaystyle Expand \; the \; following: \; 3 x^4 (2 x^2-4 x+3)\)\(\displaystyle Distribute \; 3x^4 \; over \; 2 x^2-4 x+3.\)\(\displaystyle 3 x^4 (2 x^2-4 x+3)\)
\(\displaystyle = 3 x^4\cdot 2 x^2 + 3 x^4 (-4 x)+3 x^4\cdot 3\)

\(\displaystyle = (3\times 2)x^4×x^2 + (3\times -4)x^4 x+(3\times 3)x^4\)

\(\displaystyle = 6x^4×x^2 + (-12)x^4 x+9x^4\)

\(\displaystyle = 6x^{4+2}+ (-12)x^{4+1}+9x^4\)

\(\displaystyle = 6x^6+ (-12)x^5+9x^4\)

\(\displaystyle = 6x^6-12x^5+9x^4\)
\(\displaystyle Rearrange \; terms: \;9 x^4-12 x^5+6 x^6\)

To multiply this expression, multiply the monomial times both terms in the binomial.

First we will multiply $\displaystyle 3x^2 * 4x^3$, which represents $\displaystyle 3*x*x * 4*x*x*x$. This is why we evalute this by multiplying $\displaystyle 3*4 = 12$ and adding the exponents for x to get $\displaystyle x^5$.

Now multiply $\displaystyle 3x^2 * -5 = -15x^2$.

Our answer is just $\displaystyle 12x^5 - 15x^2$.

To multiply a monomial by a polynomial, you simply multiply the monomial by each term in the polynomial. In this case that means that the solution is to multiply everything by $\displaystyle 3y$. The answer becomes $\displaystyle 3x^3y+6x^2y+12xy+24y$.

When similar bases are multiplied, their powers can be added.  Distribute the monomial through the polynomial in the parentheses.

$\displaystyle xy(x^2+x) = x^3y+x^2y$

The answer is: $\displaystyle x^3y+x^2y$

When multiplying a whole number by a polynomial, we simply multiply that number by whatever coefficient is present in front of the variables of the polynomial. We then maintain the variables in the simplified expression.

$\displaystyle 10 \times 5 = 50$

$\displaystyle 50xy$

Use the distributive property to multiply the monomial and polynomial.

$\displaystyle 3xy(2x^{2}y+3y^{2}-7)$

$\displaystyle (3xy)(2x^{2}y)+(3xy)(3y^{2})-7(3xy)$

$\displaystyle 6x^{3}y^{2}+9xy^{3}-21xy$

Multipying a monomial and trinomial boils down to distributing the monomial amongst all the parts of the trinomial as such:

$\displaystyle 5x(x^{2}-3x+1)=(5x\cdotx^{2})+(5x\cdot-3x)+(5x\cdot1)$

After some cleanup we get:

$\displaystyle 5x^{3}-15x^{2}+5x$

$\displaystyle 4r(3r^{2}+2r-12)$

All we need to do here is multiply every term within the polynomial $\displaystyle (3r^{2}+2r-12)$ by the monomial on the outside of the parentheses: $\displaystyle 4r$.

To do this we need to multiply every term by $\displaystyle 4$ and by $\displaystyle r$. Remember that when we multiply by a variable (in this case $\displaystyle r$), we need to add $\displaystyle 1$ to each of the exponents.

So this leaves us with $\displaystyle 12r^{3}+8r^{2}-48r$.

To divide this, we must pull out a common factor from the numerator and denominator.

The common factor from the numerator is only $\displaystyle x$.

The common factor from the denominator is $\displaystyle 2x$.

$\displaystyle \frac{x^2+x}{2x^2-2x} = \frac{x\cdot(x+1)}{2\cdot x\cdot (x-1)}$

The only term that will cancel is the $\displaystyle x$.  We cannot cancel the $\displaystyle x$ inside $\displaystyle x+1$ and $\displaystyle x-1$ terms because they are different entities of a quantity.

$\displaystyle \frac{x\cdot(x+1)}{2\cdot x\cdot (x-1)} = \frac{x+1}{2(x-1)} = \frac{x+1}{2x-2}$

The answer is:  $\displaystyle \frac{x+1}{2x-2}$