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:  \(\displaystyle \frac{x+1}{2x-2}\)