Begin by moving all of the related variables (and constants) next to each other. You can group these in parentheses to make it clear. This is allowed because of the associative rule for multiplication.

$\displaystyle (2 * 15 * 10)(x^{3})(y^{2})$

There are no variables to combine. We only need to combine the numerical coefficient. This gets you:

$\displaystyle (300)(x^{3})(y^{2})$ 

This is the same as:

$\displaystyle 300x^{3}y^{2}$

Begin by moving all of the related variables (and constants) next to each other. You can group these in parentheses to make it clear. This is allowed because of the associative rule for multiplication.

$\displaystyle (x * x* x^{3})(y * y)$

When multiplying variables of the same type, you add their exponents together. This gets you:

$\displaystyle (x^{1+1+3})(y^{1+1})$

 $\displaystyle (x^{5})(y^{2})$

This is the same as:

$\displaystyle x^{5}y^{2}$

Begin by moving all of the related variables (and constants) next to each other. You can group these in parentheses to make it clear. This is allowed because of the associative rule for multiplication.

$\displaystyle (a^{3} * a^{2})(x * x^{2} * x)(y^{4})$ 

When multiplying variables of the same type, you add their exponents together. This gets you:

$\displaystyle (a^{3+2})(x^{1+2+1})(y^4)$

 $\displaystyle (a^{5})(x^{4})(y^{4})$

This is the same as:

$\displaystyle a^{5}x^{4} y^{4}$

$\displaystyle xy(x^{2} + 2aby)$

Distribute the outside term into the parentheses. Multiply each term in parentheses by $\displaystyle xy$:

$\displaystyle (xy)(x^{2}) + (xy)(2aby)$

Now, for each member, move the similar variables into separate groups. You can do this because of the associative rule for multiplication:

 $\displaystyle (x*x^{2})(y) + (2)(a)(b)(x)(y*y)$

When multiplying variables of the same type, you add their exponents together. This gets you:

$\displaystyle (x^{1+2})(y)+(2)(a)(b)(x)(y^{1+1})$

$\displaystyle (x^{3})(y) + (2)(a)(b)(x)(y^2)$

This is the same as:

$\displaystyle x^{3}y + 2abxy^{2}$

$\displaystyle 2ab(5xy - 3ax + 4by)$

Distribute the outside term into the parentheses. Multiply each term in parentheses by $\displaystyle 2ab$:

 $\displaystyle (2ab)(5xy) - (2ab)(3ax)+ (2ab)(4by)$

Now, for each member, move the similar variables into separate groups. You can do this because of the associative rule for multiplication:

 $\displaystyle (2*5)(abxy) - (2*3)(a*a)(bx)+ (2*4)(a)(b*b)(y)$

When multiplying variables of the same type, you add their exponents together. This gets you:

$\displaystyle (10)(abxy)-(6)(a^{1+1})(bx)+(8)(a)(b^{1+1})(y)$

 $\displaystyle (10)(abxy) - (6)(a^{2})(bx)+ (2*4)(a)(b^{2})(y)$

This is the same as:

$\displaystyle 10abxy - 6a^{2}bx + 8ab^{2}y$

$\displaystyle 5bv(vx^{2} - 10bx - 5vb^{6})$

Distribute the outside term into the parentheses. Multiply each term in parentheses by $\displaystyle 5bv$:

$\displaystyle (5bv)(vx^{2}) - (5bv)(10bx) - (5bv)(5vb^{6})$ 

Now, for each member, move the similar variables into separate groups. You can do this because of the associative rule for multiplication:

$\displaystyle (5)(b)(v * v)(x^{2}) - (5 * 10)(b *b)(vx) - (5*5)(b *b^{6})(v*v)$

When multiplying variables of the same type, you add their exponents together. This gets you:

$\displaystyle (5)(b)(v^{1+1})(x^2)-(50)(b^{1+1})(vx)-(25)(b^{1+6})(v^{1+1})$

 $\displaystyle (5)(b)(v^{2})(x^{2}) - (50)(b^{2})(vx) - (25)(b^{7})(v^{2})$

This is the same as:

$\displaystyle 5bv^{2}x^{2} - 50b^{2}vx - 25b^{7}v^{2}$

To simplify, add the exponents:

$\displaystyle x^{2}* x^{5}=x^{7}$

Answer: $\displaystyle x^{7}$