You must multiply out the first set of parenthesis (distribute) and you get 4x – x². Then multiply out the second set and you get –3x + x². Combine like terms and you get x.

x(4 – x) – x(3 – x)

4x – x² – x(3 – x)

4x – x² – (3x – x²)

4x – x² – 3x + x² = x

Factor the numerator and denominator:

$\displaystyle \frac{3x^2+3x-6}{3x+6}=\frac{3(x+2)(x-1)}{3(x+2)}$

Cancel the factors that appear in both the numerator and the denominator:

$\displaystyle \frac{3(x+2)(x-1)}{3(x+2)}=\frac{x-1}{1}=x-1$

$\displaystyle p^{7}$ and $\displaystyle p^{4}$ cancel out, leaving $\displaystyle p^{3}$ in the numerator. 5 and 25 cancel out, leaving 5 in the denominator

$\displaystyle \frac{4x+6x^2}{2x}$

First, let us factor the numerator:

$\displaystyle 4x+6x^2=x(4+6x)=2x(2+3x)$

$\displaystyle \frac{2x(2+3x)}{2x}=2+3x$

First, mulitply the mononomial by the first term of the polynomial:

$\displaystyle \small 7n\times8n\ = 56n^2$

Second, multiply the monomial by the second term of the polynomial:

$\displaystyle \small 7n\times (-2)\ = -14n$

Add the terms together:

$\displaystyle \small 56n^2\ +\ (-14n)\ = 56n^2-14n$

Cross-cancel the coefficients by dividing both 15 and 25 by 5, and both 14 and 21 by 7:

$\displaystyle \small \small \small \small \frac{15x^{5}}{14y^{3}} \cdot \frac{21y^{2} }{25x^{3}} = \small \small \small \small \frac{3x^{5}}{2y^{3}} \cdot \frac{3y^{2} }{5x^{3}} = \frac{3x^{5} \cdot 3y^{2}}{5x^{3} \cdot 2y^{3}} = \frac{3 \cdot 3 \cdot x^{5} y^{2}}{5\cdot 2\cdot x^{3} y^{3}} = \frac{9 x^{5} y^{2}}{10 x^{3} y^{3}}$ 

Now use the quotient rule on the variables by subtracting exponents:

$\displaystyle \small \frac{9 x^{5} y^{2}}{10 x^{3} y^{3}} = \frac{9 x^{5-3}}{10 y^{3-2}} = \frac{9 x^{2}}{10 y^{1}} = \frac{9 x^{2}}{10 y}$

In this problem, you have two fractions being multiplied. You can first simplify the coefficients in the numerators and denominators. You can divide and cancel the 2 and 14 each by 2, and the 3 and 15 each by 3:

$\displaystyle \frac{2xy^2}{3yz^2} \times \frac{15y^2z}{14x^2y} = \frac{xy^2}{yz^2} \times \frac{5y^2z}{7x^2y}$

You can multiply the two numerators and two denominators, keeping in mind that when multiplying like variables with exponents, you simplify by adding the exponents together:

$\displaystyle \frac{xy^2}{yz^2} \times \frac{5y^2z}{7x^2y} = \frac{5xy^4z}{7x^2y^2z^2}$

Any variables that are both in the numerator and denominator can be simplified by subtracting the numerator's exponent by the denominator's exponent. If you end up with a negative exponent in the numerator, you can move the variable to the denominator to keep the exponent positive:

$\displaystyle \frac{5xy^4z}{7x^2y^2z^2} = \frac{5y^2}{7xz}$

To find the greatest common factor, we need to break each term into its prime factors:

$\displaystyle \small 2xy^3=2 \cdot x \cdot y \cdot y \cdot y$

$\displaystyle \small 14xy^2 = 2 \cdot 7 \cdot x \cdot y \cdot y$

$\displaystyle \small 9xy = 3 \cdot 3 \cdot x \cdot y$

Looking at which terms all three expressions have in common; thus, the GCF is $\displaystyle \small xy$. We then factor this out: $\displaystyle \small \small xy(2y^2 -14y +9)$. 

To expand, multiply 8x by both terms in the expression (3x + 7).

8x multiplied by 3x is 24x².

8x multiplied by 7 is 56x.

Therefore, 8x(3x + 7) = 24x² + 56x.

First, distribute –5 through the parentheses by multiplying both terms by –5.

$\displaystyle -5(x+15)-3x=-5x-75-3x$

Then, combine the like-termed variables (–5x and –3x).

$\displaystyle -5x-75-3x=-8x-75$