To simplify a polynomial, we have to do two things: 1) combine like terms, and 2) rearrange the terms so that they're written in descending order of exponent.

First, we combine like terms, which requires us to identify the terms that can be added or subtracted from each other. Like terms always have the same variable (with the same exponent) attached to them. For example, you can add 1 "x-squared" to 2 "x-squareds" and get 3 "x-squareds", but 1 "x-squared" plus an "x" can't be combined because they're not like terms.

Let's identify some like terms below.

$\displaystyle f(x)=\mathbf{-4x}+3x^{2}-7\mathbf{\:+\: 9x}-12x^{2}-5x^{4}$

Here you can see that -4x and 9x are like terms. When we combine (add) -4x and 9x, we get 5x. So let's write 5x instead:

$\displaystyle f(x)=5x+3x^{2}-7-12x^{2}-5x^{4}$

Let's do the same thing with the x-squared terms:

$\displaystyle f(x)=5x+\mathbf{3x^{2}}-7-\mathbf{12x^{2}}-5x^{4}$

$\displaystyle f(x)=5x-\mathbf{9x^{2}}-7-5x^{4}$

Now there are no like terms left. Our last step is to organize the terms so that x is written in descending power:

$\displaystyle f(x)=-5x^{4}-9x^{2}+5x-7$

$\displaystyle (3p^{2}+8)+(-p^{2}+5)$

This is not a FOIL problem, as we are adding rather than multiplying the terms in parenteses.

Add like terms to solve.

$\displaystyle 3p^{2}-p^{2}=2p^{2}$

$\displaystyle 8+5=13$

Combining these terms into an expression gives us our answer.

$\displaystyle 2p^{2}+13$

In simplifying this expression, be mindful of the order of operations (parenthical, division/multiplication, addition/subtraction).  

$\displaystyle 5x^{^{3}} - 14x^{2} - 10x + 3 - (4x+3) * x^{2} - (8-10x)$

Since operations invlovling parentheses occur first, distribute the factors into the parenthetical binomials. Note that the $\displaystyle x^2$ outside the first parenthetical binomial is treated as $\displaystyle -x^2$ since the parenthetical has a negative (minus) sign in front of it. Similarly, multiply the members of the expression in the second parenthetical by $\displaystyle -1$ because of the negative (minus) sign in front of it. Distributing these factors results in the following polynomial.

$\displaystyle 5x^3 - 14x^2 - 10x + 3 - 4x^3 - 3x^2 - 8 + 10x$

Now like terms can be added and subtracted. Arranging the members of the polynomial into groups of like terms can help with this. Be sure to retain any negative signs when rearranging the terms.

$\displaystyle 5x^3 - 4x^3 -14x^2 - 3x^2 - 10x + 10x + 3 - 8$

Adding and subtracting these terms results in the simplified expression below.

$\displaystyle x^3 - 17x^2 -5$

$\displaystyle \left (x^{2} + 2x + 7 \right ) \left (x^{2} + x - 6 \right )$

$\displaystyle = x^{2}\left (x^{2} + x - 6 \right ) + 2x \left (x^{2} + x - 6 \right ) + 7 \left (x^{2} + x - 6 \right )$

$\displaystyle = x^{2} \cdot x^{2} + x^{2} \cdot x - x^{2} \cdot 6 + 2x \cdot x^{2} + 2x \cdot x - 2x \cdot 6 + 7 \cdot x^{2} + 7 \cdot x - 7 \cdot 6$

$\displaystyle = x^{4} + x^{3} - 6x^{2} + 2x^{3} + 2x^{2} - 12x + 7x^{2} +7x - 42$

$\displaystyle = x^{4} + x^{3} + 2x^{3} - 6x^{2} + 2x^{2} + 7x^{2} - 12x +7x - 42$

$\displaystyle = x^{4} + 3 x^{3} +3x^{2} - 5x - 42$

$\displaystyle \small (3x+2)-(x+4)(x+1)$

Use FOIL to expand the monomials.

$\displaystyle (x+4)(x+1)$

$\displaystyle x^2+x+4x+4=x^2+5x+4$

Return this expansion to the original expression.

$\displaystyle \small (3x+2)-(x^2+5x+4)$

Distribute negative sign.

$\displaystyle \small (3x+2)-x^2-5x-4$

Combine like terms.

$\displaystyle \small -x^2+3x-5x+2-4$

$\displaystyle \small -x^2-2x-2$

$\displaystyle 18x^{2}+ 9x -3$

We can accomplish this division by re-writing the problem as a fraction.

$\displaystyle \frac{18x^2+9x-3}{9x}$

The denominator will distribute, allowing us to address each element separately.

$\displaystyle \frac{18x^{2}+ 9x -3}{9x}=\frac{18x^{2}}{9x}+\frac{ 9x}{9x}-\frac{3}{9x}$

Now we can cancel common factors to find our answer.

$\displaystyle (\frac{18}{9}* \frac{x^{2}}{x})+1-(\frac{3}{9}* \frac{1}{x})$

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

To subtract these two trinomials, you first need to flip the sign on every term in the second trinomial, since it is being subtrated:

$\displaystyle (2x^2+\frac{3}{4}x-5) - (x^2+x-10)$

$\displaystyle 2x^2+\frac{3}{4}x-5 - x^2-x+10$

Next you can combine like terms. You have two terms with $\displaystyle x^2$, two terms with $\displaystyle x$, and two terms with no variable:

$\displaystyle x^2-\frac{1}{4}x+5$

$\displaystyle \left (\frac{1}{3}x^{2}+\frac{1}{6}xy-\frac{5}{4}y^{2} \right )-\left (\frac{1}{9}x^{2}-\frac{4}{3}xy+y^{2} \right )$

Since we are only adding and subtracting (there is no multiplication or division), we can remove the parentheses.

$\displaystyle \frac{1}{3}x^{2}+\frac{1}{6}xy-\frac{5}{4}y^{2}-\left \frac{1}{9}x^{2}+\frac{4}{3}xy-y^{2} \right$

Regroup the expression so that like variables are together. Remember to carry positive and negative signs.

$\displaystyle (\frac{1}{3}x^{2}-\left \frac{1}{9}x^{2})+(\frac{1}{6}xy+\frac{4}{3}xy)-(\frac{5}{4}y^{2}-y^{2} \right)$

For all fractional terms, find the least common multiple in order to add and subtract the fractions.

$\displaystyle (\frac{3}{9}x^{2}-\left \frac{1}{9}x^{2})+(\frac{1}{6}xy+\frac{8}{6}xy)-(\frac{5}{4}y^{2}-\frac{4}{4}y^{2} \right)$

Combine like terms and simplify.

$\displaystyle \frac{2}{9}x^{2}+\frac{9}{6}xy-\frac{9}{4}y^{2}$

$\displaystyle \frac{2}{9}x^{2}+\frac{3}{2}xy-\frac{9}{4}y^{2}$

Set up this problem vertically like you would a normal multiplication problem without variables. Then, multiply the $\displaystyle -4$ term to each term in the trinomial. Next, multiply the $\displaystyle 3x$ term to each term in the trinomial (keep in mind your placeholder!).

Then combine the two, which yields:

$\displaystyle 12x^3-31x^2+47x-36$

$\displaystyle \frac{1}{2}(4x^2-5x+10)+\frac{1}{4}(7x^2-5x+12)$

First distribute the $\displaystyle \frac{1}{2}$:

$\displaystyle 2x^2-\frac{5}{2}x+5+\frac{1}{4}(7x^2-5x+12)$

Then distribute the $\displaystyle \frac{1}{4}$:

$\displaystyle 2x^2-\frac{5}{2}x+5+\frac{7}{4}x^2-\frac{5}{4}x+\frac{12}{4}$

Finally combine like terms:

$\displaystyle 2x^2-\frac{10}{4}x+5+1\frac{3}{4}x^2-\frac{5}{4}x+3$

$\displaystyle 3\frac{3}{4}x^2-\frac{15}{4}x+8$

$\displaystyle 3\frac{3}{4}x^2-3\frac{3}{4}x+8$