Remember to distribute the \(\displaystyle -5\) to each terms and the \(\displaystyle -4\) to each term then combine the like terms.

\(\displaystyle -5\cdot 8x-2y \cdot -5-4\cdot -6x -4\cdot -3y\)

\(\displaystyle =-40x+10y+24x+12y\)

\(\displaystyle =-16x+22y\)

Simplify the following expression by combining like terms:

\(\displaystyle 4x^6-5x^3+2x^6-5\)

Begin by looking for terms to combine. In this case, we only have 2 terms we can combine. Remember, we can only combine terms that have the same exponent and variable.

\(\displaystyle {\color{Green} 4x^6}-5x^3+{\color{Green} 2x^6}-5\)

In this case, the green colors are the only ones which can be combined:

\(\displaystyle 4x^6-5x^3+2x^6-5=6x^6-5x^3-5\)

So our answer is:

\(\displaystyle 6x^6-5x^3-5\)

To simplify this expression, you must remember to distribute that negative sign to all of the terms of the second parantheses. It should then look like this: \(\displaystyle 3x^2-2x+1-6x^2-11x+10\). Then, combine like terms. Your answer should be: \(\displaystyle -3x^2-13x+11\)

Simplify the following expression:

\(\displaystyle \frac{y^6(x-7)^3}{y^4z^9(x-7)^2}\)

So, we have a big fraction. Don't be intimidated, all we have to do is cancel some parts out and we'll be good.

Begin by looking for pieces that are in common to the top and bottom. I've highlighted them below:

\(\displaystyle \frac{{\color{Teal} y^6}{\color{DarkOrange} (x-7)^3}}{{\color{Teal} y^4}z^9{\color{DarkOrange} (x-7)^2}}\)

Now, the parts that are in common need to be simplified, but only with eachother. 

Starting with the y's, we simply subtract the bottom exponent from the top exponent.

Do the same with the part in parentheses.

Leave the z's alone because we only have one set of z's and it's on the bottom.

And we get the following:

\(\displaystyle \frac{y^2(x-7)}{z^9}\)

\(\displaystyle \frac{a^3c^4 (b^2a+b^2c)}{c^5b^5(a^2 -c^2)}+\frac{1}{b^2(a-c)}\)

It is often best to wrok one term at a time before adding. Looking at the first term, notice that we can immediately cancel the \(\displaystyle c^4\) in the numerator by rules of exponents. Also, we can factor out the \(\displaystyle b^2\) in the numerator and cancel it out with the \(\displaystyle b^5\) in the denominator, 

  \(\displaystyle =\frac{a^3(a+c)}{cb^3(a^2 -c^2)}+\frac{1}{b^2(a-c)}\)

Notice the difference of two squares \(\displaystyle \large a^2-c^2\) in the denominator; It can be factored as follows, 

\(\displaystyle a^2 - c^2 = (a+c)(a-c)\)

(It's recommended that you memorize the formula for the difference of two squares, you will see this often).  

Substituting this into the expression, 

\(\displaystyle \frac{a^3 (a+c)}{cb^3(a -c)(a+c)}+\frac{1}{b^2(a-c)}\)

Cancel \(\displaystyle (a+c)\) in the numerator and denominator, 

\(\displaystyle = \frac{a^3 }{cb^3(a -c)}+\frac{1}{b^2(a-c)}\)

Now we need to add the two terms. Multiply the second term above and below by \(\displaystyle cb\) so it has the same denominator as the first term, 

\(\displaystyle = \frac{a^3 }{cb^3(a -c)}+\frac{cb}{cb}\times\frac{1}{b^2(a-c)}\)

\(\displaystyle = \frac{a^3 }{cb^3(a -c)}+\frac{cb}{cb^3(a-c)}\)

Add the numerators now that they have a common denominator, 

\(\displaystyle = \frac{a^3+cb }{cb^3(a -c)}\)

To simplify, it can be easier to start by expanding the function:

\(\displaystyle f(t)=\frac{t\cdot t\cdot t\cdot t\cdot t\cdot t}{t\cdot t}\)

We can then cancel terms:

\(\displaystyle f(t)=t\cdot t\cdot t\cdot t\)

and collect them into a more simplified form:

\(\displaystyle f(t)=t^{4}\)

To start, we can extract a \(\displaystyle n^{2}\) from both terms in the numerator to give:

\(\displaystyle f(n)=\frac{n^{2}(n^{3}-1)}{n^{2}}\)

Then we cancel the \(\displaystyle n^{2}\) terms:

\(\displaystyle f(n)=n^{3}-1\)

First we need to follow the order of operations, and do the multiplication first.  When you multiply like variables together, you add their exponents:

\(\displaystyle 2x^{2} + x^{3}\)

At this point we can no longer add the two remaining terms because their exponents are different, so we're done.

When you multiply two terms with like variables you still multiply the constants together like normal.  In this case it would be:

\(\displaystyle 4 \times 3 = 12\)

The exponents would be added together for:

\(\displaystyle 2 + 3 = 5\)

Putting the constant term in front, and the exponent back where it belongs, we have:

\(\displaystyle 12x^{5}\)

\(\displaystyle 5x(5x+3)\)

Use the distributive property to simplify the expression. The distributive property states that to multiply a sum by a term, you first multiply each part of the sum by that term separately, and then add these terms together. It looks like this in practice:\(\displaystyle a(b+c)=(a*b)+(a*c)\)

This means we multiply 5x by both terms, and then add these together.

\(\displaystyle 5x*5x + 5x*3\)

Multiply the constant terms together, and the x's together to yield:

\(\displaystyle 25x^2 + 15x\)

This expression is now simplified!