\(\displaystyle xy = (1+i)(1-i)\)

\(\displaystyle = 1-i+i-i^2\)

\(\displaystyle =1-i^2\)

\(\displaystyle \text{Because}\, i^2=-1\)

\(\displaystyle =1-(-1)\)

\(\displaystyle =2\)

The most important part of this problem is to remember the order of operations: PEMDAS

First: Perform any calculations that are within parentheses.

Second: Perform any calculations that are raised to an exponent.

Third: Working from left to right, perform any multiplications or divisions.

Fourth: Working from left to right, perform any additions or subtractions

For this problem:

First we do all of the calculations inside parentheses:  \(\displaystyle \left ( 8-2\right )=6\) and \(\displaystyle \left ( 6\cdot 0.5\right ) = 3\).

Therefore, the expression becomes \(\displaystyle x = \frac{6}{3} + 3\cdot 4\).Now working from left to right, we perform any multiplications and/or divisions: \(\displaystyle \frac{6}{3} = 2\) and \(\displaystyle 3\cdot 4 = 12\).

Therefore, the expression becomes \(\displaystyle x = 2 + 12\) and we simply add the remaining numbers to get \(\displaystyle x = 14\)

To solve this problem, we simply follow our order of operations, PEMDAS:

First: Perform any calculations that are within parentheses.

Second: Perform any calculations that are raised to an exponent.

Third: Working from left to right, perform any multiplications or divisions.

Fourth: Working from left to right, perform any additions or subtractions.

 First, we evaluate our parentheses: \(\displaystyle (4-12\cdot 0.25) = (4-3) = (1)\) and \(\displaystyle (6\cdot 3 - 2) = (18 -2) = 16\).

The original expression then becomes \(\displaystyle x = 32 \cdot \frac{1}{16} = \frac{32}{16} = 2\).

To find the prime factorization of 40, write 40 as a combination of its prime factors.

\(\displaystyle 40 = 2 \times 2 \times 2 \times 5 = 2^{3} \times 5\)

The distributive property is handy to help get rid of parentheses in expressions. The distributive property says you "distribute" the multiple to every term inside the parentheses. In symbols, the rule states that 

\(\displaystyle a(b+c)=ab+ac\)

So, using this rule, we get \(\displaystyle 4(3x^2+5x-2) =4*3x^2+4*5x-4*2) = 12x^2+20x-8\)

Thus we have our answer is \(\displaystyle 12x^2+20x-8\).

We are dividing the polynomial by a monomial. In essence, we are dividing each term of the polynomial by the monomial. First I like to re-write this expression as a fraction. So,

\(\displaystyle (24x^8-2x^6+12x^2)\div x^2 = \frac{24x^8-2x^6+12x^2}{x^2}\)

So now we see the three terms to be divided on top. We will divide each term by the monomial on the bottom. To show this better, we can rewrite the equation. \(\displaystyle \frac{24x^8-2x^6+12x^2}{x^2} = \frac{24x^8}{x^2}-\frac{2x^6}{x^2}+\frac{12x^2}{x^2}\)

Now we must remember the rule for dividing variable exponents. The rule is \(\displaystyle \frac{x^a}{x^b} = x^{a-b}\)So, we can use this rule and apply it to our expression above. Then,\(\displaystyle \frac{24x^8}{x^2}-\frac{2x^6}{x^2}+\frac{12x^2}{x^2} = 24x^{8-2}-2x^{6-2}+2x^{2-2}=24x^6-2x^4+12\)

The first two factors are the product of the sum and the difference of the same two terms, so we can use the difference of squares:

\(\displaystyle (x - 4)(x+ 4) (x -7)\)

\(\displaystyle = (x^{2} - 4^{2}) (x -7)\)

\(\displaystyle = (x^{2} - 16) (x -7)\)

Now use the FOIL method:

\(\displaystyle = x^{2}\cdot x - x^{2} \cdot 7 - 16 \cdot x + 16 \cdot7\)

\(\displaystyle = x^{3} - 7x^{2} - 16x + 112\)

When dividing, focus on the first digit in the dividend with the divisor. \(\displaystyle 5\) can go into \(\displaystyle 8\) only one time. So put the \(\displaystyle 1\) on top of the \(\displaystyle 8\) and the \(\displaystyle 5\) goes under the \(\displaystyle 8\). Then, take the difference which is \(\displaystyle 3\). Then bring down the next digit in the dividend which is \(\displaystyle 5\). Next, figure out if \(\displaystyle 5\) goes into \(\displaystyle 35\) which is \(\displaystyle 7\). \(\displaystyle 5\) times \(\displaystyle 7\) is \(\displaystyle 35\) which means we get a difference of zero and so \(\displaystyle 85\) divides evenly with \(\displaystyle 5\) to give us a final answer of \(\displaystyle 17\). 

When multiplying, you can draw out a grid.

Have three rows and five columns and these should create little boxes.

Count them up individually and you should get \(\displaystyle 15.\)

When dividing, you draw out \(\displaystyle 15\) circles.

Then circle \(\displaystyle 3\) circles and that would be one set.

Once most or all the circles are covered, count out the sets.

There should be \(\displaystyle 5\).