This week in math 10 we mostly reviewed for our polynomials test on Friday. Although it was review one thing I learned this week was how to factor difference of squares.

Every difference of squares can be factored as:  = (a + b)(a – b), due to the + and – having the same value the middle term gets cancelled out.

When factoring a difference of squares there are 3 things you need to make sure of

1.You need to make sure the equation is a binomial

2.There is a negative sign ( – )

3.Both terms are perfect squares

Example:

Further Examples:

 

How to Factor:

1. Decide if the terms have anything in common or have a GCF. If so factor out the GCF
2. Since every difference of squares can be factored as  = (a + b)(a – b), to get this you just need to find what numbers squared will produce the results that you want. In this case you want to find what number squared will produce   [x (x)] and what numbers squared will produce 16 [4 (4)]
3. Check to see if the remaining factors can be factored even further

 

 

This week in math 10 we talked about the different ways of factoring polynomials. One thing that I learned this week was how to factor simple trinomials.

When factoring a simple trinomial there will always be a pattern in the equation ( + bx + c ). In order to factor the trinomial you need to find two integers that when multiplied together will be of equal value with c and when added together will be of equal value of b, in other words you list the multiples of the constant that represents c and pick the integers from those numbers. If there are no two set of integers that make this possible then the equation can’t be factored. You must also be sure to check whether the product is a positive or negative number as well as the sign in front of he sum since it will effect if the integers are positive or negative.

Example:

However, if the  has a coefficient greater than one the equation is no longer simple. In that case you need to check and see if the terms have a GCF that they can be divided by, if not then you just have to factor them out. In this case there is a GCF that the terms can be divided by. After dividing the terms by the GCF, you write the quotient inside brackets and the GCF as a coefficient outside of the brackets. Lastly, you factor as normally would factor a trinomial.

Example:

This week we learned more about polynomials, from working with multiplication questions with multiple terms to learning how to factor polynomials. This week I will explain how to solve a polynomial multiplication question that has multiple terms.

Example:

How to Simplify:

1. First, you want to lessen the amount of brackets so in this case you would multiply the brackets in each term by one another. The product will need to be written in brackets since it hasn’t been multiplied by the coefficient outside the brackets yet.
2. Next, you want to get rid of the brackets so you multiply the brackets by the coefficients that are directly beside them.
3. Lastly, you look at the products and collect any like terms to simplify it. Like terms are those that have the same variable and exponent ie. -10x, and +5x, if you were to combine these two like terms you would be left with -5x.

This week in math 10 we were reviewing polynomials. Though it was all review one thing that I learned how to do was how to make an area model.

An area model is one way to help solve a multiplication question. When multiplying polynomials in a sense you are also calculating the area of a rectangle.

Example:

How to make an area model:

1. first, you need to draw a rectangle and separate it into four squares
2. You then need to take the first term and separate it into two parts ex. 24 could also be written as 20 + 4, in this case you would separate the first term into  and +8. After you separate the first term you write the first part on the top of the first rectangle and the second half of the term on top of the rectangle right beside it.
3. Then you do the same thing to the second term but place the first half of the term (x) on the side of the first rectangle and the second half (-2) beside the rectangle directly under the first part of the term. (-2 would be written beside the rectangle under the rectangle that is beside x)
4. You then find the area of each of the individual rectangles
5. Add all of the areas together, collecting an like terms if there are any to get your answer

This week in math 10 we learned about trigonometry. We learned how to solve for missing lengths and how to identify different parts of the triangle, as well as an acronym to remember the different ratios (SOH CAH TOA).

For each equation solve for “X” and round your answer to the nearest whole number

Problem 1: 

1. First, you should label all sides of the triangle to help you write the equation (opposite, adjacent, hypotenuse)
2. Look at the lengths given and decide which of the tree signs you will be most appropriate (sin, cos, tan)
3. You need to isolate the variable in the equation. To isolate the variable you divide the sign by itself.
4. You the multiply the fraction on the right side by the sign in your equation. Since we don’t know the angle we have to invert the sign when we write it. Signs should always be written with an angle.
5. Solve the equation. Make sure to ado a degree symbol to the answer since you were solving for the angle and not side length.

 

Problem 2:

1. First, you should label all sides of the triangle to help you write the equation (opposite, adjacent, hypotenuse)
2. Look at the lengths given and decide which of the tree signs you will be most appropriate  (sin, cos, tan)

3. You need to isolate the variable in the equation. To do this you Multiply both sides of the problem by the denominator on the right side, in this case it would be 21.

4. Solve the equation.

 

Problem 3:

1. First, you should label all sides of the triangle to help you writ the equation (opposite, adjacent, hypotenuse)

2. Look at the lengths given and decide which of the tree signs you will be most appropriate (sin, cos, tan)

3. You need to get the variable on the top of the fraction so you reciprocate both sides of the equation.

4. You then want to isolate the variable so you multiply both numerators by the denominator of the fraction that contains the variable. In this case you multiply both sides by 28.

5. Solve the equation.