This week in math I had a question on rational expressions that seemed very tricky at first. The question was:

then i start with the first step

1.factoring:

The second step finding the Common denominator of them

and than write them with the common denominator all together

after all these steps now we can do the adding or anything that the question asks:

now we can see, we could still factor the numerator to 4

And the final answer will be

When working with rational expressions, we always need to check which numbers make the denominator zero.
Because division by zero is impossible → so those numbers are not allowed.

In this question:

1. First denominator:

If:

or

, the denominator becomes zero.

So these two values ​​are not allowed.

2. Second denominator:

If:

or

, the denominator becomes zero.

(We already had x = -8)

3. Final result (Domain Restrictions)

Numbers that we cannot substitute for x:

Because each of these values ​​causes the denominator to become zero.

As soon as the denominator becomes zero, the whole expression becomes meaningless.

This week  we learned a new topic that was very confusing to me at first: working with rational expressions.
The same fractions where the numerator and denominator are polynomials. I always thought they were hard, but when I went step by step I realized they were just as scary.

First: Understanding When a Number Is Not Allowed

The first thing I learned was that before doing anything, we need to see when the denominator of a fraction becomes zero.
Because if the denominator becomes zero, that value is not allowed for x or , and we need to write it off.

ex:

At first we check it if we can factor the numerator or denominator And then start the simplifying

Now we start simplifying

Since is on both sides, we simplify it and take it out

As I worked through this topic step by step, I realized that rational expressions are neither scary nor complicated; they just require discipline and precision.

This week I came across a question  in desmos activity from the quadratic function lesson that at first I had no idea how to find its equation. I was especially surprised when I saw that the answer was a fraction.
The graph of a parabola was given with the vertex at:

(-1,2)

And there was another point on the graph:

(4,7)

At first I didn’t know where to start, but then I realized that the best way was to use the vertex form (vertex form):

so then we replace the number:

Next step I used the point (4,7) to find the value of a.

7=25a+2

5=25a

and the final answer i got was

This week the new light bulb i had was about the new chapter “ Quadratic Function “

This chapter is about graphs and equations and it has many different ways to use them.

The part that we learned this week in class was about Properties of Quadratic Functions.
To start, let’s look at the parent function,
f(x) =
to see the basic shape of a parabola.

Then we looked at a few other functions, such as
f(x)= , f(x)= , f(x)= +1, f(x)= -3, f(x)= +x
and saw how numbers cause the graph to shift or change its shape.

For each function, we found the points using a table of values ​​and then drew the graph.
By drawing the graph, we learned that the vertex is the point at which the graph changes direction.

We also learned that:
• If a is positive, the parabola opens up and the vertex is the lowest point.
• If a is negative, the parabola opens down and the vertex is the highest point.
• The line of symmetry always passes through the middle of a parabola and through the vertex.

Finally, I learned that we can either use the formula
x =
to find the vertex coordinates, or we can figure it out by looking at the graph.

This lesson helped me better understand how each number in the equation affects the shape of the graph  especially how the vertex moves when we add +1 or −3 to

This week I was working on a question from Chapter 3 that we had to solve using the Zero Product Law. The question was:
9x – = 0

At first I thought I had to divide both sides of the equation or take the radical to get the answer, but then I realized that this was the wrong way to do it.
In fact, I had to factor the equation. When I factor x in both terms, the equation becomes:
x(9 – 4x) = 0

Then using the Zero Product Law, I could say that either
x = 0
or
9 – 4x = 0 x=

I finally realized that the key to solving this question was to factor first, rather than resorting to more complicated methods.

This week I had really good mistake that helped me so much about the New lessone the” quadratic method ” .

The first step to solve a equation with ‘ the Quadratic method is label the a b c than write the rest of  the formula

the formula is written like that:

The equation that i made mistakes with solving it was :

My mistake was that when I wanted to determine the coefficients of a, b, and c, I put the number 6 in place of b, even though since the equation has no term with x, that number should be c, not b.

♥

Factoring a Trinomial: Example

This week, I learned how to factor a trinomial that has three terms and a coefficient in front of the squared variable.

At first, I wasn’t sure how to handle the number in front of , but after learning the steps, it became much clearer.

Step 1: Check for a common factor (GCF)

The first thing to do is look for the greatest common factor of all three terms.

In , there is no common factor other than 1,

so we move to the next step.

Step 2: Multiply the first and last coefficients

We multiply the coefficient of (which is 3) by the constant term (which is 60):

Now we need to find two numbers that multiply to 180 and add up to 28.

Those two numbers are 18 and 10, because .

Step 3: Split the middle term

We rewrite the trinomial by splitting the middle term () into two parts,

using the numbers we found:

Step 4: Group the terms

We divide the expression into two groups to make factoring easier:

Step 5: Factor each group

From the first group, , we can take out :

From the second group, , we can take out :

Now the expression looks like this:

Step 6: Factor out the common binomial

Both parts include the factor ,

so we take it out as a common factor:

Final Answer:

At first, I thought factoring with a coefficient in front of would be confusing,

but breaking it into steps made it simple.

Now I understand that the key is to multiply the first and last numbers,

find two numbers that work for both multiplication and addition,

and then use grouping to finish the factoring.

This week in math, we were learning about factoring.

One example we worked on it and i had struggled to solve it was

+ 3k

At first, I made a mistake because I thought I should write it like this:

(k + 3)(k + k)

But that was wrong  this would actually equal + 6k not .

Then I learned the correct way that is like

To factor properly, you need to find the greatest common factor (GCF) of all terms.

In this case, both terms have a k, so the GCF is k.

When I take k out, I get:

+ 3k= k(k + 3)

Now I understand that factoring means pulling out the common factor, not just grouping terms randomly.

It was a small mistake, but it helped me understand factoring much better.

Here is some example to explain it better and better understanding

6x + 9

Step 1: Find the greatest common factor (GCF).

Both terms have a common factor of 3.

Step 2: Divide each term by 3 and write the factored form:

6x + 9 = 3(2x + 3)

The Math Mistake That Helped Me Understand Radicals Better

Sometimes I think I understand a math rule… until I actually try it. That happened to me recently when I was dividing radicals with coefficients.

The question was:

$\frac{6\sqrt{72}}{3\sqrt{8}}

At first, I solved it like this:

Wrong (but lucky) Method:
$\frac{6\sqrt{72}}{3\sqrt{8}} = 2\sqrt{9} = 6

I thought it was correct — and it actually was — but later I learned that my method wasn’t safe. It only worked by accident.

The Proper Way

I found out that you’re supposed to separate the coefficients and the radicals first:

$\frac{6\sqrt{72}}{3\sqrt{8}} = \frac{6}{3} \times \frac{\sqrt{72}}{\sqrt{8}} = 2 \times $\sqrt{\frac{72}{8}} = 2 \times \sqrt{9} = 2 \times 3 = 6

This time I understood why the answer was 6

Another Example (This Time It’s More Clear)

$\frac{5\sqrt{20}}{2\sqrt{5}}

❌ Wrong Way (what I used to do):
$\frac{5}{2} \sqrt{4} = 5

✅ Correct Way:
$\frac{5}{2} \times \sqrt{\frac{20}{5}} = \frac{5}{2} \times \sqrt{4} = 5

Again, it worked now I actually know what I’m doing.

What I Learned

Always split the outside numbers and the radicals before dividing, making this mistake helped me remember the rule better than any lesson.

This week in math, something finally clicked for me  like a light bulb turning on in my head.

I used to get confused when I saw expressions like

I didn’t know if I was supposed to add the numbers, the radicals, or do something else. But then I learned the trick:

you can only add radicals if the part under the root is the same, just like combining like terms.
For example: