This week in math we learned how to evaluate factorable expressions that contain functions, containing variables like x and y. This process is important because it allows you to solve for the restrictions of the variables as well the skill of being able to turn a complicated expression into a more neat and clean looking expression. Being able to factor expressions containing functions is important because it is a concept that we will build upon in further units and levels of math. I chose this topic because I enjoyed the process of substituting my functions with variables.
Lets look at some examples…
Example 1.
My first step is to rewrite the expression by subsituting variables where my functions are. Once you do this, you will notice my new expression is similar to the expresson + 4a + 3
Now that I have rewritten my expression, I need to find what multiples of 1 and 3, multiply together to give me a sum of 4. In this case, 3 x 1 = 3; and 1 x 1 = 1 | 3 + 1 = 4 which is what we needed to determine the factors of our expression. We add the terms across giving me (1a + 3) and 1a + 1) or in other terms (a + 3)(a + 1)
Now we want to subsitute our original values of x into a. (X + 2 + 3)(X + 2 + 1)
Finally we can add like terms, since 2+3 = 5 the first factor is (x + 5) and our second factor is (X + 3)
Example 2.
In my first step, I substitute my functions of x + 2 and y + 5 with a and b; x + 2 = a; y + 5 = b
In my next step, I look for one thing in common between both terms. I divide 32 and -18 by the GFC of 2.
In this step, I move the two that I simplified the expression with outside of the brackets, you cannot get rid of it. This leaves me with 2(16 – 9 )
I can further factor this expression because 16 and 9 are perfect squares, and and means that there is one a and one b in each set of brackets. This leaves me with 2(4a + 3b)(4a – 3b)
5. In this step, I subsitute my original functions back into my factored expression 2(4a + 3b)(4a -3b) = 2[4(x+2) + 3(y+5)] and my second term [4(x+2) – 3(y+5)
6. Now I can expand this expression and collect like terms. Distribute 4 to x + 2. 3 to y + 5. On the right side i do the same, distribute 4 to x and 2, and -3 to y and 5. My new expression is now 2(4x + 8 + 3y + 15) and on the right side (4x + 8 – 3y – 15)
7. Finally I can add like terms giving me 2(4x +3y + 23)(4x -3y -7)
Glossary:
Function : A function from a set X to a set Y assigns to each element of X exactly one element of Y.
subsitute: substituting the value of any one of the variables from one equation into the other equation.