Heading towards exams, tutoring returns to factoring and other challenging topics. The math tutor shows how to solve a factoring problem that, at high school level, might be difficult.
Suppose you are faced with the following problem:
This problem, when broken into the right steps, is not too hard. However, the steps are numerous.
Like most problems you encounter that contain x2 or higher, you must factor. At the beginning of every factoring process comes the question: Is there a common factor? (For a crash course in common factoring, look here.)
We realize that, in our case, there indeed is a common factor: -2x. (As I mention in my post about common factoring, whenever the lead term is negative, you should factor the negative out.)
After the common factor is taken out front, we arrive at
We are now faced with how to factor the “inside”: 3x4-23x2-36. Of course, the common factor has already been removed. Since the lead coefficient is a 3, rather than a 1, we must use complex trinomial factoring:
i) Examining 3x4-23x2-36, we multiply 3(-36) to get -108.
ii) We need to find two numbers that multiply to -108, but add to -23 (the middle term). We start writing down pairs of numbers that multiply to make 108.
Once your number pairs reverse to earlier combinations, you won’t get anything new.
iii) Among the numbers we’ve tabulated above, we must find the pair that, if one is negative, can also add to make -23 (the second term in 3x4-23x2-36). We notice the pair to be 4 and 27, if we make 27 negative. Once again: one of the pair has to be negative, since, as mentioned above, the pair must actually multiply to -108.
iv) We rewrite 3x4-23x2-36 with its middle term shown as the sum of 4 and -27:
v) We separate the rewritten expression into two pairs, then common factor each pair:
vi) The repeating factor (3x2+4) indicates we are successful. Now, we reorganize our factored expression into two brackets:
Step 3: We rewrite our original equation in factored form:
We now notice the term (x2-9). Being a difference of squares, it can be factored to (x+3)(x-3). Finally, we arrive at
Step 4: We need to solve the equation: we need to report the values of x that will make the left side equal to 0. We use the following reasoning:
If several values multiply to make zero, one of them must already be zero.
implies that either -2x=0, or (x+3)=0, or (x-3)=0, or (3x2+4)=0. One by one, we consider each possibility:
If -2x=0, then x=0. Therefore, x=0 is one solution.
If (x+3)=0, then x=-3. Therefore, x=-3 is a solution.
If (x-3)=0, then x=3. Therefore, x=3 is a solution.
Since, in the real numbers, x2 cannot be negative, (3x2+4) cannot be equal to zero. It yields no solutions.
Our solutions to the seemingly endless problem -6x5+46x3+72x=0 are, finally, x=0, x=3, and x=-3.
You wouldn’t see many problems this difficult on a high school exam; you might encounter one. Good luck with it!
Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.