Textbook Reference:
Section 7.5 **Solving Trigonometric Equations**

Let's take a closer look at "splitting", in the context where it's most useful: factoring.

*(Again, if you remember this from last semester, you can actually skip right to the Preview Activity, but if you need the refresher, read on.)*

## Splitting polynomials

Say we've got the following quadratic equation. \[2x^2+6x-20=0\] You have a couple of ways to solve this at your disposal:

- On one hand, you could complete the square or use the Quadratic Formula. These are guaranteed to work (in fact they're actually the same thing), but they can be a little long-winded and tedious.
- On the other hand, you could try to factor the left side. This
*isn't*guaranteed to always work, but if it does, it'll make it much easier.

By now, you've probably already noticed that all the coefficients are divisible by \(2\), so at least we can factor that out:
\[2(x^2+3x-10)=0\]
That new quadratic in the parentheses is relatively easy to factor, since its leading coefficient is \(1\). We need to find two numbers whose **product** is \(-10\) and whose **sum** is \(-7\); after some thought, you should see that \(5\) and \(-2\) will do the trick. This means we can now factor the expression as follows:
\[2(x+5)(x-2)=0\]
Cool, now it's factored. But how is this supposed to help us?

Here's the key idea behind *why* we factor:

**If a product is zero,then one of the factors must be zero.**

Since we have three things multiplied together, we now have three possibilities:

- We could have \(2=0\). That leads nowhere pretty fast, though, so we'll ignore it.
- We could have \(x+5=0\), which would mean \(x=-5\).
- We could have \(x-2=0\), which would mean \(x=2\).

Thus we end up with two solutions: \(x=-5\), or \(x=2\).

## Taking it up a notch

Let's do a harder one: \[6x^2-x-2=0\] This one doesn't have that common factor of \(2\), and even if it did we still wouldn't have a leading coefficient of \(1\).

We can still do it with a bit of guess-and-check, but personally I'm a fan of the so-called "area model" or "box method". Here's a video if you've never seen the method before (done on a different problem):

So here's how we'd set it up for our problem:

To fill in the other two boxes, we need two numbers whose product is \(6\cdot-2=-12\) and whose sum is \(1\). This leads us to \(-4\) and \(3\), so we'll use those numbers to split up the middle term:

The greatest common factor of the top row is \(2x\), so we put that to the left, and after that the rest of the table fills itself in:

Essentially we've reconstructed the original multiplication problem: \[(2x+1)(3x-2)=0\] Since we have a product of two things equaling zero, this splits into two possibilities:

- We could have \(2x+1=0\), which would mean \(x=-\dfrac12\).
- We could have \(3x-2=0\), which would mean \(x=\dfrac23\).

Thus we end up with two solutions: \(x=-\dfrac12\), or \(x=\dfrac23\).

These aren't the only ways to do factoring, but at least it should jog your memory. You'll need these techniques to be able to tackle some of the new equations we'll solve in class with circular functions.

# Preview Activity 9

*Answer these questions and submit your answers as a document
on Moodle. (Please submit as .docx or .pdf if possible.)*

- Consider this equation: \[x^2=x\]
A student solves the equation by dividing both sides by \(x\) and leaving \(x=1\).

Show that there is another solution. Why did the student miss this solution? What should they have done instead? - Consider this equation: \[x^2-6x+8=1\]
A student solves the equation by factoring the left-hand side, which gives \((x-2)(x-4)=1\). They split the equation into two equations, \(x-2=1\) and \(x-4=1\), and conclude that solutions \(x=3\) or \(x=5\).
- How can you show the student that these are not solutions of the original problem?
- Explain the student's mistake in reasoning.
- Correctly solve the equation.

- Solve this equation: \[3x^2+10x-8=0\]
If you really want to, you can just complete the square or use the Quadratic Formula to solve it, but the left-hand side is factorable, so you should try to factor it. It'll be good practice.

(If you need a bunch more practice because your factoring is rusty, here's some optional extra practice with answers provided. -
Answer
**AT LEAST**one of the following questions:- What was something you found interesting about this reading?
- What was an "a-ha" moment you had while doing this reading?
- What was the muddiest point of this reading for you?
- What question(s) do you have about anything you've read?

## No comments :

## Post a Comment