# Multiplying Polynomials to Find Values of 'k'

1. May 21, 2012

### AnTiFreeze3

1. The problem statement, all variables and given/known data
Find the value or values of k that make the equation true.

2. Relevant equations
Starting Equation:

(2x + k)(x - 2k) = 2x2 + 9x - 18

3. The attempt at a solution

(2x + k)(x - 2k) = 2x2 + 9x - 18

2x2 - 3xk - 2k2 = 2x2 + 9x - 18 <- I just foiled the left side

Sadly enough, this is where I got stuck. I'm currently reviewing for my final, and this is what we went over at the very beginning of the quarter, so these types of problems aren't very fresh in my mind. The book isn't helpful at all because all it explains how to do is multiply binomials, and doesn't elaborate on anything beyond that. My teacher, who was short on time when I asked, said that I either needed to factor the right side to get both sides to be similar, or FOIL the left side to get both sides similar, but either he didn't explain it well enough or it didn't click for me on what to do next to solve for k.

Any help would be appreciated, thanks.

2. May 21, 2012

### Villyer

If you were solving a 'regular' quadratic equation, what would your first step be?

Edit: Actually, this can be set up much more simply.
No matter the value of k, no new x term can be introduced. For example, the -3xk term cannot turn into a constant and the -2k2 cannot become a linear term.
From this, do you see any possibilities to set up a system of equations that would allow you to solve for k? (The system will be very simple, and probably isn't even needed but if you look at it as a system future problems might become easier.)

3. May 21, 2012

### AnTiFreeze3

Set everything equal to zero and then either factor, or if I couldn't factor, use the quadratic forumala.

4. May 21, 2012

### scurty

Right. So in this case the variable you are solving for is k, not x. Your answer for k will be in terms of x.

Edit: And what Villyer is suggesting by his edit is the equate the coefficients of the $x^2$, $x$, and constant terms and set up a system of linear equations.

5. May 21, 2012

### Villyer

See my edit to that old post.
The set equal to 0 method I suggested also works, because the 2x2 terms will end up cancelling, but can be very annoying if those terms don't cancel and the quadratic formula is necessary.

6. May 21, 2012

### AnTiFreeze3

I guess where the confusion comes from is that I still have two variables in the problem, so I'm unsure what to do with it.

Assuming that what I put up there was still correct, I continued off of that to get everything equal to zero, and got this:

2k2 + 3xk + 9x - 18 = 0

Sorry if I'm missing something here, but I don't remember having to solve quadratics with more than one variable.

Could I literally do something as sneaky as divide the 3xk and 9x by x, and also divide 0 by x, since that wouldn't change the right side, but would also get rid of the x's and leave me with only k so that I could then solve for it?

7. May 21, 2012

### micromass

Staff Emeritus
The trick is that your equation

$$2k^2 + 3xk + 9x - 18 = 0$$

must be true for ALL x in this context. So since it must be true for all x, it must be true for a specific x you choose.

So why not choose x very specifically and see what you get. For example, what do you get if you choose x=0?

8. May 21, 2012

### AnTiFreeze3

When you're referring to me setting it up much more simply, are you implying that I would set up a system of linear equations using:

2x + k = ?
x - 2k = ?

I'm feeling slightly lost (not because anybody is doing anything wrong, just because I've never been taught to use systems of equations to solve quadratics before) so if I'm not making sense, forgive me.

9. May 21, 2012

### AnTiFreeze3

If x = 0, then k = 3, right? If it's asking for the value or values of k that make the equation true, will I need to set it up as an inequality?

EDIT: k = ±3 .... rookie mistake.

Last edited: May 21, 2012
10. May 21, 2012

### micromass

Staff Emeritus
If $k^2=9$, then k=3 isn't the only solution...

11. May 21, 2012

### Villyer

2x2 - 3xk - 2k2 = 2x2 + 9x - 18

Taking this, the 2x2 terms cancel, leaving - 3xk - 2k2 = 9x - 18

Then the get to a 'system' you can equate the coefficients of the different x terms.
So you can set the terms without an x equal, and you can also set the terms with an x equal, since the value of k cannot change the power of the x term.

Of course, any other method mentioned here is also perfectly valid, and you should take the route you feel most comfortable with, because it is never good to be trying a new method on the day of your final.
I would show you a more complete answer in order to better illustrate what I am trying to say, but I believe that to be against pf's rules.

Edit: If you use micromass's method, I would avoid choosing 0 as the test point, as it result in dropping the 3xk term and possibly affecting your results. Also, while some values of k work for x being a certain point (i.e. x = 0), the same values of k may not work for another point (i.e. x = 1).

Last edited: May 21, 2012
12. May 21, 2012

### AnTiFreeze3

Edited my post micromass.... I guess I wasn't thinking too much about that.

Going off of villyer's way, I'd be left with:

2k2 = -18
-3xk = 9x

As I've worked out before, 2k2 = 18 ends up being k = ± 3(thanks for catching that earlier, micro)

-3xk = 9x
___ ___

-3x -3x

k = -3.

So, the answer ends up being k = -3, correct?

.... that took a lot more effort and energy than I think it should have, but I appreciate the help from everybody.

Last edited: May 21, 2012
13. May 21, 2012

### Villyer

k = -3 is the answer I came to as well.

If you were using micromass' suggestion, but plugged in 1 instead of 0, then the problem reduces to 2k2 + 3k - 9, resulting in the answers k = -3 and k = 3/2.
Note that x = 0 resulted in k = 3 and k = -3.

I just bring this up as a note of caution: Check your final answers! Some may be extemporaneous.