Tuesday, June 5, 2007

Review for Final - Units 5-10

8 comments:

Unknown said...

How do u factor something when x^2 has a number in front of it...like 4x^2

Mr. Chamberlain said...

Everyone's favorite topic... FACTORING!! Woo-hoo!!

As we know, good candidates for factoring are quadratics with a leading coefficient of +1. But that shouldn't totally discourage us from factoring other quadratics since, after all, factoring is such fun!

Sometimes the best way to see how something is taken apart (i.e. factoring) is to put it together (i.e. multiply/simplify binomials).

Go ahead and multiply (simplify) these:

(2x - 4)(2x + 3)
this will FOIL to:
4x^2 + 2x - 12 ... right?

how about
(4x + 3)(x + 2)
this FOILs to:
4x^2 + 11x + 6 ... right?

So now, you PLAINLY see that expressions such as the two products (aka simplified expressions) above can be factored. It takes a little patience and trial and error, but it CAN BE DONE.

In the case of 4x^2, IF the expression can be factored, it will either be
(4x +/- a)(x +/- b) ... or ...
(2x +/- a)(2x +/- b)

... there is no other way to cleanly produce 4x^2. Now it's just a matter of tinkering with various a's and b's until you get a hit.

Since the leading coefficient is not +1, you lose the major advantage of the "diamond problems" but the "a" and "b" still have be factors of the constant term from the original quadratic.

Try it and let me know how you do.

Anonymous said...

derek--
just to sumarize mr.c's response, when you factor, basically you have to just play around with the numbers and get them to FOIL out to the original question. The answer to you example would be:

(2x^2)(2x^2) because as you hopefully know (I still have trouble knowing this lol) (2x^2)(2x^2) = 4x^2

Anonymous said...

okay now that i just answered derek's question, i have a question of my own now:

how do you know whether when you factoring if it is prime or not?

also, when you are figuring out the x-coordinates for a parabola, how do you know if there is no real solution or many solutions?

Mr. Chamberlain said...

Well, if you can't factor (with integer coefficients), then you have a prime polynomial. Of course, we have been working mainly with quadratics.

For example:
5x^+30 can be factored to 5(x^2 + 6)

but
5x^2+31 cannot be factored... therefore it is prime.

x^2 + 5x + 6 can be factored
x^2 + 5x + 7 cannot... ... therefore it is prime.

If you just want to know whether a quadratic equation has roots (aka solutions, aka x-intercepts, aka zeroes) you can simply evaluate the discriminant (b^2 - 4ac). Remember that little gem... it's tucked inside the square root symbol of the quadratic formula.

If the discriminant is positive, you have two roots, if it's zero you have one root (a "touch" - witness x^2 - 6x + 9), if it's negative, the parabola is a "floater"... it does not "take root" in the x-axis (whether opening up or down).

It'll make more sense after the final exam!! ;)

Don't fret... it's not a heavy topic on the exam!!

Anonymous said...

okay and now i have yet another question =[.

I totally forgot...i tried to look at my work from in class but that didn't help me...how do I do the substitution method?

Anonymous said...

wen you hav a y equals u put that in were you have a (y) like 2x+3y=

Mr. Chamberlain said...

Just what the previous guy said...

To solve linear systems (2 or more linear equations), one method is substitution. This is a good method when one of the equations is in "y=" or "x=" form and the other isn't.

Such as:

3x + 2y = 12
y = x + 1

In this case, you can substitute "2x+2" for y in the first equation... like so:

3x + 2(x+1) = 12
3x + 2x + 2 = 12
5x + 2 = 12
5x = 10
x=2

Then, once you have the x value, you just sub it into one (or both!) of the original equations, to find the y value.

3(2) + 2y = 12
6 + 2y = 12
2y = 6
y=3

So the solution for the system is (2,3). Graph the linear system on the graphing calculator so you can actually see the point of intersection.

What FUN!!