Factoring and Quadratics

This Tuesday we worked on problems with an x**2 coefficient of more than one. After going over the homework, which were mostly problems with an x**2 coefficient of one. Lisa gave us this problem to try out:

20x**2 + 39x + 18

While she told us that we could try to trial and error the factors, there was a much more efficient way of solving the problem.

By setting up a table like the one above and plugging in AxC and B you can find the two numbers that add up to B.

       A + B + C

After finding the two terms that work, Lisa told us to then substitute them in for B, like so:

20x**2 + 39x +18

becomes

20x**2 + 15x + 24x +18

We figured out from here that this problem is now perfectly set up for factoring by grouping.

5x(4x + 3) 6(4x + 3)

(4x + 3)(5x +6)

*poof like magic*

On Wednesday we spent a few minutes going over some similar problems.

Such as:

Find the Common GCF of

z**3 - 4z = z(z**2 - 4) = z(z - 2)(z + 2)

z**3 - 2z**2 = z**2(z - 2)

z**3 - z**2 - 2z = z(z**2 - z - 2) = z(z - 2)(z + 1)

All have z(z - 2)

After going over the homework Lisa introduced us to quadratics by drawing a picture and asking us a question.

She asked us "What does the triangle have to equal for this to be true? What about the square?".

After thinking it over we decided that one or both of them had to equal zero.

Lisa then took it up a notch and showed us a problem like this:

(x + 3)(x + 5) = 0

She then asked us what x would have to equal for this to be true

We decided that since x is + 3 it must equal -3 to have the equation equal zero

and since it is also + 5, x must also equal -5 for this to be true.

Now that we had our answer we had two ways to write it either;

x = -3,-5

or

{-3,-5}

After completing more simple quadratic problems, we began on word problem like this:

Find two consecutive odd integers whose product is 99

(although many of us guessed this one we still had to write it out)

We first came up with a simple equation

x(x + 2) = 99

We made this equation because its a product so its multiplying, and its two consecutive odd integers so x and x +2

Now that we had our equation we went into solving it

x(x + 2) = 99

x**2 + 2x = 99

     -99         -99

x**2 - 2x - 99 = 0

(x + 11)(x - 9)

{-11,-9}

However we are not done yet. Since -11 and 9 are not x and (x + 2) we know that we must take them separately and have 9 and 11 as one set and -11 and -9 as the other.

Hope this was helpful!