HELLO, Mr. Tarrou! Ok, today we are going to be solving polynomial inequalities. Now
most of these polynomials are going to be quadratic polynomials. So you are going to
see a lot of things that look familiar like factoring and the quadratic formula. But there
is one extra step that we are going to have to take a look at. If you want to solve a
polynomial inequality the first thing that we are going to have to do is set the polynomial
equal to zero. If you have a square term and a single degree term you cannot just get that
variable alone. You got to get that equation equal to zero and then factor or use the quadratic
equation...or if it is a higher order polynomial you might be doing some synthetic division
or factoring by grouping. You want to solve it for zero. These values in an equation are
your x intercepts, your real solutions. But in equalities these values that you get, I
want to call them solutions, but they are critical values that you are going to use
to break up the number line into intervals and then.... So if you have two solutions
there is going to be three intervals in the number line. Then you are going to pick a
number, this is where the extra inequality work comes in, you are going to pick a number
within each interval and substitute back into the original inequality, or maybe a version
you fiddled with just a little bit, to determine the final answer to your inequality. So let's
see what that looks like. We are going to start off with 2 x squared minus x minus 28
is great than or equal to zero. Now this quadratic equation has a leading coefficient that is
not one and the last constant is a rather large number with a lot of different factor.
There is 1 and 28, 2 and 14, 4 and 7... You can be doing trial and error a lot on your
factoring. So let's remember that if you would like to more easily do your factoring you
are going start by taking your first and last term and multiply them together. 2 times negative
28 is negative 56. Then you are looking for factors of negative 56 that add to this middle
term of negative 1. Well that is going to be negative 8 and 7. So negative eight times
seven does give us...they are factors of negative 56 that we got by multiplying the leading
coefficient and the constant, and more importantly they add to the middle term of negative one.
I am going to use these two values to rewrite the polynomial, this quadratic. We get 2 x
squared minus 8x plus 7x minus 28 is greater than or equal to 0. You are now going to factor
by grouping now that we have made this quadratic look different. It just looks different, negative
8 plus 7 still negative 1. What comes out of these two terms, these first two terms?
They both share a 2 and they both share an x, so we can take out 2x. 2 x squared divided
by 2 x is 1x and negative 8 x divided by 2 x is negative 4. Now I need this x minus 4
to show up again in the other grouping, so what can take out of 7 x minus 28 and get
x minus 4 to show up again? That is going to be positive 7. So I am going to divide
out a positive 7. 7 x divided by 7 is x. Negative 28 divided by 7 is negative 4. And now we
have the same factor show up twice, the x minus 4 is in both of these terms. So we are
going to pull it out again and get x minus 4 times 2 x plus 7 is greater than or equal
to 0. If this were an equation we would be very close to our final answer. We would just
set this equal to zero, this equal to zero and if one of these is equal to zero then
the entire equation is equal to zero. But this is an inequality. For these two numbers
to multiply together to give us a number bigger than zero they either need to be both positive
or both negative. We cannot just say well... I do have to do this step. x minus 4 is equal
to 0 and 2 x plus 7 is equal to 0. So my solutions are going to be, actually not my solutions...
my critical values are going to be x is four and after I subtract both sides by 7 and divide
both sides by two I will have x is equal to negative 3.5. That would be done if this were
an equation, but this is an inequality. So what I have to do now is use those critical
values to break up the number line into intervals. I am going to need to erase these directions
to finish my example here so make sure you stop the tape, stop the video, and copy those
notes. I just showed my age, "Stop the tape"!? Negative 3.5 and 4. So the last thing here
says again to pick a number out of each interval. We have one, two, and three intervals. See
which intervals work to keep this quadratic greater than or equal to zero. (While I am
erasing, you can also not that this is a parabola opening up and you have the x intercepts so
you can finish this problem with a sketch. Below the x axis are negative y values and
above positive y values.) Stop the tape on YouTube...haha... Let's say, let's do the
check in a different color... We are going to look at anything less than negative 3.5
will be fine so let's say negative four. Anything between negative 3.5 and 4 like 0 will be
fine. And anything greater than 4 will work, let's say 5. So I am going to plug each of
these numbers, one number from each of these intervals back into the original inequality...the
ones that work, work... and those that don't, that is not part of the answer. We are going
to have 2 times negative 4 squared minus negative 4 minus 28 and we are going to make sure that
is greater than or equal to zero. That is going to be 2 times 16 plus 4 minus 28 is
greater than or equal to 0. That is 32 plus 4 minus 28. I do believe that is going to
work. So this negative four in this section of the number line works so that interval
is ok. Now let's try 0. 2 times 0 squared minus 0 minus 28 is greater than or equal
to zero. 0 and 0 and negative 28. Negative twenty-eight is not greater than 0. So when
I tried a number in this second interval it failed, so that is not going to be a part
of my final answer. AND then finally let's try 5. 2 times 5 squared minus 5 minus 28
is greater than or equal to 0. That is 2 times 25, because 5 squared is 25, minus 5 minus
28 is greater than or equal to 0. Well that is 50 minus 5 is 45, and 45 minus 28 is definitely
going to be a positive answer. So our last interval works as well. My final answer for
what makes this inequality not only equal to 0, what makes it equal to 0 is 4 and negative
3.5, but also makes it positive are these two intervals of the number line make it positive.
My final answer x has to be less than or equal to, from the equal sign here, negative 3.5
or x has to be greater than or equal to 4. In interval notation that is going to look
like negative infinity to negative 3.5 with a square bracket because of the equal sign,
or again picking up four and we can use this because 4 makes it equal to 0 up to infinity.
BAM!!! Example number. Let's see if we can get another one done before my time is up.
Ok. Alright then. Let's take a look at this example. x cubed minus x squared minus 4 x
plus 4 is greater than 0. This is a third degree and to solve third degrees either they
need to be easily factorable by grouping or you need to work with synthetic division to
finish that up. Now this will factor by grouping like you just saw me do a second ago. X squared
will come out of these two terms and leave you will x minus one. These two terms both
have a 4 but I need x minus 1` to show up so I am going to take out a negative 4, because
this is plus a constant and this has minus a constant. Take out a negative 4. Now that
I have got two terms that both have an x minus 1 I am going to factor that out. X minus 1
times x squared minus 4. This is a difference of squares so this is actually factorable
again. So, now we have x minus 1 times x plus 2 times x minus 2, the difference of squares
pattern, is greater than 0. My critical values on my number line are going to be x equals
1, negative 2, and positive 2. Again if this were an equation we would be done, but it
is not. We are going to have these values on the number line so negative 2, positive
1, positive 2. I am going to have to check out a number in this interval say...I don't
know...negative 3. Let's try a number between negative 2 and 1 like 0. We are going to have
to plug in a number that is a decimal like maybe 1.5, and then a number that is bigger
than 2 like 3. Let's see which intervals work. Back up to this line and we have negative
3 cubed minus negative 3 squared minus 4 times negative 3 plus 4 is greater than 0. That
is going to be negative 27 minus 9 plus 12 plus 4 is greater than 0. That is going to
be negative 36 plus 12 plus 4. I don't think that is going to come out to be greater than
0, so section does not pass the test. Let's try 0. That is going to be easy...0, 0, 0,
plus four is greater than 0. If I show that I am going to have 0 - 0 - 0 + 4 > 0 Well
that is going to be pretty nice, right? Four is greater than zero so this section does
work out. Now I am kind of pausing because I am realizing that I am going to work out
a decimal here and I don't have a calculator. Alright!!! Let's take a look at 1.5. 1.5 cubed
minus 1.5 squared minus 4 times 1.5 plus 4 is greater than 0. That is going to be...
1.5 raised to the third is 3.375 minus 1.5 squared...I believe I know what that is but
I don't want to make a mistake on the internet...that is 2.25, minus that is going to be four and
another half so that is minus six plus 4 is greater than 0 This is going to be just a
hair over one and this is going to be negative two, so this is going to be negative number
is greater than 0 and that is false. So this interval does not work. I have two minutes
left to finish this up. Let's take a look at the value of three. We have got 3 cubed
is 27...3, 9, 27. 3 squared is 9. Negative 4 times 3 is negative 12 plus 4 is greater
than zero. This is going to be what?... 18 minus 12 plus 4 is greater than 0. That is
going to be positive 6 plus a positive 4 is still positive so that works out. My last
interval works. My final answer is going to be x has to be greater than negative two and
less than 1 or x can be greater than 2. I am out of time. THANK YOU VERY MUCH!!! WHOOO!
Solving polynomial inequalities.