Identifying graph based on roots - Engineering Mathematics Video Lecture

Use the real 0's of the
polynomial function y
equal to x to the third plus
3x squared plus x plus 3
to determine which of the
following could be its graph.
So there's several ways
of trying to approach it.
One, we could just look at what
the 0's of these graphs are
or what they appear to be and
then see if this function is
actually 0 when x
is equal to that.
So for example, in graph
A-- and first of all,
as always, I encourage
you to pause this video
and try it before I show
you how to solve it.
So I'm assuming you've
given a go at it.
So let's look at this
first graph here.
Its 0, it clearly has a
0 right at this point.
And just by trying to
inspect this graph,
it looks like this is at
x is equal to negative 3,
if I were to estimate.
So that looks like the
point negative 3, 0.
So let's see, if we substitute
x equals negative 3 here,
whether we get y equaling 0.
So let's see, negative 3
to the third power plus 3
times negative 3 squared
plus negative 3 plus 3.
What does this give us?
This gives us negative 27.
This gives us positive 27.
This gives, of
course, negative 3.
This is plus 3.
These two cancel out.
These two cancel out.
This does indeed equal 0.
So this was actually
pretty straightforward.
Graph A does indeed work.
You could try
graph B right here,
and you would have to
verify that we have a 0 at,
this looks like negative 2.
Another one, this looks like at
1, another one that looks at 3.
And since we already know that
A is the answer, none of these--
if you were to input x equals
negative 2, x equals 1,
or x equals 3 into this function
definition right over here,
you should not get 0.
And you'll see that
this doesn't work.
Same thing for this one.
If you tried 4 or
7 for your x's, you
should not get 0
over here, because we
see that the real function
does not equal 0 at 4 or 7.
Another giveaway that this is
not going to be the function
is that you are going to
have a total of three roots.
Let me write this down.
So you're going to have
a total of three roots.
Now, , those three roots could
be real or complex roots.
And the big key is complex
roots come in pairs.
So you might have a situation
with three real roots.
And this is an example
with three real roots,
although we know
this actually isn't
the function right over here.
Or if you have one
complex root, you're
going to have
another complex root.
So if you have
any complex roots,
the next possibility is one
real and two complex roots.
And this right over
here has two real roots.
That's not a possibility.
That would somehow imply that
you have only one complex root,
which that is not a possibility.
Now another way that you could
have thought about this--
and this would have
been the longer way.
But let's say you didn't
have the graphs here for you,
and someone asked you to
just find the roots-- well,
you could have attempted
to factor this.
And this one actually
is factorable.
y is equal to x to the third
plus 3x squared plus x plus 3.
As mentioned in previous videos,
factoring things of a degree
higher than 2, there is
something of an art to it.
But oftentimes, if
someone expects you to,
you might be able to group
things in interesting ways,
especially when you see
that several terms have
some common factors.
So for example, these first
two terms right over here
have the common
factor x squared.
So if you were to
factor that out,
you would get x squared
times x plus 3, which is neat
because that looks a lot
like the second two terms.
We could write that as
plus 1 times x plus 3.
And then you can factor
the x plus 3 out.
We could factor
the x plus 3 out,
and we would get x plus
3 times x squared plus 1.
And now, your 0's
are going to happen,
or this whole y--
remember this is
equal to y-- y is
going to equal 0
if either one of these
factors is equal to 0.
So when does x plus 3 equal 0?
Well, subtract 3
from both sides.
That happens when x is
equal to negative 3.
When does x squared plus
1 equal 0, I should say?
Well, when x squared
is equal to negative 1.
Well, there's no real
x's, no real valued x's.
There's no real
number x's such that x
squared is equal to negative 1.
x is going to be
an imaginary-- or I
guess I'll just say it in
more general terms-- it's
going to be complex valued.
So once again, you
see you're going
to have a pair of
complex roots, and you
have one real root at x
is equal to negative 3.