NARRATOR: Exercise six.
Solve the equation x cubed + 2x squared + 3x plus 6 equals 0.
So we first need to factorise the left-hand side
in order to be able to use the null factor law,
and in factorising a cubic, we need to first identify one factor.
And we do that by looking for one root of this equation -
so we're looking for a number that we can substitute into the polynomial
in order to make it 0.
So I'm going to start with the value of the polynomial at 1.
And that's gonna give me 1 + 2 + 3 + 6
which is in fact 12, so not 0.
So therefore this graph goes through the point (1, 12).
That point isn't an x-intercept
and that's not useful to us in the factorisation.
I'm also noticing from this
that all the terms in my polynomial are positive,
and so I'm gonna need to be substituting in a negative number
in order to get 0.
I also know that the numbers I substitute in
are gonna all have to be ...
... if they're going to give me 0, are going to need to be factors of 6
because the only way the brackets can expand out
to give a constant term of 6 at the end
is using factors of 6 in those brackets.
So let's first have a look at P at –1.
So that's gonna give me -1 + 2 – 3 + 6,
which gives me 4, which, again, isn't 0.
So we know we need to stick with the negative,
so we'll try –2.
That gives me –8 + 8 – 6 and + 6
which is indeed 0.
So that tells me that (x + 2) is a factor
in order to give me one of my x-intercepts at x = –2.
So now that I've identified one factor,
I need to divide my cubic by that factor
to determine the other factors.
So I use long division for this process.
There are different ways to go about doing this.
There's synthetic division.
You can also do it in a sort of more linear fashion
without actually writing out any kind of division.
At the end of the day, you do all the same mathematical steps
to determine the factor.
It's just about what you choose to write down.
So I'm dividing the cubic by the linear factor,
so what we're looking at here is we've got our cubic equation
x cubed + 2x squared + 3x + 6,
and we know that that's equal to (x + 2)
multiplied by some other factor.
And obviously we can see
from this linear right up here on the right
that the first term in that bracket there
is going to have to be x squared.
And what we're doing to get that x squared is we're asking ourselves,
"Well, what would I need to multiply x by in order to get x cubed?"
And that's what we do in the long division as well.
"What would I need to multiply x by to get x cubed?"
Well, that's x squared.
And then I'm gonna expand that out,
just as we would when we're expanding out the brackets -
we would do x times x squared and also 2 times x squared.
So we're gonna do x squared times (x + 2)
which is gonna give me x cubed + 2x squared.
And then we subtract this.
And we've chosen x squared in order that the x cubed will cancel out,
and in fact we have an interesting situation here
where we've also got the 2x squareds cancelling out.
So 2x squared – 2x squared gives me 0.
And I'm gonna write "x squared" here
just to maintain the sort of place value of my columns.
Then I bring down the next term, + 3x,
and we repeat the process.
So what would I need to multiply x by to get 0x squared?
Well, that would be 0x,
so we're not going to have an x term in our factor.
And then we expand out.
So 0x times 0x, that's 0x squared.
0x times 2, 0x.
Then we take away and we've got 3x.
Bring down the next term,
which is + 6.
What will we need to
multiply x by to get 3x?
Well, that's 3.
Expand out, so 3x + 6.
Write that down here.
And subtract, which leaves us with 0.
And that's what we expect
when we've divided by a factor,
we know that we're going to get no remainder.
So what we've identified here is
that x cubed + 2x squared + 3x + 6
is equivalent to (x + 2)
multiplied by (x squared + 3).
So now we're looking to solve this equation equal to 0.
So we're looking to solve that equal to 0.
And we can now use the null factor law.
So we know that x + 2 might equal 0
or x squared + 3 might equal 0.
So we get x = –2,
which is the solution that we identified back at the beginning.
And we also have x squared equals –3.
So we'll be trying to square root a negative number here,
so we're going to get no solutions from this factor.
So in fact we have a cubic equation with just one solution
at x = –2.