www.mathcentre.ac.uk/.../Solving%20Cubic%20Equations.mp4

The general strategy for solving
a cubic equation is to reduce it

to a quadratic and then solve
the quadratic by the usual means

either by Factorizing or using

the formula. A cubic equation
has the form a X cubed

plus BX squared plus CX plus
D equals not.

It must have a Turman X cubed,
or it wouldn't be a cubic.

But any or all of BC&D can be 0.

So for instance X cubed.

Minus six X squared plus
11X minus six equals note

is a cubic.

So is 4X cubed plus 57
equals not.

So is. X cubed

+9 X equals not.

Just as a quadratic equation may
have two real roots.

So a cubic equation
possibly has three.

But unlike a quadratic equation
which main have no real

solution, a cubic equation
always has at least one real

root. I'll explain why later.

If a cubic does have 3
Routes, 2 or even all three

of them may be repeated.

This gives
us four

possibilities.

The equation X cubed minus
six X squared plus 11X.

Minus 6 equals 0 Factorizes
2X minus one times X

minus two times X minus
three equals 0.

This equation has three real
roots, all different.

Solutions X equals 1 X equals 2
or X equals 3.

I'd like to show you the graph
of the curve Y equals X cubed

minus six X squared plus 11X
minus six. I'm not very good at

drawing freehand graphs.

So here's one I prepared
earlier.

Notice that it starts slow down
to the left.

Because as X gets larger,
negative, so does X cubed and it

finishes high to the right
because there's X gets large and

positive. So does X cubed.

And the curve crosses the X axis
three times, once where X equals

1 once, where X equals 2 and
once where X equals 3. This

gives us our three separate

solutions. The
equation X

cubed. Minus five

X squared. Plus 8X
minus 4 equals 0

Factorizes 2X minus one
times X minus two all

squared, and that is
equal to 0.

In this case we have. We do have
3 routes, but two of them are

the same. We have X minus 2
squared, so we only actually

have two solutions. Again, I'll
show you the graph of Y equals X

cubed minus five X squared plus
8X minus 40 equals 0.

Again, the curve starts load
to the left and goes high to

the right. It crosses the X
axis once and then just

touches it. So we have our two
roots, X equals 1 and X equals

2 and it touches at the
repeated root X equals 2.

The equation X cubed
minus three X squared

plus 3X minus one

equals note. The left
hand side factorizes to X minus

one or cubed equals not.

So there there are three
factors. They're all the same

and we only have a single
solution. X equals 1.

The corresponding curve is Y
equals X cubed, minus three X

squared plus 3X minus one.

And it looks like this.

As with all the Cubix I've
shown you so far, it starts

slow down on the left and
goes high up to the right.

Notice that the curve does
cross the X axis at the point X

equals 1, but it is also a
tangent. X axis is a tangent to

the curve at this point,
indicating the three repeated

roots.

Now look at the equation
X cubed plus X squared

plus X. Minus
3 equals 0.

This expression Factorizes 2X

minus one. X squared plus 2X
plus three, so we can put this

equal to 0.

The quadratic X squared +2 X +3
equals not has no real

solutions. So the only
solution to the cubic

equation is to put X minus
one equal to 0, giving this

single real solution X equals
1.

The graph Y equals X cubed plus
X squared plus 6 - 3 looks like

this. And you can
see that it only

crosses the X axis
in one place.

From the graphs that I've shown

you. You can see why a cubic
equation always has at least

one real root.

The graph either starts large,
negative, and finishes large

positive. If the coefficient of
X cubed is positive or it will

start large positive and
finished down here. Large

negative if the coefficient of X
cubed is negative, the graph of

a cubic must cross the X axis,
giving you one real root. So any

problem you get that involves
solving a cubic equation will

have a real solution.

Now let's move on to
how we solve cubics.

Like a quadratic, cubic should
always be rearranged into the

form X cubed plus BX squared
plus CX plus D equals 0.

The equation X squared plus 4X
minus one equals 6 over X is a

cubic, but I wouldn't like to
try and solve it in this

particular form. We need to
multiply through by X, giving us

X cubed plus four X squared
minus X equals 6.

And then we subtract 6 from both
sides, giving us X cubed.

Plus four X squared minus
X minus 6 equals 0.

When solving cubics, it
helps if you know or

think you know one route
to start with. For

instance, take the
equation X cubed.

Minus five X squared
minus 2X plus 24

equals not. Given that
X equals minus two is a

solution. There is a theorem
called the factor theorem which

I'm not going to attempt to
prove here that says that if X

equals minus two is a solution

of this equation. Then X
+2 is a

factor. Of this whole
expression.

This means that X cubed minus
five X squared minus 2X plus 24

is equal to X +2.

Times some quadratic which
will call X squared plus

8X Plus B.

And then all that is
equal to, not.

So our task now is to find A&B
and we do this by a process

called synthetic division.

This involves looking
at the coefficients of.

The original expression.

So for instance, the coefficient
of X cubed is one.

The coefficient of X squared
is minus 5. The coefficient

of X is minus 2.

And the constant is 24.

And we just right in
that we're

synthetically dividing
by minus 2.

I leave a line.

And then bring the one down one
times minus two is minus 2.

Minus 5 plus minus 2.

Is minus 7.

Minus Seven times minus two is

14. 14 plus minus two

is 12. 12 times minus

2. Is minus 24?

And 24 plus minus 24 is 0.

The zero tells us that X +2 is
indeed a factor, and the numbers

we have here give us the
coefficients of the quadratic. A

is equal to minus Seven and B is
equal to 12. So the quadratic

that we're looking for is X
squared minus 7X.

Plus 12. And synthetic division
is explained fully in the

accompanying notes. So we've
reduced our cubic 2X plus

two times X squared minus
7X plus 12 equals zero

X squared minus 7X plus
12 can be factorized into

X minus three times X
minus four. So we have

X +2 times X minus
three times X minus 4.

Equals 0. Giving us.

X equals minus 2.

3 or 4.

If you don't know a
route, it's always worth trying

a few simple values. Let's
solve X cubed.

Minus.

7X.

Minus 6. Equals 0.

The simplest value should try is

one. 1 - 7.

Minus 6 - 12 so that doesn't
work. Let's try minus 1 - 1

cubed is minus 1 + 7 - 6 is
0, so minus one is a route.

Which means that X plus
one is a factor.

If minus one is a route, we
can synthetically divide through

this expression by minus one.

Coefficient of X squared, sorry
coefficient of X cubed is one.

The coefficient of X squared is
0, there's no Turman X squared.

The coefficient of X is minus
Seven and the constant is minus

6 and were synthetically
dividing by minus one.

I bring the one down one times
minus one is minus one.

Minus one and zero is minus one.

Minus one times minus one is

one. Minus 7 Plus
One is minus 6.

Minus six times minus one is 6
- 6 at 6 is 0.

These numbers give us the
coefficients of the quadratic

equation, and we have X squared.

Minus X minus 6 equals
0 and we now need to

solve this equation.

So we have X

plus one. Times X
squared.

Minus X minus 6 equal to zero
X squared minus X minus 6,

factorizes to X minus three X
+2, so we have X Plus one

times X, minus three times X +2
equals 0, and the three

solutions to the cubic equation
RX equals minus 2 - 1.

Or three.

Sometimes you may be able to
spot a factor.

In the equation X cubed.

Minus four X squared
minus 9X plus 36

equals 0.

The coefficient of X squared is
minus four times the coefficient

of X cubed.

And the constant is minus four

times. The coefficient of X.

This means that we're going to
be able to take out X minus 4 as

a factor from that those two
terms and those two terms.

Those two terms are X squared
times X minus four X squared

times, X is X cubed X squared
times minus four is minus

four X squared.

Those two terms are minus.

9 times X minus 4.

Minus nine times X is minus
9X and minus nine times minus

four is plus 36 and all that
is equal to 0.

And we can now factorize again
because we have a common factor

X minus four, so we have X
squared minus nine times X minus

4 equals 0.

X squared minus nine is the
difference of two squares, so we

can write that in as X plus
three times X minus three times

X minus 4 equals 0.

Giving a solutions X equals
minus 3, three or four.

You may have noticed that in
each example that we've done.

Every root. There's a factor of
the constant term in the

original equation. For instance,
three and four, both divided

into 36. As long as the
coefficient of X cubed is one,

this must be the case.

Because. The constant is simply
the product of the roots 3 * 3 *

4 is equal to 36.

This gives us another possible

approach. Consider the
equation X cubed

minus six X

squared. Minus six X
minus 7 equals 0.

Now it's possible for every
solution to be irrational, but

if there is a rational solution,
then because the coefficient of

X cubed is one, it's going to be
an integer, and it's going to be

a factor of 7.

This only leaves us with four
possibilities, 1 - 1 Seven and

minus Seven, so we can try each
of them in turn.

You can see fairly quickly that
one and minus one don't work, so

let's try 7.

Rather than substituting 7 into
this expression and having to

workout 7 cubed and so on, what
I'm going to do is synthetically

divide by 7, because if Seven is
a route I'll end up with a 0 and

What's more, the division will
give me the quadratic that I'm

looking for. So let's
synthetically divide this

expression by 7.

The coefficient of X cubed is

one. Coefficient of X squared is

minus 6. The coefficient of X is
minus 6 and the constant is

minus Seven and were
synthetically dividing by 7.

Bring the one down 1 * 7 is
Seven 7 - 6 is one.

1 * 7 is 7 Seven and minus six

is one. 1 * 7 is 7.

7 and minus 70.

So 7 is indeed a route, and
the resulting quadratic is

X squared plus X plus one
equals not.

Now you'll find if you try to
solve it that the quadratic

equation X squared plus X plus

one equals not. Has no
real solutions, so the

only possible solution to
this cubic is X equals 7.

Certain basic identity's which
you may wish to learn can

help. In fact, rising both
cubics and quadratics. I'll

just give you 1 example.

We have the equation X cubed
plus three X squared.

Plus 3X plus
one equals 0.

1331 is the standard expansion
of X plus one cubed.

So the solution to this equation
is X plus one cubed equals 0.

We have The root minus
One X equals minus one

repeated three times.

If you can't find a factor
by these methods, then draw an

accurate graph of the cubic

expression. The points where it
crosses the X axis.

Will give you the solutions to

the equation. But their accuracy
will be limited to the accuracy

of your graph.

You might indeed find the graph
crosses the X axis at a point

that would suggest a factor. For
instance, if you draw craft a

graph that appears to cross the
X axis would say.

X equals 1/2. Then it's worth
trying to find out if X minus

1/2 is indeed a factor.

Let's look at the equation.

X cubed

Plus four X squared.

Plus X minus
5 equals 0.

Now this equation won't yield
affected by any of the methods

that I've discussed, so it's
time to draw a graph.

And this is the graph of Y
equals X cubed plus four X

squared plus 6 - 5.

It crosses the X axis at three
places, 1 near X equals minus 3,

one near X equals minus 2.

And one near X equals 1.

Pending on how accurate your
graph is, you may be able to

pinpoint it a bit closer than
this, and we get solutions.

X is approximately equal to

minus 3.2.
Minus 1.7

and not .9.

These solutions may be
accurate enough for your

needs, but if you require more
accurate answers then you

should use a numerical
algorithm using the

approximate answers. If you
obtain from the graph as a

starting point.