>> Here we are going to show
that a given value is a zero of P
of x and then find all other
zeros of P of x. We're given P
of x equals x cubed minus 3 x
squared minus 6x plus 18,
and the given zero is 3.
So we're going to divide,
using 3 as our divisor,
and we'll be dividing
into x cubed minus 3 x
squared minus 6x plus 18.
We use the coefficients,
in synthetic division:
1 will be the leading
coefficient in the quotient,
3 times 1 is 3, negative 3
plus 3 is 0, 0 times 3 is 0,
0 plus negative 6 is negative 6,
and negative 6 times
3 is negative 18.
Adding 18 with negative
18 we get 0, as expected,
because P of 3 must equal 0.
So at this point we
know that our P of x
is x minus 3 times a quadratic.
The second factor is a
second degree factor
because x cubed divided.. or a
third degree polynomial divided
by first degree polynomial,
will have to be a
second degree polynomial.
Therefore, in our quotient,
negative 6 is the x to the 0,
0 is the coefficient
of x to the first,
and 1 is the coefficient
of x squared.
Now, that means that
our polynomial..
our quadratic polynomial.. is
x squared plus 0x minus 6.
In terms of the quadratic
formula, the 1 would be the a,
the 0 would be the b, and the
negative 6 would be the c.
So if we use the quadratic
formula we would get x equals
the opposite of b plus
or minus the square root
of, b squared minus
4ac, all over 2a.
And so we're going to have plus
or minus the square root of 24,
notice that the two negatives
make a positive, all over 2.
Now this can be simplified
as plus
or minus the square root 4,
times the square root 6, over 2;
or plus or minus 2
square root 6, all over 2.
The 2's cancel to give us plus
or minus the square root of 6.
So the other zeros,
are square root 6
and minus square root of 6.
These are in addition
to 3, the given zero.
Now in this case we had
another option as far
as finding the zeros
of x squared minus 6.
Since x squared minus 6 equals
0, we'd have x squared equals 6,
and using the square
root principle,
we would have also got plus
or minus the square root of 6.
But in general we'll be
using the quadratic formula.
Alright, so let's go
to the second example,
which is a fourth
degree polynomial,
and it doesn't matter which
of the zeros we start with.
Let's start with one third,
[Background speaker] take them
in order, and there's
no missing 0's.
So we have 3, negative 7,
negative 37, negative
17, and 10.
3 is going to be the leading
term in this first quotient..
the leading coefficient in the
first term of this quotient,
3 times one third is 1,
negative 7 plus 1 is negative 6,
negative 6 times one
third is negative 2,
negative 37 plus
negative 2 is negative 39,
one third of negative
39 is negative 13.
When we add negative 17, we
get negative 30, and one third
of negative 30 is negative 10.
When we add the 10 and
the negative 10, we get 0.
So P of one third is 0,
that does not surprise us,
because one third
we were told is a zero.
Now, at this point we
have the coefficients
of a third degree
polynomial as the quotient.
The.. it would be 3 x to the
third minus 6 x squared,
minus 39x, minus 30.
But instead of writing
that out what we're going
to do is we're going to use the
quotient as the next dividend
when we divide by
the negative 2.
Remember this time we were
given two zeros and so we now..
we know that P of x equals x
minus one third times, 3 x cubed,
minus 6 x squared, minus 39x,
minus 30. And now we're going
to go to the next
step, where we're going
to also have a factor of x minus
negative 2, which is x plus 2,
and a second degree polynomial.
So the quotient for this second
division will be second degree.
So 3 will be the leading
coefficient in the quadratic,
times negative 2 is negative 6.
Add negative 6 with negative 6,
gives us negative 12. Negative
12 times negative 2 is positive
24. Add negative 39 is
negative 15 times negative 2,
is positive 30.
And now this confirms that
P of negative 2 is also 0.
We're left with the second
degree polynomial 3 x squared,
minus 12x, minus 15.
So in terms of the
quadratic formula 3 is our a,
negative 12 is our b,
and negative 15 is our c.
So we would have that x equals
12 plus or minus the square root
of b squared.. which is 144..
minus 4 times a, times c,
all over 2a, which will be 6.
So this is 12 plus or minus,
and it's the square root
of 144 plus 4 times 45,
which is 180, all over 6.
Which is 12 plus or minus
the square root we have..
in the 1's place of 4, 10's
would be a 2.. 324 all over 6.
Now, 324 is not a number
that we readily recognize,
but let's factor out. We
see that it divisible by 4.
So let's factor out the square
root of 4, which would be 2,
and when we do that 4
goes into 32, 8 times,
and 4 goes into 4 once.
So it was 4 times 81.
So the square root of 4 times
the square root of 81 is going
to be 2 times 9, or 18.
So 12 plus or minus 18 over 6.
This gives us the zeros, the
additional zeros, are 30 over 6,
and negative 6 over 6.
These reduce to 5
and negative 1.
