Synthetic division is a
shorthand, or shortcut, method of polynomial
division in the
special case of dividing by a linear factor -- and it only
works in this case. Synthetic division is generally used, however, not
for dividing out factors but for finding zeroes (or roots) of polynomials.
More about this later.

If you are given, say,
the polynomial equation y
= x^{2} + 5x + 6,
you can factor the polynomial as y
= (x + 3)(x + 2). Then
you can find the zeroes of y
by setting each factor equal to zero and solving. You will find that x
= 2 and x
= 3 are the two zeroes
of y.

You can, however, also
work backwards from the zeroes to find the originating polynomial. For
instance, if you are given that x
= 2 and x
= 3 are the zeroes
of a quadratic, then you know that x
+ 2 = 0, so x
+ 2 is a factor, and
x
+ 3 = 0, so x
+ 3 is a factor. Therefore,
you know that the quadratic must be of the form y
= a(x + 3)(x + 2).

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(The extra number "a"
in that last sentence is in there because, when you are working
backwards from
the zeroes, you don't know toward which quadratic you're working. For
any non-zero value of "a",
your quadratic will still have the same zeroes. But the issue of the value
of "a"
is just a technical consideration; as long as you see the relationship
between the zeroes and the factors, that's all you really need to know
for this lesson.)

Anyway, the above is a
long-winded way of saying that, if xn is a factor,
then x
= nis
a zero, and ifx = n
is a zero, then xn is a factor.
And this is the fact you use when you do synthetic division.

Let's look again at the
quadratic from above: y
= x^{2} + 5x + 6.
From the Rational
Roots Test, you
know that
1, 2, 3, and
6 are possible zeroes
of the quadratic. (And, from the factoring above, you know that the zeroes
are, in fact, 3
and 2.)
How would you use synthetic division to check the potential zeroes? Well,
think about how long
polynomial divison
works. If we guess that
x = 1 is a zero,
then this means that x
1 is a factor of
the quadratic. And if it's a factor, then it will divide out evenly; that
is, if we divide x^{2}
+ 5x + 6 by
x
1, we would get a
zero remainder. Let's check:

As expected (since we know
that x
1 is not a factor),
we got a non-zero remainder. What does this look like in synthetic division?
Copyright Elizabeth Stapel 2002-2011 All Rights Reserved

First, write the
coefficients ONLY inside an upside-down division symbol:

Make sure you leave
room inside, underneath the row of coefficients, to write another
row of numbers later.

Put the test zero,
x
= 1, at the left:

Take the first number
inside, representing the leading coefficient, and carry it down,
unchanged, to below the division symbol:

Multiply
this carry-down value by the test zero, and carry the result up
into the next column:

Add down
the column:

Multiply the previous
carry-down value by the test zero, and carry the new result up into
the last column:

Add down
the column:

This last carry-down
value is the remainder.

Comparing, you can see
that we got the same result from the synthetic division, the same quotient
(namely, 1x
+ 6) and
the same remainder at the end (namely, 12),
as when we did the long division:

The results are formatted
differently, but you should recognize that each format provided us with
the result, being a quotient of x
+ 6, and a remainder
of 12.

Content Continues Below

You already know
(from the factoring above) that x
+ 3 is a factor
of the polynomial, and therefore that x
= 3 is a zero.

Now compare the results of long division and synthetic division
when we use the factor x
+ 3 (for the
long division) and the zero x
= 3 (for the
synthetic division):

As you can see above, while
the results are formatted differently, the results are otherwise the same:

In the long division, I
divided by the factor x
+ 3, and arrived at
the result of x
+ 2 with a remainder
of zero. This means that x
+ 3 is a factor, and
that x
+ 2 is left after factoring
out the x
+ 3. Setting the factors
equal to zero, I get that x
= 3 and x
= 2 are the zeroes
of the quadratic.

In the synthetic division,
I divided by x
= 3, and arrived at
the same result of x
+ 2 with a remainder
of zero. Because the remainder is zero, this means that x
+ 3 is a factor and
x
= 3 is a zero. Also,
because of the zero remainder, x
+ 2 is the remaining
factor after division. Setting this equal to zero, I get that x
= 2 is the other zero
of the quadratic.

I will return to this relationship
between factors and zeroes throughout what follows; the two topics are
inextricably intertwined.

You can use the Mathway widget below to
practice synthetic division. Try the entered exercise, or type in
your own exercise. Then click the "paper-airplane" button to compare your answer
to Mathway's. (Or skip the widget and continue
with the lesson.)