# Use Synthetic Division to Solve What is the Quotient

**Use Synthetic Division to Solve What is the Quotient**

## Constructed Division: The Process

## What is constructed division?

Synthetic division is a shorthand, or shortcut, method of polynomial sectionalization in the special example of dividing by a linear factor — and it

*only*

works in this case. Synthetic partitioning is by and large used, however, not for dividing out factors but for finding zeroes (or roots) of polynomials. More about this later.

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## How are polynomial zeroes and factors related?

If you are given, say, the polynomial equation

*y*

=*10*

^{2} + 5x + vi, you can factor the polynomial equally

*y*

= (*x* + iii)(*x* + 2). Then you can find the zeroes of

*y*

by setting each gene equal to zero and solving. You will find that the two zeroes of the polynomial are

*10* = −two

and

*ten* = −3.

You can, yet, besides work backwards from the zeroes to discover the originating polynomial. For instance, if y’all are given that

*x* = −2

and

*x* = −iii

are the zeroes of a quadratic, and so you know that

*x* + 2 = 0, so

*x* + 2

is a factor, and

*x* + 3 = 0, and then

*x* + iii

is a factor. Therefore, you know that the quadratic must exist of the form

*y*

=

*a*(*x* + 3)(*ten* + ii).

(The actress number “

*a*

” in that last sentence is in there considering, when you are working backwards from the zeroes, you don’t know toward which quadratic y’all’re working. For any non-zero value of “

*a*

“, your quadratic volition notwithstanding have the same zeroes. But the issue of the value of “

*a*

” is merely a technical consideration; as long as you lot see the relationship between the zeroes and the factors, that’s all you lot really need to know for this lesson.)

Anyhow, the above is a long-winded style of saying that, if

*x* −*north*

is a factor, and so

*10* =*due north*

is a nada, and if

*x* =*north*

is a nil, and then

*x* −*north*

is a factor. And this is the fact you use when you lot do synthetic division.

Let’s look again at the quadratic from above:

*y*

=*ten*

^{2} + 5*x* + 6. From the Rational Roots Test, we know that

± 1, 2, 3,

and

6

are possible zeroes of the quadratic. (And, from the factoring above, we know that the zeroes are, in fact,

−3

and

−2.) How would y’all use synthetic division to check the potential zeroes?

Well, think near how long polynomial divison works. If I were to guess that

*x* = 1

is a nix, so this means that

*x* − one

is a cistron of the quadratic. And if it’s a cistron, and then information technology will carve up out evenly; that is, if we dissever

*x*

^{two} + 5*x* + 6

past

*10* − one, we would go a cypher residuum. Let’southward cheque:

Equally expected (since nosotros know that

*x* − 1

is non a factor), nosotros got a non-nada residual. What does this look like in synthetic segmentation?

## How do you practice synthetic sectionalization?

First, take the polynomial (in our example,

*ten*

^{2} + 5*x* + 6), and write the coefficients ONLY inside an upside-downwards segmentation-type symbol:

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

Put the examination zero, in our case

*10* = 1, at the left, next to the (top) row of numbers:

Accept the commencement number that’south on the inside, the number that represents the polynomial’s leading coefficient, and carry it down, unchanged, to below the division symbol:

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

Add downward the column:

Multiply the previous acquit-downwards value by the test zero, and deport the new result up into the last column:

Add down the column:

This last comport-down value is the remainder.

Comparing, yous can see that we got the same issue from the synthetic partition, the same caliber (namely,

i*x* + 6) and the same residue at the stop (namely,

12), as when we did the long division:

The results are formatted differently, merely you should recognize that each format provided united states of america with the same outcome, being a quotient of

*x* + 6, and a rest of

12.

Nosotros already know (from the factoring to a higher place) that

*10* + three

is a factor of the polynomial, and therefore that

*10* = −3

is a zero.

Now allow’s compare the results of long division and synthetic sectionalisation when we employ the factor

*x* + 3

(for the long partition) and the nada

*x* = −three

(for the synthetic division):

Equally you can encounter above, while the results are formatted differently, the results are otherwise the same:

In the long division, I divided by the cistron

*x* + 3, and arrived at the issue of

*ten* + 2

with a balance of zero. This means that

*x* + 3

is a cistron, and that

*x* + 2

is left after factoring out the

*10* + 3. Setting the factors equal to zero, I get that

*ten* = −iii

and

*10* = −2

are the zeroes of the quadratic.

In the synthetic sectionalisation, I divided past

*x* = −iii, and arrived at the aforementioned result of

*x* + 2

with a residual of null. Considering the remainder is zero, this ways that

*x* + 3

is a factor and

*x* = −three

is a zero. Also, because of the aught rest,

*x* + 2

is the remaining factor after partitioning. Setting this equal to zero, I become that

*ten* = −2

is the other aught of the quadratic.

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

## Use Synthetic Division to Solve What is the Quotient

Source: https://www.purplemath.com/modules/synthdiv.htm