Synthetic Division
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Key Questions

Answer:
Synthetic division is a way to divide a polynomial by a linear expression.
Explanation:
Suppose our problem is this:
#y=x^3+2x^2+3x6# Now, the main use of synthetic division is to find the roots or solutions to an equation.
The process for this serves to cut down on the gessing you have to do to find a value of x that makes the equation equal 0.
First, list the possible rational roots, by listing the factors of the constant (6) over the list of the factors of the lead coefficient (1).
#+# (1,2,3,6)/1Now, you can begin trying numbers. First, you simplify the equation to just the coefficients:
)Â¯Â¯1Â¯Â¯Â¯2Â¯Â¯Â¯Â¯3Â¯Â¯Â¯Â¯6Â¯Â¯And now, plug your possible rational roots in, one at a time, until one works. (I suggest doing 1 and 1 first, since they're the easiest)
1 )Â¯Â¯1Â¯Â¯Â¯2Â¯Â¯Â¯Â¯3Â¯Â¯Â¯Â¯6Â¯Â¯
#color(white)Â¯Â¯# Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯1 .First bring down the lead number (1)
1 )Â¯Â¯1Â¯Â¯Â¯2Â¯Â¯Â¯Â¯3Â¯Â¯Â¯Â¯6Â¯Â¯
#color(white)Â¯Â¯# Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯
#color(white)00# 12 . Now multiply that number by the divisor (1)
1 )Â¯Â¯1Â¯Â¯Â¯2Â¯Â¯Â¯Â¯3Â¯Â¯Â¯Â¯6Â¯Â¯
#color(white)Â¯Â¯# Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯
#color(white)00# 13 . Now place the product underneath the second number (2)
1 )Â¯Â¯1Â¯Â¯Â¯2Â¯Â¯Â¯Â¯3Â¯Â¯Â¯Â¯6Â¯Â¯
#color(white)ddots# #color(white)00# 1
#color(white)Â¯Â¯# Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯
#color(white)00# 1#color(white)00# 4 . Now add the two numbers together (2&1) and move the sum down
1 )Â¯Â¯1Â¯Â¯Â¯2Â¯Â¯Â¯Â¯3Â¯Â¯Â¯Â¯6Â¯Â¯
#color(white)ddots# #color(white)00# 1
#color(white)Â¯Â¯# Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯
#color(white)sum# 1#color(white)00# 35 . Now multiply the sum (3) by the divisor (1) and move it underneath the next value in the dividend
1 )Â¯Â¯1Â¯Â¯Â¯2Â¯Â¯Â¯3Â¯Â¯6Â¯Â¯
#color(white)ddots# #color(white)00# 1#color(white)00# 3
#color(white)Â¯Â¯# Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯
#color(white)sum# 1#color(white)00# 36 . Now add the two values together (3&3) and move the sum down
1 )Â¯Â¯1Â¯Â¯Â¯2Â¯Â¯Â¯3Â¯Â¯6Â¯Â¯
#color(white)ddots# #color(white)00# 1#color(white)00# 3
#color(white)Â¯Â¯# Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯
#color(white)sum# 1#color(white)00# 3#color(white)00# 67 . Now multiply the new sum (6) with the divisor (1) and move it underneath the next value in the dividend
1 )Â¯Â¯1Â¯Â¯Â¯2Â¯Â¯Â¯3Â¯Â¯6Â¯Â¯
#color(white)ddots# #color(white)00# 1#color(white)00# 3#color(white)00# 6
#color(white)Â¯Â¯# Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯
#color(white)sum# 1#color(white)00# 3#color(white)00# 68 . Now add together the two values (6&6) and move that sum down
1 )Â¯Â¯1Â¯Â¯Â¯2Â¯Â¯Â¯3Â¯Â¯6Â¯Â¯
#color(white)ddots# #color(white)00# 1#color(white)00# 3#color(white)00# 6
#color(white)Â¯Â¯# Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯
#color(white)sum# 1#color(white)00# 3#color(white)00# 6#color(white)00# 08 . Now you have the equation, 0=
#x^2+3x+6# , with the sums you found being the coeffiecients1 )Â¯Â¯1Â¯Â¯Â¯2Â¯Â¯Â¯3Â¯Â¯6Â¯Â¯
#color(white)ddots# #color(white)00# 1#color(white)00# 3#color(white)00# 6
#color(white)Â¯Â¯# Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯
#color(white)sum# 1#color(white)00# 3#color(white)00# 6#color(white)00# 0 
Here is a reasonable PreCalculus example of synthetic division to illustrate the concept.
Let's say you had:
#2x^4  3x^3  5x^2 + 3x + 8# Like Joan said, there is a trial and error aspect to this.
Look at all the coefficients, and think about what a common factor might be.
 If you don't get a zero remainder, then the factor doesn't really work and you should try again.
 If the possible factors are all used up, maybe it isn't factorable.
Here, factors you might try include the ones that correspond to the fourth order coefficient (
#2# ) and the zeroth order coefficient (#8# ).#8# has factors of#1, 2, 4# , and#8# .#2# has factors of#1# and#2# .
So the possible factors might be said to be
#pmp/q# , where#p# consists of the factors of the highest degree coefficient, and#q# consists of the factors of the zeroth degree coefficient.You can therefore have factors of:
#pm[1, 2, 4, 8, 1/2]# So you can try all of these (
#2/2# ,#4/2# , and#8/2# are duplicates). Recall that if#a# is used as what is written in the synthetic division process on the left corner, it corresponds to#x+a# .We will use
#1# here. I tend to try#1# and#1# first, and go up in value, and try the fractions last.#ul(1)" "2" "3" "5" "" "3" "" "8# Drop down the
#2# , and multiply by the#1# to get#2# .#ul(1 )" "2" "3" "5" "" "3" "" "8#
#ul(" "" "" "" "2" "" "" "" "" "" "" "" ")#
#" "" "color(white)(.)2# Add
#3# and#2# , then multiply the resultant#5# by#1# again.#ul(1 )" "2" "3" "5" "" "3" "" "8#
#ul(" "" "" "" "2" "" "5" "" "" "" "" "" ")#
#" "" "color(white)(.)2" "5# Repeat until you're done.
Add
#3# and#2# , then multiply the resultant#1# by#1# again.#ul(1 )" "2" "3" "5" "" "3" "" "8#
#ul(" "" "" "" "2" "" "5" "color(white)(.)" "0color(white)(.)" "3" ")#
#" "" "color(white)(.)2" "5" "" "0" "" "3" "" "5# Your answer here happens to be this, where 2 corresponds to
#2x^3# , since you divided a fourthorder polynomial by a firstorder polynomial.So, one way to express the result is:
#(2x^4  3x^3  5x^2 + 3x + 8)/(x+1)# #= color(blue)(overbrace(2x^3  5x^2 + 0x + 3)^"Quotient Term" + overbrace(5/(x+1))^"Remainder Term")# where the
#5/(x+1)# was written by saying that the last value below the horizontal bar (below#2, 5, 0, 3# ), being#5# , is divided by the#x pm a# equation such that#x pm a = 0# . So,#x+1# indicates that the factor we have just used is#1# .(Naturally, if the remainder is
#0# , you do not have the remainder fraction at the end.) 
The remainder in synthetic division could be written as a fraction or with R written in front of it. If writing as a fraction, the remainder is in the numerator of the fraction and the divisor is in the denominator.
For example:
Dividing
#x^2 + 3x  12 # by#x  3# :When you use Synthetic Division, the answer is
#x + 6# with a remainder of 6.
Here are two ways you can write the answer:
x + 6 R 6

#x + 6 + 6/(x  3)#


Please see the video below for a detail explanation.

Answer:
You use synthetic division when dividing polynomials to find the zeroes. By checking your remainder in your preferred method (synthetic division or long division) , you can use the remainder theorem.
Explanation:
Thus, Synthetic division is a shortcut instead of using the long division of polynomial. Again, you can check the remainder by using the remainder theorem. If the answer is zero, then it is a factor theorem. Don't be confused about factor theorem and remainder theorem, the two has the same method.
E.g.
#find f(3).#
#f(x) = x^3 â€“ 5x^2 + 3x + 7#
#x=3#
then by checking, apply the remainder theorem
#f(3)=(3)^35(3)^2+3(3)+7 # substitute the given value to x.
#=2# Site (http://www.mesacc.edu/~scotz47781/mat120/notes/divide_poly/remainder/remainder_thm.html)