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Algebraic long division
Divide 2x³ + 3x² - x + 1 by x + 2
x + 2 is the
divisor
x  2 2x 3  3x 2  x  1
The quotient
will be here.
2x³ + 3x² - x + 1
is the dividend
Algebraic long division
First divide the first term of the dividend, 2x³,
by x (the first term of the divisor).
This gives 2x².
This will be the
first term of
the quotient.
2x 2
x  2 2x 3  3x 2  x  1
Algebraic long division
Now multiply
2x² by x + 2
and subtract
2x 2
x  2 2x 3  3x 2  x  1
2x 3  4x 2
x 2
Algebraic long division
2x 2
x  2 2x 3  3x 2  x  1
Bring down the
next term, -x.
2x 3  4x 2
x 2 x
Algebraic long division
Now divide –x²,
2x 2 x
the first term of x  2 2x 3  3x 2  x  1
–x² - x, by x, the
3
2
2
x

4
x
first term of the
divisor
x 2 x
which gives –x.
Algebraic long division
2x 2 x
x  2 2x 3  3x 2  x  1
2x 3  4x 2
Multiply –x by x + 2
and subtract
x 2 x
x 2  2x
x
Algebraic long division
2x 2 x
x  2 2x 3  3x 2  x  1
2x 3  4x 2
Bring down the
next term, 1
x 2 x
x 2  2x
x 1
Algebraic long division
2x 2 x 1
x  2 2x 3  3x 2  x  1
Divide x, the first
term of x + 1, by x,
the first term of
the divisor
which gives 1
2x 3  4x 2
x 2 x
x 2  2x
x 1
Algebraic long division
2x 2 x 1
x  2 2x 3  3x 2  x  1
2x 3  4x 2
x 2 x
x 2  2x
Multiply x + 2 by 1
and subtract
x 1
x 2
1
Algebraic long division
2x 2 x 1
x  2 2x 3  3x 2  x  1
The quotient is
2x² - x + 1
2x 3  4x 2
The remainder is –1.
x 2 x
x 2  2x
x 1
x 2
1
You try one.
x  3 2x  10x  12
2
Do the next two
Synthetic Division
There is a shortcut for long division as long as the divisor is x – k
where k is some number. (Can't have any powers on x).
Set divisor = 0 and
3
2
1 x  6 x  8x  2
solve. Put answer
here.
x3
x + 3 = 0 so x = - 3
-3
1
6
8
-2
up3these
Bring
number
down
below
Addupthese
line up
- 3 firstAdd
- 9theseAdd
Multiply
Multiply
these
these and
and
2 +3 x - 1
This is the remainder
1
x
1
put
answer
put answer
above
line
above line
Put
variables
back
in (one
was of
divided
outthe
in
Sonext
the
Listanswer
all
coefficients
is:
(numbers
in xfront
x's) and
in
in next process so first number is one less power than
2 top. If a term is missing, put in a 0.
constant along the
column
column original problem).
1
x  3x  1 
x3
Let's try another Synthetic Division
Set divisor = 0 and
solve. Put answer
here.
4
1
0 x3
0x
1 x  4x  6
4
2
x4
x - 4 = 0 so x = 4
0
-4
0
6
up48
Bring
number
down
these
below
Add
upthese
line
Add
up these up
4 firstAdd
16theseAdd
192
Multiply
Multiply
Multiply
these
these and
and
3 + 4 x2 + 12 x + 48 198
This is the
these
and
1
x
put
answer
put answer
remainder
put
answer
above
line
above line Now put variables back in (remember one x was
above
lineanswer
Sonext
the
List
all coefficients
is:
(numbers in front of x's) and the
in
in next divided out 3in process2so first number is one less
in next
constant along the top. Don't forget the 0's for missing
column
column power than original problem so x3).
column
terms.
198
x  4 x  12 x  48 
x4
Let's try a problem where we factor the polynomial
completely given one of its factors.
4 x 3  8 x 2  25 x  50
-2
4
factor : x  2
You want to divide
the factor into the
polynomial so set
divisor = 0 and solve
for first number.
8 -25 -50
up
Bring
number
down
below
Addupthese
line up
- 8 firstAdd
0theseAdd
50these
Multiply
Multiply
these
No remainder so x + 2
these and
and
2
4 x + 0 x - 25
0
put
IS a factor because it
put answer
answer
above
line
divided in evenly
above line
Put
variables
back
in
(one
x
was
divided
outthe
in
Sonext
the
Listanswer
all coefficients
is the divisor
(numbers
times in
thefront
quotient:
of x's) and
in
in next
process
sothe
first
number
is one
less power
You could
check
this
byIf a term
constant
along
top.
is missing,
putthan
in a 0.
column
2
column
original
problem).
multiplying
them out
and getting
original polynomial
x  24 x
 25

You try one
3x  17 x  15 x  25
x 5
3
2
Try another one
10 x  50 x  800
x6
4
3
Do #’s 5 & 7
REMAINDER THEOREM
When polynomial f(x) is divided by x – a,
the remainder is f(a)
f(x) = 2x2 – 3x + 4
Divide the polynomial by x – 2
2
2
-3
4
2 1
4
2
6
Find f(2)
f(2) = 2(2)2 – 3(2) + 4
f(2) = 8 – 6 + 4
f(2) = 6
REMAINDER THEOREM
When synthetic division is used to
evaluate a function, it is called
SYNTHETIC SUBSTITUTION.
Try this one:
Remember – Some terms are missing
f(x) = 3x5 – 4x3 + 5x - 3
Find f(-3)
Do #’s 8 & 9
FACTOR THEOREM
The binomial x – a is a factor of the
polynomial f(x) if and only if f(a) = 0.
REMAINDER AND FACTOR
THEOREMS
Is x – 2 a factor of x3 – 3x2 – 4x + 12
2
1
-3
2
1 -1
-4 12
-2 -12
-6 0
Yes, it is a factor,
since f(2) = 0.
Can you find the two remaining factors?
REMAINDER AND FACTOR
THEOREMS
(x + 3)( ? )( ? ) = x3 – x2 – 17x - 15
Find the two unknown ( ? ) quantities.
Find all the zeros. One factor has
been given.
f ( x)  x  9 x  23x  15; x  5
3
2
Find all the zeros. One factor has
been given.
f ( x)  x  x  14 x  24; x  3
3
2
Find all the zeros. One factor has
been given.
f ( x)  x 4  3x3  13x 2  15x; x  3
You try #’s 10 - 13