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Algebraic long division
Divide 2x³ + 3x² - x + 1 by x + 2
3 22 2 3 1x x x x x + 2 is the divisor
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).
3 22 2 3 1x x x x
22xThis gives 2x². This will be the first term of the quotient.
Algebraic long division
Now multiply 2x² by x + 2
3 22 2 3 1x x x x 3 22 4x x
22x
2xand subtract
Algebraic long division
Bring down the next term, -x.
3 22 2 3 1x x x x 3 22 4x x
22x
2x x
Algebraic long division
Now divide –x², the first term of –x² - x, by x, the first term of the divisor
3 22 2 3 1x x x x 3 22 4x x
22x
2x x
x
which gives –x.
Algebraic long division
Multiply –x by x + 2
3 22 2 3 1x x x x 3 22 4x x
22x
2x x
x
2 2x x
xand subtract
Algebraic long division
Bring down the next term, 1 x
3 22 2 3 1x x x x 3 22 4x x
22x
2x x
x
2 2x x
1
Algebraic long division
Divide x, the first term of x + 1, by x, the first term of the divisor
13 22 2 3 1x x x x 3 22 4x x
22x
2x x
x
2 2x x
x 1which gives 1
Algebraic long division
Multiply x + 2 by 1
3 22 2 3 1x x x x 3 22 4x x
22x
2x x
x
2 2x x
x
1
12x 1and subtract
Algebraic long division
The remainder is –1.
3 22 2 3 1x x x x 3 22 4x x
22x
2x x
x
2 2x x
x
1
12x 1
The quotient is 2x² - x + 1
You try one.
23 2 10 12x x x
Do the next two
List all coefficients (numbers in front of x's) and the constant along the top. If a term is missing, put in a 0.
1Set divisor = 0 and solve. Put answer here.
x + 3 = 0 so x = - 3
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).
3
286 23
x
xxx
- 3 1 6 8 -2
1
Bring first number down below lineMultiply these and
put answer above line
in next column
- 3 Add these up
3Multiply these and
put answer above line
in next column
- 9 Add these up
- 1
3
1
Multiply these and
put answer above line
in next column
Add these up
This is the remainder
Put variables back in (one x was divided out in process so first number is one less power than original problem).
x2 + x
So the answer is:
3
1132
xxx
List all coefficients (numbers in front of x's) and the constant along the top. Don't forget the 0's for missing terms.
1Set divisor = 0 and solve. Put answer here.
x - 4 = 0 so x = 4
Let's try another Synthetic Division
4
64 24
x
xx
4 1 0 - 4 0 6
1
Bring first number down below lineMultiply these and
put answer above line
in next column
4 Add these up
4Multiply these and
put answer above line
in next column
16 Add these up
12
48
48
Multiply these and
put answer above line
in next column
Add these up
This is the remainder
Now put variables back in (remember one x was divided out in process so first number is one less power than original problem so x3).
x3 + x2 + x +
So the answer is:
4
19848124 23
xxxx
0 x3 0 x
Multiply these and
put answer above line
in next column
192
198
Add these up
List all coefficients (numbers in front of x's) and the constant along the top. If a term is missing, put in a 0.
You want to divide the factor into the polynomial so set divisor = 0 and solve for first number.
Let's try a problem where we factor the polynomial completely given one of its factors.
502584 23 xxx
- 2 4 8 -25 -50
4
Bring first number down below lineMultiply these and
put answer above line
in next column
- 8 Add these up
0Multiply these and
put answer above line
in next column
0 Add these up
- 25
50
0
Multiply these and
put answer above line
in next column
Add these up
No remainder so x + 2 IS a factor because it
divided in evenlyPut variables back in (one x was divided out in process so first number is one less power than original problem).
x2 + x
So the answer is the divisor times the quotient:
2542 2 xx
2 :factor x
You could check this by multiplying them out and
getting original polynomial
You try one
3 23 17 15 25
5
x x x
x
Try another one4 310 50 800
6
x x
x
Do #’s 5 & 7
REMAINDER THEOREMWhen polynomial f(x) is divided by x – a,
the remainder is f(a)
f(x) = 2x2 – 3x + 4
2 2 -3 4
2 4 1
2 6
Divide the polynomial by x – 2
Find f(2)
f(2) = 2(2)2 – 3(2) + 4f(2) = 8 – 6 + 4f(2) = 6
REMAINDER THEOREM
f(x) = 3x5 – 4x3 + 5x - 3
Find f(-3)
Try this one:Remember – Some terms are missing
When synthetic division is used to evaluate a function, it is called
SYNTHETIC SUBSTITUTION.
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 -4 12
1 2
-1 -2 -6
-12 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.
3 2( ) 9 23 15; 5f x x x x x
Find all the zeros. One factor has been given.
3 2( ) 14 24; 3f x x x x x
Find all the zeros. One factor has been given.
4 3 2( ) 3 13 15 ; 3f x x x x x x
You try #’s 10 - 13
