Synthetic Division
See how to do the synthetic division
to factor a polynomial
and solve a polynomial equation/inequality.
12 examples and their solutions.
Synthetic Division
Example
x3 - 7x + 11x - 2
Solution x3 - 7x + 11x - 2
10-711 - [1]
2 - [2]
↓
10-711
224-6
12-35 - [3]
(given) = x2 + 2x - 3 + 5x - 2 - [4]
10-711 - [1]
2 - [2]
↓
10-711
224-6
12-35 - [3]
(given) = x2 + 2x - 3 + 5x - 2 - [4]
[1]
Numerator: 1x3 + 0x2 - 7x + 11
Write the coefficients:
1, 0, -7, 11.
Write the coefficients:
1, 0, -7, 11.
[2]
2: zero of x - 2
[3]
↓: (+)
↗: ×2
Start from the left top 1.
↓: 1 + 0 = 1
↗: 1 × 2 = 2
↓: 0 + 2 = 2
↗: 2 × 2 = 4
↓: -7 + 4 = -3
↗: -3 × 2 = -6
↓: 11 + (-6) = 5
Draw an L shape form
that covers 5.
↗: ×2
Start from the left top 1.
↓: 1 + 0 = 1
↗: 1 × 2 = 2
↓: 0 + 2 = 2
↗: 2 × 2 = 4
↓: -7 + 4 = -3
↗: -3 × 2 = -6
↓: 11 + (-6) = 5
Draw an L shape form
that covers 5.
[4]
1, 2, -3: coefficients of the quotient
→ (quotient) = x2 + 2x - 3
2: zero of x - 2
5: remainder
→ (quotient) = x2 + 2x - 3
2: zero of x - 2
5: remainder
Close
Example
2x4 + x3 - 5x2 + 3x + 4x + 1
Solution 2x4 + x3 - 5x2 + 3x + 4x + 1
21-534
-1-214-7
2-1-47-3 - [1]
(given) = 2x3 - x2 - 4x + 7 - 3x + 1
21-534
-1-214-7
2-1-47-3 - [1]
(given) = 2x3 - x2 - 4x + 7 - 3x + 1
[1]
↓: (+)
↗: ×(-1)
Start from the left top 2.
↓: 2 + 0 = 2
↗: 2 × (-1) = -2
↓: 1 + (-2) = -1
↗: -1 × (-1) = 1
↓: -5 + 1 = -4
↗: -4 × (-1) = 4
↓: 3 + 4 = 7
↗: 7 × (-1) = -7
↓: 4 + (-7) = -3
↗: ×(-1)
Start from the left top 2.
↓: 2 + 0 = 2
↗: 2 × (-1) = -2
↓: 1 + (-2) = -1
↗: -1 × (-1) = 1
↓: -5 + 1 = -4
↗: -4 × (-1) = 4
↓: 3 + 4 = 7
↗: 7 × (-1) = -7
↓: 4 + (-7) = -3
Close
Remainder Theorem
Formula
The remainder of f(x)x - a
→ f(a)
By using the remainder theorem,→ f(a)
you can find the remainder of f(x)/(x - a)
without doing the whole division.
Just find f(a).
(a: zero of x - a)
Example
x3 - 7x + 11x - 2
Remainder?
Solution Remainder?
x3 - 7x + 11x - 2
(remainder) = f(2)
= 23 - 7⋅2 + 11
= 8 - 14 + 11
= 8 - 3
= 5
(remainder) = f(2)
= 23 - 7⋅2 + 11
= 8 - 14 + 11
= 8 - 3
= 5
Close
Example
(2x4 + x3 - 5x2 + 3x + 4) ÷ (x + 1)
Remainder?
Solution Remainder?
(2x4 + x3 - 5x2 + 3x + 4) ÷ (x + 1)
(remainder) = f(-1)
= 2⋅(-1)4 + (-1)3 - 5⋅(-1)2 + 3⋅(-1) + 4
= 2⋅1 + (-1) - 5⋅1 + 3⋅(-1) + 4
= 2 - 1 - 5 - 3 + 4
= 1 - 8 + 4
= -7 + 4
= -3
(remainder) = f(-1)
= 2⋅(-1)4 + (-1)3 - 5⋅(-1)2 + 3⋅(-1) + 4
= 2⋅1 + (-1) - 5⋅1 + 3⋅(-1) + 4
= 2 - 1 - 5 - 3 + 4
= 1 - 8 + 4
= -7 + 4
= -3
Close
Synthetic Substitution
Formula
f(x)
a...
f(a)
When doing the synthetic division,a...
f(a)
the remainder is the right bottom number.
And by the remainder theorem,
the remainder of f(x)/(x - a) is f(a).
So (right bottom number) = (remainder) = f(a).
So you can find f(a)
by finding the remainder of the synthetic division.
When f(x) is complex,
this method will save your time.
Example
f(x) = x4 - 9x3 + 15x2 + 3x - 62
f(7)?
Solution f(7)?
1-9153-62
77-14770
1-21108 = f(7)
∴ f(7) = 8
77-14770
1-21108 = f(7)
∴ f(7) = 8
Factor Theorem
Theorem
If f(a) = 0,
then f(x) = (x - a)(quotient).
If f(a) = 0,then f(x) = (x - a)(quotient).
then, by the remainder theorem,
the remainder of f(x)/(x - a) is 0.
So f(x) = (x - a)(quotient).
This is the factor theorem.
If f(a) = 0, f(b) = 0, ... ,
then f(x) = (x - a)(x - b)(quotient).
So, if a and b are the zeros of f(x),then f(x) = (x - a)(x - b)(quotient).
then f(x) = (x - a)(x - b)(quotient).
Formula
f(x)
a...
0
b...
(quotient)0
f(x) = (x - a)(x - b)(quotient)
Let's mix the factor theorema...
0
b...
(quotient)0
f(x) = (x - a)(x - b)(quotient)
and the synthetic division.
To factor f(x),
do the synthetic division
and find the zeros [a, b, ...]
that makes the remainder 0.
Example
x3 + 3x2 - 16x + 12
Solution 13-1612
114-12
14-120 - [1]
2212
160 - [2]
f(x) = (x - 1)(x - 2)(x + 6)
114-12
14-120 - [1]
2212
160 - [2]
f(x) = (x - 1)(x - 2)(x + 6)
[1]
Pick a number
that seems to make the remainder 0.
1 seems to be good.
[when (sum of the coefficients) = 0]
The remainder is 0.
So you picked the right number.
that seems to make the remainder 0.
1 seems to be good.
[when (sum of the coefficients) = 0]
The remainder is 0.
So you picked the right number.
[2]
1, 4, -12
→ x2 + 4x - 12
This seems to be factorable.
So do the synthetic division again.
Pick a number
that seems to make the remainder 0.
2 seems to be good.
The remainder is 0.
So you picked the right number.
→ x2 + 4x - 12
This seems to be factorable.
So do the synthetic division again.
Pick a number
that seems to make the remainder 0.
2 seems to be good.
The remainder is 0.
So you picked the right number.
Close
Example
x4 - 2x3 - 4x2 - 2x + 3
Solution 1-2-4-23
11-1-5-3
1-1-5-30
-1-123
1-2-30
-1-1-3
1-30
f(x) = (x - 1)(x + 1)2(x - 3)
11-1-5-3
1-1-5-30
-1-123
1-2-30
-1-1-3
1-30
f(x) = (x - 1)(x + 1)2(x - 3)
Close
Example
x4 + x3 - 5x2 + x - 6
Solution 11-51-6
22626
13130
-3-30-3
1010 - [1]
f(x) = (x - 2)(x + 3)(x2 + 1)
22626
13130
-3-30-3
1010 - [1]
f(x) = (x - 2)(x + 3)(x2 + 1)
[1]
1, 0, 1
→ x2 + 1
x2 + 1 is always (+).
(x2 ≥ 0
x2 + 1 ≥ 1)
So x2 + 1 has no zeros.
So x2 + 1 is not factorable.
So set (quotient) = (x2 + 1).
To check if a quadratic expression has zero(s),
find the discriminant D.
→ x2 + 1
x2 + 1 is always (+).
(x2 ≥ 0
x2 + 1 ≥ 1)
So x2 + 1 has no zeros.
So x2 + 1 is not factorable.
So set (quotient) = (x2 + 1).
To check if a quadratic expression has zero(s),
find the discriminant D.
Close
Polynomial Equation
Example
x4 + 4x3 - 3x2 - 10x + 8 = 0
Solution 14-3-10+8
1152-8
152-80
1168
1680
-2-2-8
140
(x - 1)2(x + 2)(x + 4) = 0
x = 1, -2, -4
= -4, -2, 1
1152-8
152-80
1168
1680
-2-2-8
140
(x - 1)2(x + 2)(x + 4) = 0
x = 1, -2, -4
= -4, -2, 1
Close
Polynomial Inequality
Formula
(x - a)odd(x - a)even
then y = f(x) passes through the x-axis
at x = a.
If f(x) = (x - a)even(quotient),
then y = f(x) bounces off the x-axis
at x = a.
Use this formula
when graphing y = f(x) on the x-axis
to solve a polynomial inequality.
Example
x4 - x2 < 0
Solution x4 - x2 < 0
x2(x2 - 1) < 0
x2(x + 1)(x - 1) < 0 - [1]
(x + 1)x2(x - 1) < 0
x2(x2 - 1) < 0
x2(x + 1)(x - 1) < 0 - [1]
(x + 1)x2(x - 1) < 0
[2]
Draw y = (x + 1)x2(x - 1)
on the x-axis.
The highest order term is x4.
It's coefficient is 1: (+).
So start drawing the graph
from the right top of the x-axis.
(x - 1) = (x - 1)1 = (x - 1)odd
So the graph passes through the x-axis
at x = 1.
x2 = xeven
So the graph bounces off the x-axis
at x = 0.
(x + 1) = (x + 1)1 = (x + 1)odd
So the graph passes through the x-axis
at x = -1.
on the x-axis.
The highest order term is x4.
It's coefficient is 1: (+).
So start drawing the graph
from the right top of the x-axis.
(x - 1) = (x - 1)1 = (x - 1)odd
So the graph passes through the x-axis
at x = 1.
x2 = xeven
So the graph bounces off the x-axis
at x = 0.
(x + 1) = (x + 1)1 = (x + 1)odd
So the graph passes through the x-axis
at x = -1.
↓
-1 < x < 0, 0 < x < 1
[3]
(x + 1)x2(x - 1) < 0
So color the region
where the graph is below the x-axis.
The inequality sign, <, does not include '='.
So draw empty circles on x = -1, 0, 1.
So color the region
where the graph is below the x-axis.
The inequality sign, <, does not include '='.
So draw empty circles on x = -1, 0, 1.
Close
Example
x3 + x2 - 10x + 8 ≥ 0
Solution 11-108
112-8
12-80
228
140
(x - 1)(x - 2)(x + 4) ≥ 0
(x + 4)(x - 1)(x - 2) ≥ 0
112-8
12-80
228
140
(x - 1)(x - 2)(x + 4) ≥ 0
(x + 4)(x - 1)(x - 2) ≥ 0
[1]
Draw y = (x + 4)(x - 1)(x - 2)
on the x-axis.
The highest order term is x3.
It's coefficient is 1: (+).
So start drawing the graph
from the right top of the x-axis.
(x - 2) = (x - 2)1 = (x - 2)odd
So the graph passes through the x-axis
at x = 2.
(x - 1) = (x - 1)1 = (x - 1)odd
So the graph passes through the x-axis
at x = 1.
(x + 4) = (x + 4)1 = (x + 4)odd
So the graph passes through the x-axis
at x = -4.
on the x-axis.
The highest order term is x3.
It's coefficient is 1: (+).
So start drawing the graph
from the right top of the x-axis.
(x - 2) = (x - 2)1 = (x - 2)odd
So the graph passes through the x-axis
at x = 2.
(x - 1) = (x - 1)1 = (x - 1)odd
So the graph passes through the x-axis
at x = 1.
(x + 4) = (x + 4)1 = (x + 4)odd
So the graph passes through the x-axis
at x = -4.
↓
-4 ≤ x ≤ 1, x ≥ 2
[2]
(x + 4)(x - 1)(x - 2) ≥ 0
So color the region
where the graph is above the x-axis.
The inequality sign, ≥, does include '='.
So draw full circles on x = -4, 1, 2.
So color the region
where the graph is above the x-axis.
The inequality sign, ≥, does include '='.
So draw full circles on x = -4, 1, 2.
Close
Example
x4 + x3 - 5x2 + 3x ≤ 0
Solution x4 + x3 - 5x2 + 3x ≤ 0
x(x3 + x2 - 5x + 3) ≤ 0
11-53
112-3
12-30
113
130
x(x - 1)2(x + 3) ≤ 0
(x + 3)x(x - 1)2 ≤ 0
x(x3 + x2 - 5x + 3) ≤ 0
11-53
112-3
12-30
113
130
x(x - 1)2(x + 3) ≤ 0
(x + 3)x(x - 1)2 ≤ 0
[1]
Draw y = (x + 3)x(x - 1)2
on the x-axis.
The highest order term is x4.
It's coefficient is 1: (+).
So start drawing the graph
from the right top of the x-axis.
(x - 1)2 = (x - 1)even
So the graph bounces off the x-axis
at x = 1.
x = x1 = xodd
So the graph passes through the x-axis
at x = 0.
(x + 3) = (x + 3)1 = (x + 3)odd
So the graph passes through the x-axis
at x = -3.
on the x-axis.
The highest order term is x4.
It's coefficient is 1: (+).
So start drawing the graph
from the right top of the x-axis.
(x - 1)2 = (x - 1)even
So the graph bounces off the x-axis
at x = 1.
x = x1 = xodd
So the graph passes through the x-axis
at x = 0.
(x + 3) = (x + 3)1 = (x + 3)odd
So the graph passes through the x-axis
at x = -3.
↓
-3 ≤ x ≤ 0, x = 1
[2]
(x + 3)x(x - 1)2 ≤ 0
So color the region
where the graph is below the x-axis.
The inequality sign, ≤, does include '='.
So draw full circles on x = -3, 0, 1.
So color the region
where the graph is below the x-axis.
The inequality sign, ≤, does include '='.
So draw full circles on x = -3, 0, 1.
Close
Example
x4 - x < 0
Solution x4 - x < 0
x(x3 - 1) < 0
x(x - 1)(x2 + x + 1) < 0 - [1]
D = 12 - 4⋅1⋅1
= 1 - 4
= -3 < 0 - [2]
x(x - 1) < 0 - [3]
0 < x < 1
x(x3 - 1) < 0
x(x - 1)(x2 + x + 1) < 0 - [1]
D = 12 - 4⋅1⋅1
= 1 - 4
= -3 < 0 - [2]
x(x - 1) < 0 - [3]
0 < x < 1
[2]
(x2 + x + 1) seems to be always (+).
To find out,
find the discriminant D.
D = -3 < 0
So y = x2 + x + 1
don't meet with the x-axis.
So (x2 + x + 1) is always (+).
Quadratic Function: Number of Zeros
To find out,
find the discriminant D.
D = -3 < 0
So y = x2 + x + 1
don't meet with the x-axis.
So (x2 + x + 1) is always (+).
Quadratic Function: Number of Zeros
[3]
x(x - 1)(x2 + x + 1) < 0
÷ (x2 + x + 1) to both sides.
→ x(x - 1) < 0.
÷ (x2 + x + 1) to both sides.
→ x(x - 1) < 0.
Close