# 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

x

Solution ^{3}- 7x + 11x - 2 x

10-711 - [1]

2 - [2]

↓

10-711

224-6

12-35 - [3]

(given) = x

^{3}- 7x + 11x - 210-711 - [1]

2 - [2]

↓

10-711

224-6

12-35 - [3]

(given) = x

^{2}+ 2x - 3 + 5x - 2 - [4][1]

Numerator: 1x

Write the coefficients:

1, 0, -7, 11.

^{3}+ 0x^{2}- 7x + 11Write 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) = x

2: zero of x - 2

5: remainder

→ (quotient) = x

^{2}+ 2x - 32: zero of x - 2

5: remainder

Close

### Example

2x

Solution ^{4}+ x^{3}- 5x^{2}+ 3x + 4x + 1 2x

21-534

-1-214-7

2-1-47-3 - [1]

(given) = 2x

^{4}+ x^{3}- 5x^{2}+ 3x + 4x + 121-534

-1-214-7

2-1-47-3 - [1]

(given) = 2x

^{3}- x^{2}- 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

x

Remainder?

Solution ^{3}- 7x + 11x - 2Remainder?

x

(remainder) = f(2)

= 2

= 8 - 14 + 11

= 8 - 3

= 5

^{3}- 7x + 11x - 2(remainder) = f(2)

= 2

^{3}- 7⋅2 + 11= 8 - 14 + 11

= 8 - 3

= 5

Close

### Example

(2x

Remainder?

Solution ^{4}+ x^{3}- 5x^{2}+ 3x + 4) ÷ (x + 1)Remainder?

(2x

(remainder) = f(-1)

= 2⋅(-1)

= 2⋅1 + (-1) - 5⋅1 + 3⋅(-1) + 4

= 2 - 1 - 5 - 3 + 4

= 1 - 8 + 4

= -7 + 4

= -3

^{4}+ x^{3}- 5x^{2}+ 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

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) = x

f(7)?

Solution ^{4}- 9x^{3}+ 15x^{2}+ 3x - 62f(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

x

Solution ^{3}+ 3x^{2}- 16x + 12 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

→ x

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.

→ x

^{2}+ 4x - 12This 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

x

Solution ^{4}- 2x^{3}- 4x^{2}- 2x + 3 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)

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

x

Solution ^{4}+ x^{3}- 5x^{2}+ x - 6 11-51-6

22626

13130

-3-30-3

1010 - [1]

f(x) = (x - 2)(x + 3)(x

22626

13130

-3-30-3

1010 - [1]

f(x) = (x - 2)(x + 3)(x

^{2}+ 1)[1]

1, 0, 1

→ x

x

(x

x

So x

So x

So set (quotient) = (x

To check if a quadratic expression has zero(s),

find the discriminant D.

→ x

^{2}+ 1x

^{2}+ 1 is always (+).(x

^{2}≥ 0x

^{2}+ 1 ≥ 1)So x

^{2}+ 1 has no zeros.So x

^{2}+ 1 is not factorable.So set (quotient) = (x

^{2}+ 1).To check if a quadratic expression has zero(s),

find the discriminant D.

Close

## Polynomial Equation

### Example

x

Solution ^{4}+ 4x^{3}- 3x^{2}- 10x + 8 = 0 14-3-10+8

1152-8

152-80

1168

1680

-2-2-8

140

(x - 1)

x = 1, -2, -4

= -4, -2, 1

1152-8

152-80

1168

1680

-2-2-8

140

(x - 1)

^{2}(x + 2)(x + 4) = 0x = 1, -2, -4

= -4, -2, 1

Close

## Polynomial Inequality

### Formula

(x - a)

^{odd}(x - a)

^{even}

^{odd}(quotient),

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

x

Solution ^{4}- x^{2}< 0 x

x

x

(x + 1)x

[2]

^{4}- x^{2}< 0x

^{2}(x^{2}- 1) < 0x

^{2}(x + 1)(x - 1) < 0 - [1](x + 1)x

^{2}(x - 1) < 0[2]

Draw y = (x + 1)x

on the x-axis.

The highest order term is x

It's coefficient is 1: (+).

So start drawing the graph

from the right top of the x-axis.

(x - 1) = (x - 1)

So the graph passes through the x-axis

at x = 1.

x

So the graph bounces off the x-axis

at x = 0.

(x + 1) = (x + 1)

So the graph passes through the x-axis

at x = -1.

^{2}(x - 1)on the x-axis.

The highest order term is x

^{4}.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.

x

^{2}= x^{even}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)x

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.

^{2}(x - 1) < 0So 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

x

Solution ^{3}+ x^{2}- 10x + 8 ≥ 0 11-108

112-8

12-80

228

140

(x - 1)(x - 2)(x + 4) ≥ 0

(x + 4)(x - 1)(x - 2) ≥ 0

[1]

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 x

It's coefficient is 1: (+).

So start drawing the graph

from the right top of the x-axis.

(x - 2) = (x - 2)

So the graph passes through the x-axis

at x = 2.

(x - 1) = (x - 1)

So the graph passes through the x-axis

at x = 1.

(x + 4) = (x + 4)

So the graph passes through the x-axis

at x = -4.

on the x-axis.

The highest order term is x

^{3}.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

x

Solution ^{4}+ x^{3}- 5x^{2}+ 3x ≤ 0 x

x(x

11-53

112-3

12-30

113

130

x(x - 1)

(x + 3)x(x - 1)

[1]

^{4}+ x^{3}- 5x^{2}+ 3x ≤ 0x(x

^{3}+ x^{2}- 5x + 3) ≤ 011-53

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)

on the x-axis.

The highest order term is x

It's coefficient is 1: (+).

So start drawing the graph

from the right top of the x-axis.

(x - 1)

So the graph bounces off the x-axis

at x = 1.

x = x

So the graph passes through the x-axis

at x = 0.

(x + 3) = (x + 3)

So the graph passes through the x-axis

at x = -3.

^{2}on the x-axis.

The highest order term is x

^{4}.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 = x

^{1}= x^{odd}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)

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.

^{2}≤ 0So 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

x

Solution ^{4}- x < 0 x

x(x

x(x - 1)(x

D = 1

= 1 - 4

= -3 < 0 - [2]

x(x - 1) < 0 - [3]

- [4]

0 < x < 1

^{4}- x < 0x(x

^{3}- 1) < 0x(x - 1)(x

^{2}+ x + 1) < 0 - [1]D = 1

^{2}- 4⋅1⋅1= 1 - 4

= -3 < 0 - [2]

x(x - 1) < 0 - [3]

0 < x < 1

[2]

(x

To find out,

find the discriminant D.

D = -3 < 0

So y = x

don't meet with the x-axis.

So (x

Quadratic Function: Number of Zeros

^{2}+ x + 1) seems to be always (+).To find out,

find the discriminant D.

D = -3 < 0

So y = x

^{2}+ x + 1don't meet with the x-axis.

So (x

^{2}+ x + 1) is always (+).Quadratic Function: Number of Zeros

[3]

x(x - 1)(x

÷ (x

→ x(x - 1) < 0.

^{2}+ x + 1) < 0÷ (x

^{2}+ x + 1) to both sides.→ x(x - 1) < 0.

Close