The document discusses sign charts and inequalities. It provides examples of determining the sign (positive or negative) of expressions for given values of x. Specifically, it explains how to: 1. Factor polynomials or rational expressions to determine sign. 2. Construct a sign chart by solving for f=0, marking those values on a number line, and testing points in each segment. 3. Use a sign chart to indicate where an expression like f=x^2 - 3x - 4 is equal to 0, positive, or negative based on the value of x.

1. Inequalities
http://www.lahc.edu/math/precalculus/math_260a.html

2. Sign–Charts and Inequalities I
Given an expression f, it’s important to identify when
the output is positive (f > 0) and when the output is
negative (f < 0) when f is evaluated with a value x.

3. Example A. Determine whether the outcome is + or –
for x2 – 2x – 3 if x = –3/2, –1/2.
Sign–Charts and Inequalities I
Given an expression f, it’s important to identify when
the output is positive (f > 0) and when the output is
negative (f < 0) when f is evaluated with a value x.

4. Sign–Charts and Inequalities I
For polynomials or rational expressions,
factor them to determine the signs of their outputs.
Given an expression f, it’s important to identify when
the output is positive (f > 0) and when the output is
negative (f < 0) when f is evaluated with a value x.
Example A. Determine whether the outcome is + or –
for x2 – 2x – 3 if x = –3/2, –1/2.

5. In factored form x2 – 2x – 3 = (x – 3)(x + 1)
Sign–Charts and Inequalities I
For polynomials or rational expressions,
factor them to determine the signs of their outputs.
Given an expression f, it’s important to identify when
the output is positive (f > 0) and when the output is
negative (f < 0) when f is evaluated with a value x.
Example A. Determine whether the outcome is + or –
for x2 – 2x – 3 if x = –3/2, –1/2.

6. In factored form x2 – 2x – 3 = (x – 3)(x + 1)
Hence, for x = –3/2:
(–3/2 – 3)(–3/2 + 1)
Sign–Charts and Inequalities I
For polynomials or rational expressions,
factor them to determine the signs of their outputs.
Given an expression f, it’s important to identify when
the output is positive (f > 0) and when the output is
negative (f < 0) when f is evaluated with a value x.
Example A. Determine whether the outcome is + or –
for x2 – 2x – 3 if x = –3/2, –1/2.

7. In factored form x2 – 2x – 3 = (x – 3)(x + 1)
Hence, for x = –3/2:
(–3/2 – 3)(–3/2 + 1) is (–)(–) = + .
Sign–Charts and Inequalities I
For polynomials or rational expressions,
factor them to determine the signs of their outputs.
Given an expression f, it’s important to identify when
the output is positive (f > 0) and when the output is
negative (f < 0) when f is evaluated with a value x.
Example A. Determine whether the outcome is + or –
for x2 – 2x – 3 if x = –3/2, –1/2.

8. In factored form x2 – 2x – 3 = (x – 3)(x + 1)
Hence, for x = –3/2:
(–3/2 – 3)(–3/2 + 1) is (–)(–) = + .
And for x = –1/2:
(–1/2 – 3)(–1/2 + 1)
Sign–Charts and Inequalities I
For polynomials or rational expressions,
factor them to determine the signs of their outputs.
Given an expression f, it’s important to identify when
the output is positive (f > 0) and when the output is
negative (f < 0) when f is evaluated with a value x.
Example A. Determine whether the outcome is + or –
for x2 – 2x – 3 if x = –3/2, –1/2.

9. In factored form x2 – 2x – 3 = (x – 3)(x + 1)
Hence, for x = –3/2:
(–3/2 – 3)(–3/2 + 1) is (–)(–) = + .
And for x = –1/2:
(–1/2 – 3)(–1/2 + 1) is (–)(+) = – .
Sign–Charts and Inequalities I
For polynomials or rational expressions,
factor them to determine the signs of their outputs.
Given an expression f, it’s important to identify when
the output is positive (f > 0) and when the output is
negative (f < 0) when f is evaluated with a value x.
Example A. Determine whether the outcome is + or –
for x2 – 2x – 3 if x = –3/2, –1/2.

10. Sign–Charts and Inequalities I
Example B. Determine whether the outcome is
x2 – 2x – 3
x2 + x – 2
if x = –3/2, –1/2.+ or – for

11. Example B. Determine whether the outcome is
x2 – 2x – 3
x2 + x – 2
x2 – 2x – 3
x2 + x – 2
In factored form =
(x – 3)(x + 1)
(x – 1)(x + 2)
Sign–Charts and Inequalities I
if x = –3/2, –1/2.+ or – for

12. x2 – 2x – 3
x2 + x – 2
In factored form =
(x – 3)(x + 1)
(x – 1)(x + 2)
Hence, for x = –3/2:
(x – 3)(x + 1)
(x – 1)(x + 2)
=
(–)(–)
(–)(+)
Sign–Charts and Inequalities I
Example B. Determine whether the outcome is
x2 – 2x – 3
x2 + x – 2
if x = –3/2, –1/2.+ or – for

13. x2 – 2x – 3
x2 + x – 2
In factored form =
(x – 3)(x + 1)
(x – 1)(x + 2)
Hence, for x = –3/2:
(x – 3)(x + 1)
(x – 1)(x + 2)
=
(–)(–)
(–)(+)
< 0
Sign–Charts and Inequalities I
Example B. Determine whether the outcome is
x2 – 2x – 3
x2 + x – 2
if x = –3/2, –1/2.+ or – for

14. x2 – 2x – 3
x2 + x – 2
In factored form =
(x – 3)(x + 1)
(x – 1)(x + 2)
Hence, for x = –3/2:
(x – 3)(x + 1)
(x – 1)(x + 2)
=
(–)(–)
(–)(+)
< 0
For x = –1/2:
(x – 3)(x + 1)
(x – 1)(x + 2)
=
(–)(+)
(–)(+)
Sign–Charts and Inequalities I
Example B. Determine whether the outcome is
x2 – 2x – 3
x2 + x – 2
if x = –3/2, –1/2.+ or – for

15. x2 – 2x – 3
x2 + x – 2
In factored form =
(x – 3)(x + 1)
(x – 1)(x + 2)
Hence, for x = –3/2:
(x – 3)(x + 1)
(x – 1)(x + 2)
=
(–)(–)
(–)(+)
< 0
For x = –1/2:
(x – 3)(x + 1)
(x – 1)(x + 2)
=
(–)(+)
(–)(+)
> 0
Sign–Charts and Inequalities I
Example B. Determine whether the outcome is
x2 – 2x – 3
x2 + x – 2
if x = –3/2, –1/2.+ or – for

16. x2 – 2x – 3
x2 + x – 2
In factored form =
(x – 3)(x + 1)
(x – 1)(x + 2)
Hence, for x = –3/2:
(x – 3)(x + 1)
(x – 1)(x + 2)
=
(–)(–)
(–)(+)
< 0
For x = –1/2:
(x – 3)(x + 1)
(x – 1)(x + 2)
=
(–)(+)
(–)(+)
> 0
This leads to the sign charts of formulas. The sign–
chart of a formula gives the signs of the outputs.
Sign–Charts and Inequalities I
Example B. Determine whether the outcome is
x2 – 2x – 3
x2 + x – 2
if x = –3/2, –1/2.+ or – for

17. Here is an example, the sign chart of f = x – 1:
1
f = 0 + +– – – – x – 1
Sign–Charts and Inequalities I

18. Here is an example, the sign chart of f = x – 1:
1
f = 0 + +– – – – x – 1
The "+" indicates the region where the output is
positive i.e. if 1 < x.
Sign–Charts and Inequalities I

19. Here is an example, the sign chart of f = x – 1:
1
f = 0 + +– – – – x – 1
The "+" indicates the region where the output is
positive i.e. if 1 < x. Likewise, the "–" indicates the
region where the output is negative, i.e. x < 1.
Sign–Charts and Inequalities I

20. Construction of the sign–chart of f.
Here is an example, the sign chart of f = x – 1:
1
f = 0 + +– – – – x – 1
The "+" indicates the region where the output is
positive i.e. if 1 < x. Likewise, the "–" indicates the
region where the output is negative, i.e. x < 1.
Sign–Charts and Inequalities I

21. Construction of the sign–chart of f.
I. Solve for f = 0 (and denominator = 0) if there is any
denominator.
Here is an example, the sign chart of f = x – 1:
1
f = 0 + +– – – – x – 1
The "+" indicates the region where the output is
positive i.e. if 1 < x. Likewise, the "–" indicates the
region where the output is negative, i.e. x < 1.
Sign–Charts and Inequalities I

22. Construction of the sign–chart of f.
I. Solve for f = 0 (and denominator = 0) if there is any
denominator.
II. Draw the real line, mark off the answers from I.
Here is an example, the sign chart of f = x – 1:
1
f = 0 + +– – – – x – 1
The "+" indicates the region where the output is
positive i.e. if 1 < x. Likewise, the "–" indicates the
region where the output is negative, i.e. x < 1.
Sign–Charts and Inequalities I

23. Construction of the sign–chart of f.
I. Solve for f = 0 (and denominator = 0) if there is any
denominator.
II. Draw the real line, mark off the answers from I.
III. Sample each segment for signs by testing a point
in each segment.
Here is an example, the sign chart of f = x – 1:
1
f = 0 + +– – – – x – 1
The "+" indicates the region where the output is
positive i.e. if 1 < x. Likewise, the "–" indicates the
region where the output is negative, i.e. x < 1.
Sign–Charts and Inequalities I

24. Construction of the sign–chart of f.
I. Solve for f = 0 (and denominator = 0) if there is any
denominator.
II. Draw the real line, mark off the answers from I.
III. Sample each segment for signs by testing a point
in each segment.
Here is an example, the sign chart of f = x – 1:
1
f = 0 + +– – – – x – 1
The "+" indicates the region where the output is
positive i.e. if 1 < x. Likewise, the "–" indicates the
region where the output is negative, i.e. x < 1.
Fact: The sign stays the same for x's in between the
values from step I (where f = 0 or f is undefined.)
Sign–Charts and Inequalities I

25. Example C. Let f = x2 – 3x – 4 , use the sign–
chart to indicate when is f = 0, f > 0, and f < 0.
Sign–Charts and Inequalities I

26. Example C. Let f = x2 – 3x – 4 , use the sign–
chart to indicate when is f = 0, f > 0, and f < 0.
Solve x2 – 3x – 4 = 0
(x – 4)(x + 1) = 0  x = 4 , –1
Sign–Charts and Inequalities I

27. Example C. Let f = x2 – 3x – 4 , use the sign–
chart to indicate when is f = 0, f > 0, and f < 0.
Solve x2 – 3x – 4 = 0
(x – 4)(x + 1) = 0  x = 4 , –1
Mark off these points on a line:
(x–4)(x+1)
4–1
Sign–Charts and Inequalities I

28. Example C. Let f = x2 – 3x – 4 , use the sign–
chart to indicate when is f = 0, f > 0, and f < 0.
Solve x2 – 3x – 4 = 0
(x – 4)(x + 1) = 0  x = 4 , –1
Mark off these points on a line:
(x–4)(x+1)
Select points to sample in each segment:
4–1
Sign–Charts and Inequalities I

29. Example C. Let f = x2 – 3x – 4 , use the sign–
chart to indicate when is f = 0, f > 0, and f < 0.
Solve x2 – 3x – 4 = 0
(x – 4)(x + 1) = 0  x = 4 , –1
Mark off these points on a line:
(x–4)(x+1)
4–1
Select points to sample in each segment:
Test x = – 2,
–2
Sign–Charts and Inequalities I

30. Example C. Let f = x2 – 3x – 4 , use the sign–
chart to indicate when is f = 0, f > 0, and f < 0.
Solve x2 – 3x – 4 = 0
(x – 4)(x + 1) = 0  x = 4 , –1
Mark off these points on a line:
(x–4)(x+1)
4–1
Select points to sample in each segment:
Test x = – 2,
get – * – = + .
Hence the segment
is positive. Draw +
sign over it.
–2
Sign–Charts and Inequalities I

31. Example C. Let f = x2 – 3x – 4 , use the sign–
chart to indicate when is f = 0, f > 0, and f < 0.
Solve x2 – 3x – 4 = 0
(x – 4)(x + 1) = 0  x = 4 , –1
Mark off these points on a line:
(x–4)(x+1) + + + + +
4–1
Select points to sample in each segment:
Test x = – 2,
get – * – = + .
Hence the segment
is positive. Draw +
sign over it.
–2
Sign–Charts and Inequalities I

32. Example C. Let f = x2 – 3x – 4 , use the sign–
chart to indicate when is f = 0, f > 0, and f < 0.
Solve x2 – 3x – 4 = 0
(x – 4)(x + 1) = 0  x = 4 , –1
Mark off these points on a line:
(x–4)(x+1) + + + + +
0 4–1
Select points to sample in each segment:
Test x = – 2,
get – * – = + .
Hence the segment
is positive. Draw +
sign over it.
–2
Test x = 0,
get – * + = –.
Hence this segment
is negative.
Put – over it.
Sign–Charts and Inequalities I

33. Example C. Let f = x2 – 3x – 4 , use the sign–
chart to indicate when is f = 0, f > 0, and f < 0.
Solve x2 – 3x – 4 = 0
(x – 4)(x + 1) = 0  x = 4 , –1
Mark off these points on a line:
(x–4)(x+1) + + + + + – – – – –
0 4–1
Select points to sample in each segment:
Test x = – 2,
get – * – = + .
Hence the segment
is positive. Draw +
sign over it.
–2
Test x = 0,
get – * + = –.
Hence this segment
is negative.
Put – over it.
Sign–Charts and Inequalities I

34. Example C. Let f = x2 – 3x – 4 , use the sign–
chart to indicate when is f = 0, f > 0, and f < 0.
Solve x2 – 3x – 4 = 0
(x – 4)(x + 1) = 0  x = 4 , –1
Mark off these points on a line:
(x–4)(x+1) + + + + + – – – – –
0 4–1
Select points to sample in each segment:
Test x = – 2,
get – * – = + .
Hence the segment
is positive. Draw +
sign over it.
–2
Test x = 0,
get – * + = –.
Hence this segment
is negative.
Put – over it.
Test x = 5,
get + * + = +.
Hence this segment
is positive.
Put + over it.
5
Sign–Charts and Inequalities I

35. Example C. Let f = x2 – 3x – 4 , use the sign–
chart to indicate when is f = 0, f > 0, and f < 0.
Solve x2 – 3x – 4 = 0
(x – 4)(x + 1) = 0  x = 4 , –1
Mark off these points on a line:
(x–4)(x+1) + + + + + – – – – – + + + + +
0 4–1
Select points to sample in each segment:
Test x = – 2,
get – * – = + .
Hence the segment
is positive. Draw +
sign over it.
–2
Test x = 0,
get – * + = –.
Hence this segment
is negative.
Put – over it.
Test x = 5,
get + * + = +.
Hence this segment
is positive.
Put + over it.
5
Sign–Charts and Inequalities I

36. Example D. Make the sign chart of f =
(x – 3)
(x – 1)(x + 2)
Sign–Charts and Inequalities I

37. Example D. Make the sign chart of f =
(x – 3)
(x – 1)(x + 2)
The root for f = 0 is from the zero of the numerator
which is x = 3.
Sign–Charts and Inequalities I

38. Example D. Make the sign chart of f =
(x – 3)
(x – 1)(x + 2)
The root for f = 0 is from the zero of the numerator
which is x = 3. The zeroes of the denominator
x = 1, –2 are the values where f is undefined (UDF).
Sign–Charts and Inequalities I

39. Example D. Make the sign chart of f =
(x – 3)
(x – 1)(x + 2)
The root for f = 0 is from the zero of the numerator
which is x = 3. The zeroes of the denominator
x = 1, –2 are the values where f is undefined (UDF).
Mark these values on a real line.
Sign–Charts and Inequalities I

40. Example D. Make the sign chart of f =
(x – 3)
(x – 1)(x + 2)
The root for f = 0 is from the zero of the numerator
which is x = 3. The zeroes of the denominator
x = 1, –2 are the values where f is undefined (UDF).
Mark these values on a real line.
(x – 3)
(x – 1)(x + 2) –2 1 3
UDF UDF f=0
Sign–Charts and Inequalities I

41. Example D. Make the sign chart of f =
Select a point to sample in each segment:
(x – 3)
(x – 1)(x + 2)
The root for f = 0 is from the zero of the numerator
which is x = 3. The zeroes of the denominator
x = 1, –2 are the values where f is undefined (UDF).
Mark these values on a real line.
(x – 3)
(x – 1)(x + 2) –2 1 3
UDF UDF f=0
–3 0 2 4
Sign–Charts and Inequalities I

42. Example D. Make the sign chart of f =
Select a point to sample in each segment:
Test x = –3,
we've a
(x – 3)
(x – 1)(x + 2)
The root for f = 0 is from the zero of the numerator
which is x = 3. The zeroes of the denominator
x = 1, –2 are the values where f is undefined (UDF).
Mark these values on a real line.
(x – 3)
(x – 1)(x + 2) –2 1 3
UDF UDF f=0
–3
( – )
( – )( – )
= –
segment.
0 2 4
Sign–Charts and Inequalities I

43. Example D. Make the sign chart of f =
Select a point to sample in each segment:
Test x = –3,
we've a
(x – 3)
(x – 1)(x + 2)
The root for f = 0 is from the zero of the numerator
which is x = 3. The zeroes of the denominator
x = 1, –2 are the values where f is undefined (UDF).
Mark these values on a real line.
(x – 3)
(x – 1)(x + 2) –2 1 3
UDF UDF f=0
–3
( – )
( – )( – )
= –
segment.
0 2 4
Test x = 0,
we've a
( – )
( – )( + )
= +
segment.
Sign–Charts and Inequalities I

44. Example D. Make the sign chart of f =
Select a point to sample in each segment:
Test x = –3,
we've a
(x – 3)
(x – 1)(x + 2)
The root for f = 0 is from the zero of the numerator
which is x = 3. The zeroes of the denominator
x = 1, –2 are the values where f is undefined (UDF).
Mark these values on a real line.
(x – 3)
(x – 1)(x + 2) –2 1 3
UDF UDF f=0
–3
( – )
( – )( – )
= –
segment.
0 2 4
Test x = 0,
we've a
( – )
( – )( + )
= +
segment.
Test x = 2,
we've a
( – )
( + )( + )
segment.
= –
Sign–Charts and Inequalities I

45. Example D. Make the sign chart of f =
Select a point to sample in each segment:
Test x = –3,
we've a
(x – 3)
(x – 1)(x + 2)
The root for f = 0 is from the zero of the numerator
which is x = 3. The zeroes of the denominator
x = 1, –2 are the values where f is undefined (UDF).
Mark these values on a real line.
(x – 3)
(x – 1)(x + 2) –2 1 3
UDF UDF f=0
–3
( – )
( – )( – )
= –
segment.
0 2 4
Test x = 0,
we've a
( – )
( – )( + )
= +
segment.
Test x = 2,
we've a
( – )
( + )( + )
segment.
= –
Test x = 4,
we've a
( + )
( + )( + )
segment.
= +
– – – – + + + – – – + + + +
Sign–Charts and Inequalities I

46. The easiest way to solve a polynomial or rational
inequality is to use the sign–chart.
Sign–Charts and Inequalities I

47. Example E. Solve x2 – 3x > 4
The easiest way to solve a polynomial or rational
inequality is to use the sign–chart.
Sign–Charts and Inequalities I

48. Example E. Solve x2 – 3x > 4
The easiest way to solve a polynomial or rational
inequality is to use the sign–chart. To do this,
I. set one side of the inequality to 0,
Sign–Charts and Inequalities I

49. Example E. Solve x2 – 3x > 4
The easiest way to solve a polynomial or rational
inequality is to use the sign–chart. To do this,
I. set one side of the inequality to 0,
Setting one side to 0, we have x2 – 3x – 4 > 0
Sign–Charts and Inequalities I

50. Example E. Solve x2 – 3x > 4
The easiest way to solve a polynomial or rational
inequality is to use the sign–chart. To do this,
I. set one side of the inequality to 0,
II. factor the expression
Setting one side to 0, we have x2 – 3x – 4 > 0
Sign–Charts and Inequalities I

51. Example E. Solve x2 – 3x > 4
The easiest way to solve a polynomial or rational
inequality is to use the sign–chart. To do this,
I. set one side of the inequality to 0,
II. factor the expression
Setting one side to 0, we have x2 – 3x – 4 > 0 or
(x – 4)(x + 1) > 0.
Sign–Charts and Inequalities I

52. Example E. Solve x2 – 3x > 4
The easiest way to solve a polynomial or rational
inequality is to use the sign–chart. To do this,
I. set one side of the inequality to 0,
II. factor the expression and draw the sign–chart,
Setting one side to 0, we have x2 – 3x – 4 > 0 or
(x – 4)(x + 1) > 0.
Sign–Charts and Inequalities I

53. Example E. Solve x2 – 3x > 4
The easiest way to solve a polynomial or rational
inequality is to use the sign–chart. To do this,
I. set one side of the inequality to 0,
II. factor the expression and draw the sign–chart,
Setting one side to 0, we have x2 – 3x – 4 > 0 or
(x – 4)(x + 1) > 0. The roots are x = –1, 4.
Sign–Charts and Inequalities I

54. Example E. Solve x2 – 3x > 4
4–1
The easiest way to solve a polynomial or rational
inequality is to use the sign–chart. To do this,
I. set one side of the inequality to 0,
II. factor the expression and draw the sign–chart,
Draw the sign–chart, sample the points x = –2, 0, 5
(x – 4)(x + 1)
Setting one side to 0, we have x2 – 3x – 4 > 0 or
(x – 4)(x + 1) > 0. The roots are x = –1, 4.
Sign–Charts and Inequalities I

55. Example E. Solve x2 – 3x > 4
0 4–1–2 5
The easiest way to solve a polynomial or rational
inequality is to use the sign–chart. To do this,
I. set one side of the inequality to 0,
II. factor the expression and draw the sign–chart,
Draw the sign–chart, sample the points x = –2, 0, 5
(x – 4)(x + 1)
Setting one side to 0, we have x2 – 3x – 4 > 0 or
(x – 4)(x + 1) > 0. The roots are x = –1, 4.
Sign–Charts and Inequalities I

56. Example E. Solve x2 – 3x > 4
0 4–1–2 5
The easiest way to solve a polynomial or rational
inequality is to use the sign–chart. To do this,
I. set one side of the inequality to 0,
II. factor the expression and draw the sign–chart,
Draw the sign–chart, sample the points x = –2, 0, 5
(x – 4)(x + 1)
+ + +
Setting one side to 0, we have x2 – 3x – 4 > 0 or
(x – 4)(x + 1) > 0. The roots are x = –1, 4.
Sign–Charts and Inequalities I

57. Example E. Solve x2 – 3x > 4
0 4–1–2 5
The easiest way to solve a polynomial or rational
inequality is to use the sign–chart. To do this,
I. set one side of the inequality to 0,
II. factor the expression and draw the sign–chart,
Draw the sign–chart, sample the points x = –2, 0, 5
(x – 4)(x + 1)
+ + + – – – – – –
Setting one side to 0, we have x2 – 3x – 4 > 0 or
(x – 4)(x + 1) > 0. The roots are x = –1, 4.
Sign–Charts and Inequalities I

58. Example E. Solve x2 – 3x > 4
0 4–1–2 5
The easiest way to solve a polynomial or rational
inequality is to use the sign–chart. To do this,
I. set one side of the inequality to 0,
II. factor the expression and draw the sign–chart,
Draw the sign–chart, sample the points x = –2, 0, 5
(x – 4)(x + 1)
+ + + – – – – – – + + + +
Setting one side to 0, we have x2 – 3x – 4 > 0 or
(x – 4)(x + 1) > 0. The roots are x = –1, 4.
Sign–Charts and Inequalities I

59. Example E. Solve x2 – 3x > 4
0 4–1–2 5
The easiest way to solve a polynomial or rational
inequality is to use the sign–chart. To do this,
I. set one side of the inequality to 0,
II. factor the expression and draw the sign–chart,
III. read off the answer from the sign chart.
Draw the sign–chart, sample the points x = –2, 0, 5
(x – 4)(x + 1)
+ + + – – – – – – + + + +
Setting one side to 0, we have x2 – 3x – 4 > 0 or
(x – 4)(x + 1) > 0. The roots are x = –1, 4.
Sign–Charts and Inequalities I

60. Example E. Solve x2 – 3x > 4
0 4–1
The solutions are the + regions:
–2 5
The easiest way to solve a polynomial or rational
inequality is to use the sign–chart. To do this,
I. set one side of the inequality to 0,
II. factor the expression and draw the sign–chart,
III. read off the answer from the sign chart.
Draw the sign–chart, sample the points x = –2, 0, 5
(x – 4)(x + 1)
+ + + – – – – – – + + + +
Setting one side to 0, we have x2 – 3x – 4 > 0 or
(x – 4)(x + 1) > 0. The roots are x = –1, 4.
Sign–Charts and Inequalities I

61. Example E. Solve x2 – 3x > 4
0 4–1
The solutions are the + regions: (–∞, –1) U (4, ∞)
–2 5
4–1
The easiest way to solve a polynomial or rational
inequality is to use the sign–chart. To do this,
I. set one side of the inequality to 0,
II. factor the expression and draw the sign–chart,
III. read off the answer from the sign chart.
Draw the sign–chart, sample the points x = –2, 0, 5
(x – 4)(x + 1)
+ + + – – – – – – + + + +
Setting one side to 0, we have x2 – 3x – 4 > 0 or
(x – 4)(x + 1) > 0. The roots are x = –1, 4.
Sign–Charts and Inequalities I

62. Example E. Solve x2 – 3x > 4
0 4–1
The solutions are the + regions: (–∞, –1) U (4, ∞)
–2 5
4–1
Note: The empty dot means those numbers are excluded.
The easiest way to solve a polynomial or rational
inequality is to use the sign–chart. To do this,
I. set one side of the inequality to 0,
II. factor the expression and draw the sign–chart,
III. read off the answer from the sign chart.
Draw the sign–chart, sample the points x = –2, 0, 5
(x – 4)(x + 1)
+ + + – – – – – – + + + +
Setting one side to 0, we have x2 – 3x – 4 > 0 or
(x – 4)(x + 1) > 0. The roots are x = –1, 4.
Sign–Charts and Inequalities I

63. Example F. Solve x – 2
2 <
x – 1
3
Sign–Charts and Inequalities I

64. Example F. Solve x – 2
2 <
x – 1
3
Set the inequality to 0, x – 2
2
x – 1
3
< 0
Sign–Charts and Inequalities I

65. Example F. Solve x – 2
2 <
x – 1
3
Set the inequality to 0, x – 2
2
x – 1
3
< 0
Put the expression into factored form,
Sign–Charts and Inequalities I

66. Example F. Solve x – 2
2 <
x – 1
3
Set the inequality to 0, x – 2
2
x – 1
3
< 0
Put the expression into factored form,
x – 2
2
x – 1
3
=
(x – 2)(x – 1)
2(x – 1) – 3(x – 2)
Sign–Charts and Inequalities I

67. Example F. Solve x – 2
2 <
x – 1
3
Set the inequality to 0, x – 2
2
x – 1
3
< 0
Put the expression into factored form,
x – 2
2
x – 1
3
=
(x – 2)(x – 1)
2(x – 1) – 3(x – 2)
=
(x – 2)(x – 1)
– x + 4
Sign–Charts and Inequalities I

68. Example F. Solve x – 2
2 <
x – 1
3
Set the inequality to 0, x – 2
2
x – 1
3
< 0
Put the expression into factored form,
x – 2
2
x – 1
3
=
(x – 2)(x – 1)
2(x – 1) – 3(x – 2)
=
(x – 2)(x – 1)
– x + 4
Hence the inequality is (x – 2)(x – 1)
– x + 4 < 0
Sign–Charts and Inequalities I

69. Example F. Solve x – 2
2 <
x – 1
3
Set the inequality to 0, x – 2
2
x – 1
3
< 0
Put the expression into factored form,
x – 2
2
x – 1
3
=
(x – 2)(x – 1)
2(x – 1) – 3(x – 2)
=
(x – 2)(x – 1)
– x + 4
Hence the inequality is (x – 2)(x – 1)
– x + 4 < 0
It has a root at x = 4, and it’s undefined at x = 1, 2.
Sign–Charts and Inequalities I

70. Example F. Solve x – 2
2 <
x – 1
3
Set the inequality to 0, x – 2
2
x – 1
3
< 0
Put the expression into factored form,
x – 2
2
x – 1
3
=
(x – 2)(x – 1)
2(x – 1) – 3(x – 2)
=
(x – 2)(x – 1)
– x + 4
Hence the inequality is (x – 2)(x – 1)
– x + 4 < 0
Draw the sign chart by sampling x = 0, 3/2, 3, 5
It has a root at x = 4, and it's undefined at x = 1, 2.
Sign–Charts and Inequalities I

71. Example F. Solve x – 2
2 <
x – 1
3
Set the inequality to 0, x – 2
2
x – 1
3
< 0
Put the expression into factored form,
x – 2
2
x – 1
3
=
(x – 2)(x – 1)
2(x – 1) – 3(x – 2)
=
(x – 2)(x – 1)
– x + 4
Hence the inequality is (x – 2)(x – 1)
– x + 4 < 0
Draw the sign chart by sampling x = 0, 3/2, 3, 5
It has a root at x = 4, and it's undefined at x = 1, 2.
41 2
UDF UDF
(x – 2)(x – 1)
– x + 4
Sign–Charts and Inequalities I

72. Example F. Solve x – 2
2 <
x – 1
3
Set the inequality to 0, x – 2
2
x – 1
3
< 0
Put the expression into factored form,
x – 2
2
x – 1
3
=
(x – 2)(x – 1)
2(x – 1) – 3(x – 2)
=
(x – 2)(x – 1)
– x + 4
Hence the inequality is (x – 2)(x – 1)
– x + 4 < 0
Draw the sign chart by sampling x = 0, 3/2, 3, 5
It has a root at x = 4, and it's undefined at x = 1, 2.
410 523/2 3
UDF UDF
(x – 2)(x – 1)
– x + 4
Sign–Charts and Inequalities I

73. Example F. Solve x – 2
2 <
x – 1
3
Set the inequality to 0, x – 2
2
x – 1
3
< 0
Put the expression into factored form,
x – 2
2
x – 1
3
=
(x – 2)(x – 1)
2(x – 1) – 3(x – 2)
=
(x – 2)(x – 1)
– x + 4
Hence the inequality is (x – 2)(x – 1)
– x + 4 < 0
Draw the sign chart by sampling x = 0, 3/2, 3, 5
It has a root at x = 4, and it's undefined at x = 1, 2.
410 5
+ + +
23/2 3
UDF UDF
(x – 2)(x – 1)
– x + 4
Sign–Charts and Inequalities I

74. Example F. Solve x – 2
2 <
x – 1
3
Set the inequality to 0, x – 2
2
x – 1
3
< 0
Put the expression into factored form,
x – 2
2
x – 1
3
=
(x – 2)(x – 1)
2(x – 1) – 3(x – 2)
=
(x – 2)(x – 1)
– x + 4
Hence the inequality is (x – 2)(x – 1)
– x + 4 < 0
Draw the sign chart by sampling x = 0, 3/2, 3, 5
It has a root at x = 4, and it's undefined at x = 1, 2.
410 5
+ + + – –
23/2 3
UDF UDF
(x – 2)(x – 1)
– x + 4
Sign–Charts and Inequalities I

75. Example F. Solve x – 2
2 <
x – 1
3
Set the inequality to 0, x – 2
2
x – 1
3
< 0
Put the expression into factored form,
x – 2
2
x – 1
3
=
(x – 2)(x – 1)
2(x – 1) – 3(x – 2)
=
(x – 2)(x – 1)
– x + 4
Hence the inequality is (x – 2)(x – 1)
– x + 4 < 0
Draw the sign chart by sampling x = 0, 3/2, 3, 5
It has a root at x = 4, and it's undefined at x = 1, 2.
410 5
+ + + – – + + + +
23/2 3
UDF UDF
(x – 2)(x – 1)
– x + 4
Sign–Charts and Inequalities I

76. Example F. Solve x – 2
2 <
x – 1
3
Set the inequality to 0, x – 2
2
x – 1
3
< 0
Put the expression into factored form,
x – 2
2
x – 1
3
=
(x – 2)(x – 1)
2(x – 1) – 3(x – 2)
=
(x – 2)(x – 1)
– x + 4
Hence the inequality is (x – 2)(x – 1)
– x + 4 < 0
Draw the sign chart by sampling x = 0, 3/2, 3, 5
It has a root at x = 4, and it's undefined at x = 1, 2.
410 5
+ + + – – + + + + – – – –
23/2 3
UDF UDF
(x – 2)(x – 1)
– x + 4
Sign–Charts and Inequalities I

77. Example F. Solve x – 2
2 <
x – 1
3
Set the inequality to 0, x – 2
2
x – 1
3
< 0
Put the expression into factored form,
x – 2
2
x – 1
3
=
(x – 2)(x – 1)
2(x – 1) – 3(x – 2)
=
(x – 2)(x – 1)
– x + 4
Hence the inequality is (x – 2)(x – 1)
– x + 4 < 0
Draw the sign chart by sampling x = 0, 3/2, 3, 5
It has a root at x = 4, and it's undefined at x = 1, 2.
410 5
+ + + – – + + + + – – – –
23/2 3
UDF UDF
(x – 2)(x – 1)
– x + 4
The answer are the shaded negative regions,
i.e. (1, 2) U [4 ∞).
Sign–Charts and Inequalities I

78. Sign-Charts and Inequalities
Exercise A. Draw the sign–charts of the following
formulas.
1. (x – 2)(x + 3)
4. (2 – x)(x + 3) 5. –x(x + 3)
7. (x + 3)2
9. x(2x – 1)(3 – x)
12. x2(2x – 1)2(3 – x)
13. x2(2x – 1)2(3 – x)2 14. x2 – 2x – 3
16. 1 –15. x4 – 2x3 – 3x2
(x – 2)
(x + 3)2.
(2 – x)
(x + 3)3.
–x
(x + 3)6.
8. –4(x + 3)4
x
(3 – x)(2x – 1)10.
11. x2(2x – 1)(3 – x)
1
x + 3
17. 2 – 2
x – 2
18. 1
2x + 1 19. –1
x + 3
– 1 2
x – 2
20. –2
x – 4
1
x + 2

79. Sign-Charts and Inequalities
Exercise B. Use the sign–charts method to solve the
following inequalities.
1. (x – 2)(x + 3) > 0
3. (2 – x)(x + 3) ≥ 0
8. x2(2x – 1)2(3 – x) ≤ 0
9. x2 – 2x < 3
14. 1 <13. x4 > 4x2
(2 – x)
(x + 3)2.
–x
(x + 3)4.
7. x2(2x – 1)(3 – x) ≥ 0
1
x 15. 2 2
x – 2
16. 1
x + 3
2
x – 2
17. >2
x – 4
1
x + 2
5. x(x – 2)(x + 3)
x
(x – 2)(x + 3)6. ≥ 0
10. x2 + 2x > 8
11. x3 – 2x2 < 3x 12. 2x3 < x2 + 6x
≥
≥ 0
≤ 0
≤ 0
≤ 18. 1 < 1
x2

80. C. Solve the inequalities, use the answers from Ex.1.3.
Inequalities