3. Factoring Trinomials I
Trinomials (three-term) are polynomials of the form
ax2 + bx + c where a, b, and c are numbers. The product of
two binomials is a trinomials:
(#x + #)(#x + #) ax2 + bx + c
4. Factoring Trinomials I
Trinomials (three-term) are polynomials of the form
ax2 + bx + c where a, b, and c are numbers. The product of
two binomials is a trinomials:
(#x + #)(#x + #) ax2 + bx + c
Hence, to factor a trinomial, we write the trinomial as a
product of two binomials, if possible,
5. Factoring Trinomials I
Trinomials (three-term) are polynomials of the form
ax2 + bx + c where a, b, and c are numbers. The product of
two binomials is a trinomials:
(#x + #)(#x + #) ax2 + bx + c
Hence, to factor a trinomial, we write the trinomial as a
product of two binomials, if possible, that is:
ax2 + bx + c (#x + #)(#x + #)
6. Factoring Trinomials I
Trinomials (three-term) are polynomials of the form
ax2 + bx + c where a, b, and c are numbers. The product of
two binomials is a trinomials:
(#x + #)(#x + #) ax2 + bx + c
Hence, to factor a trinomial, we write the trinomial as a
product of two binomials, if possible, that is:
ax2 + bx + c (#x + #)(#x + #)
We start with the case where a = 1, or trinomials
of the form x2 + bx + c.
7. Factoring Trinomials I
To factor x2 + bx + c, we note hat
if (x + u)(x + v)
Trinomials (three-term) are polynomials of the form
ax2 + bx + c where a, b, and c are numbers. The product of
two binomials is a trinomials:
(#x + #)(#x + #) ax2 + bx + c
Hence, to factor a trinomial, we write the trinomial as a
product of two binomials, if possible, that is:
ax2 + bx + c (#x + #)(#x + #)
We start with the case where a = 1, or trinomials
of the form x2 + bx + c.
8. Factoring Trinomials I
To factor x2 + bx + c, we note hat
if (x + u)(x + v)
= x2 + ux + vx + uv
Trinomials (three-term) are polynomials of the form
ax2 + bx + c where a, b, and c are numbers. The product of
two binomials is a trinomials:
(#x + #)(#x + #) ax2 + bx + c
Hence, to factor a trinomial, we write the trinomial as a
product of two binomials, if possible, that is:
ax2 + bx + c (#x + #)(#x + #)
We start with the case where a = 1, or trinomials
of the form x2 + bx + c.
9. Factoring Trinomials I
To factor x2 + bx + c, we note hat
if (x + u)(x + v)
= x2 + ux + vx + uv
= x2 + (u + v)x + uv
Trinomials (three-term) are polynomials of the form
ax2 + bx + c where a, b, and c are numbers. The product of
two binomials is a trinomials:
(#x + #)(#x + #) ax2 + bx + c
Hence, to factor a trinomial, we write the trinomial as a
product of two binomials, if possible, that is:
ax2 + bx + c (#x + #)(#x + #)
We start with the case where a = 1, or trinomials
of the form x2 + bx + c.
10. Factoring Trinomials I
To factor x2 + bx + c, we note hat
if (x + u)(x + v)
= x2 + ux + vx + uv
= x2 + (u + v)x + uv
= x2 + bx + c,
Trinomials (three-term) are polynomials of the form
ax2 + bx + c where a, b, and c are numbers. The product of
two binomials is a trinomials:
(#x + #)(#x + #) ax2 + bx + c
Hence, to factor a trinomial, we write the trinomial as a
product of two binomials, if possible, that is:
ax2 + bx + c (#x + #)(#x + #)
We start with the case where a = 1, or trinomials
of the form x2 + bx + c.
11. Factoring Trinomials I
To factor x2 + bx + c, we note hat
if (x + u)(x + v)
= x2 + ux + vx + uv
= x2 + (u + v)x + uv
= x2 + bx + c,
we need to have u and v where uv = c,
Trinomials (three-term) are polynomials of the form
ax2 + bx + c where a, b, and c are numbers. The product of
two binomials is a trinomials:
(#x + #)(#x + #) ax2 + bx + c
Hence, to factor a trinomial, we write the trinomial as a
product of two binomials, if possible, that is:
ax2 + bx + c (#x + #)(#x + #)
We start with the case where a = 1, or trinomials
of the form x2 + bx + c.
12. Factoring Trinomials I
To factor x2 + bx + c, we note hat
if (x + u)(x + v)
= x2 + ux + vx + uv
= x2 + (u + v)x + uv
= x2 + bx + c,
we need to have u and v where uv = c, and u + v = b.
Trinomials (three-term) are polynomials of the form
ax2 + bx + c where a, b, and c are numbers. The product of
two binomials is a trinomials:
(#x + #)(#x + #) ax2 + bx + c
Hence, to factor a trinomial, we write the trinomial as a
product of two binomials, if possible, that is:
ax2 + bx + c (#x + #)(#x + #)
We start with the case where a = 1, or trinomials
of the form x2 + bx + c.
13. Factoring Trinomials I
To factor x2 + bx + c, we note hat
if (x + u)(x + v)
= x2 + ux + vx + uv
= x2 + (u + v)x + uv
= x2 + bx + c,
we need to have u and v where uv = c, and u + v = b. If this
can’t be done, then the trinomial is prime (not factorable).
Trinomials (three-term) are polynomials of the form
ax2 + bx + c where a, b, and c are numbers. The product of
two binomials is a trinomials:
(#x + #)(#x + #) ax2 + bx + c
Hence, to factor a trinomial, we write the trinomial as a
product of two binomials, if possible, that is:
ax2 + bx + c (#x + #)(#x + #)
We start with the case where a = 1, or trinomials
of the form x2 + bx + c.
15. Factoring Trinomials I
Example A.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6
16. Factoring Trinomials I
Example A.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
17. Factoring Trinomials I
Example A.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3)
18. Factoring Trinomials I
Example A.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2x + 3x
19. Factoring Trinomials I
Example A.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2x + 3x = 5x,
20. Factoring Trinomials I
Example A.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2x + 3x = 5x,
so x2 + 5x + 6 = (x + 2)(x + 3)
21. Factoring Trinomials I
b. Factor x2 – 5x + 6
Example A.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2x + 3x = 5x,
so x2 + 5x + 6 = (x + 2)(x + 3)
22. Factoring Trinomials I
b. Factor x2 – 5x + 6
We want (x + u)(x + v) = x2 – 5x + 6,
Example A.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2x + 3x = 5x,
so x2 + 5x + 6 = (x + 2)(x + 3)
23. Factoring Trinomials I
b. Factor x2 – 5x + 6
We want (x + u)(x + v) = x2 – 5x + 6, so we need u and v
where uv = 6
Example A.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2x + 3x = 5x,
so x2 + 5x + 6 = (x + 2)(x + 3)
24. Factoring Trinomials I
b. Factor x2 – 5x + 6
We want (x + u)(x + v) = x2 – 5x + 6, so we need u and v
where uv = 6 and u + v = –5.
Example A.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2x + 3x = 5x,
so x2 + 5x + 6 = (x + 2)(x + 3)
25. Factoring Trinomials I
b. Factor x2 – 5x + 6
We want (x + u)(x + v) = x2 – 5x + 6, so we need u and v
where uv = 6 and u + v = –5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3)
Example A.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2x + 3x = 5x,
so x2 + 5x + 6 = (x + 2)(x + 3)
26. Factoring Trinomials I
b. Factor x2 – 5x + 6
We want (x + u)(x + v) = x2 – 5x + 6, so we need u and v
where uv = 6 and u + v = –5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and –2 – 3 = –5,
Example A.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2x + 3x = 5x,
so x2 + 5x + 6 = (x + 2)(x + 3)
27. Factoring Trinomials I
b. Factor x2 – 5x + 6
We want (x + u)(x + v) = x2 – 5x + 6, so we need u and v
where uv = 6 and u + v = –5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and –2 – 3 = –5,
so x2 – 5x + 6 = (x – 2)(x – 3).
Example A.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2x + 3x = 5x,
so x2 + 5x + 6 = (x + 2)(x + 3)
28. Factoring Trinomials I
c. Factor x2 + 5x – 6
b. Factor x2 – 5x + 6
We want (x + u)(x + v) = x2 – 5x + 6, so we need u and v
where uv = 6 and u + v = –5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and –2 – 3 = –5,
so x2 – 5x + 6 = (x – 2)(x – 3).
Example A.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2x + 3x = 5x,
so x2 + 5x + 6 = (x + 2)(x + 3)
29. Factoring Trinomials I
c. Factor x2 + 5x – 6
We want (x + u)(x + v) = x2 + 5x – 6, so we need uv = –6
b. Factor x2 – 5x + 6
We want (x + u)(x + v) = x2 – 5x + 6, so we need u and v
where uv = 6 and u + v = –5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and –2 – 3 = –5,
so x2 – 5x + 6 = (x – 2)(x – 3).
Example A.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2x + 3x = 5x,
so x2 + 5x + 6 = (x + 2)(x + 3)
30. Factoring Trinomials I
c. Factor x2 + 5x – 6
We want (x + u)(x + v) = x2 + 5x – 6, so we need uv = –6 and
u + v = 5.
b. Factor x2 – 5x + 6
We want (x + u)(x + v) = x2 – 5x + 6, so we need u and v
where uv = 6 and u + v = –5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and –2 – 3 = –5,
so x2 – 5x + 6 = (x – 2)(x – 3).
Example A.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2x + 3x = 5x,
so x2 + 5x + 6 = (x + 2)(x + 3)
31. Factoring Trinomials I
c. Factor x2 + 5x – 6
We want (x + u)(x + v) = x2 + 5x – 6, so we need uv = –6 and
u + v = 5.
Since -6 = (–1)(6) = (1)(–6) = (–2)(3) =(2)(–3)
b. Factor x2 – 5x + 6
We want (x + u)(x + v) = x2 – 5x + 6, so we need u and v
where uv = 6 and u + v = –5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and –2 – 3 = –5,
so x2 – 5x + 6 = (x – 2)(x – 3).
Example A.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2x + 3x = 5x,
so x2 + 5x + 6 = (x + 2)(x + 3)
32. Factoring Trinomials I
c. Factor x2 + 5x – 6
We want (x + u)(x + v) = x2 + 5x – 6, so we need uv = –6 and
u + v = 5.
Since -6 = (–1)(6) = (1)(–6) = (–2)(3) =(2)(–3) and –1 + 6 = 5,
b. Factor x2 – 5x + 6
We want (x + u)(x + v) = x2 – 5x + 6, so we need u and v
where uv = 6 and u + v = –5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and –2 – 3 = –5,
so x2 – 5x + 6 = (x – 2)(x – 3).
Example A.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2x + 3x = 5x,
so x2 + 5x + 6 = (x + 2)(x + 3)
33. Factoring Trinomials I
c. Factor x2 + 5x – 6
We want (x + u)(x + v) = x2 + 5x – 6, so we need uv = –6 and
u + v = 5.
Since -6 = (–1)(6) = (1)(–6) = (–2)(3) =(2)(–3) and –1 + 6 = 5,
so x2 + 5x – 6 = (x – 1)(x + 6).
b. Factor x2 – 5x + 6
We want (x + u)(x + v) = x2 – 5x + 6, so we need u and v
where uv = 6 and u + v = –5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and –2 – 3 = –5,
so x2 – 5x + 6 = (x – 2)(x – 3).
Example A.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2x + 3x = 5x,
so x2 + 5x + 6 = (x + 2)(x + 3)
36. Factoring Trinomials I
Observations About Signs
Given that x2 + bx + c = (x + u)(x + v) so that uv = c,
we observe the following.
1. If c is positive, then u and v have same sign.
In particular,
if b is also positive, then both are positive.
37. Factoring Trinomials I
Observations About Signs
Given that x2 + bx + c = (x + u)(x + v) so that uv = c,
we observe the following.
1. If c is positive, then u and v have same sign.
In particular,
if b is also positive, then both are positive.
From the examples above
x2 + 5x + 6 = (x + 2)(x + 3)
38. Factoring Trinomials I
Observations About Signs
Given that x2 + bx + c = (x + u)(x + v) so that uv = c,
we observe the following.
1. If c is positive, then u and v have same sign.
In particular,
if b is also positive, then both are positive.
if b is negative, then both are negative.
From the examples above
x2 + 5x + 6 = (x + 2)(x + 3)
39. {
Factoring Trinomials I
Observations About Signs
Given that x2 + bx + c = (x + u)(x + v) so that uv = c,
we observe the following.
1. If c is positive, then u and v have same sign.
In particular,
if b is also positive, then both are positive.
if b is negative, then both are negative.
From the examples above
x2 + 5x + 6 = (x + 2)(x + 3)
x2 – 5x + 6 = (x – 2)(x – 3)
40. {
Factoring Trinomials I
Observations About Signs
Given that x2 + bx + c = (x + u)(x + v) so that uv = c,
we observe the following.
1. If c is positive, then u and v have same sign.
In particular,
if b is also positive, then both are positive.
if b is negative, then both are negative.
From the examples above
x2 + 5x + 6 = (x + 2)(x + 3)
x2 – 5x + 6 = (x – 2)(x – 3)
2. If c is negative, then u and v have opposite signs.
41. {
Factoring Trinomials I
Observations About Signs
Given that x2 + bx + c = (x + u)(x + v) so that uv = c,
we observe the following.
1. If c is positive, then u and v have same sign.
In particular,
if b is also positive, then both are positive.
if b is negative, then both are negative.
From the examples above
x2 + 5x + 6 = (x + 2)(x + 3)
x2 – 5x + 6 = (x – 2)(x – 3)
2. If c is negative, then u and v have opposite signs. The
one with larger absolute value has the same sign as b.
42. {
Factoring Trinomials I
Observations About Signs
Given that x2 + bx + c = (x + u)(x + v) so that uv = c,
we observe the following.
1. If c is positive, then u and v have same sign.
In particular,
if b is also positive, then both are positive.
if b is negative, then both are negative.
From the examples above
x2 + 5x + 6 = (x + 2)(x + 3)
x2 – 5x + 6 = (x – 2)(x – 3)
2. If c is negative, then u and v have opposite signs. The
one with larger absolute value has the same sign as b.
From the example above
x2 – 5x – 6 = (x – 6)(x + 1)
45. Factoring Trinomials I
Example B.
a. Factor x2 + 4x – 12
We need u and v having opposite signs such that uv = –12,
u + v = +4.
46. Factoring Trinomials I
Example B.
a. Factor x2 + 4x – 12
We need u and v having opposite signs such that uv = –12,
u + v = +4. Since -12 = (-1)(12) = (-2)(6) = (-3)(4)…
47. Factoring Trinomials I
Example B.
a. Factor x2 + 4x – 12
We need u and v having opposite signs such that uv = –12,
u + v = +4. Since -12 = (-1)(12) = (-2)(6) = (-3)(4)…
They must be –2 and 6
48. Factoring Trinomials I
Example B.
a. Factor x2 + 4x – 12
We need u and v having opposite signs such that uv = –12,
u + v = +4. Since -12 = (-1)(12) = (-2)(6) = (-3)(4)…
They must be –2 and 6 hence x2 + 4x – 12 = (x – 2)(x + 6).
49. Factoring Trinomials I
Example B.
a. Factor x2 + 4x – 12
We need u and v having opposite signs such that uv = –12,
u + v = +4. Since -12 = (-1)(12) = (-2)(6) = (-3)(4)…
They must be –2 and 6 hence x2 + 4x – 12 = (x – 2)(x + 6).
b. Factor x2 – 8x – 12
50. Factoring Trinomials I
Example B.
a. Factor x2 + 4x – 12
We need u and v having opposite signs such that uv = –12,
u + v = +4. Since -12 = (-1)(12) = (-2)(6) = (-3)(4)…
They must be –2 and 6 hence x2 + 4x – 12 = (x – 2)(x + 6).
b. Factor x2 – 8x – 12
We need u and v such that uv = –12, u + v = –8 with
u and v having opposite signs.
51. Factoring Trinomials I
Example B.
a. Factor x2 + 4x – 12
We need u and v having opposite signs such that uv = –12,
u + v = +4. Since -12 = (-1)(12) = (-2)(6) = (-3)(4)…
They must be –2 and 6 hence x2 + 4x – 12 = (x – 2)(x + 6).
b. Factor x2 – 8x – 12
We need u and v such that uv = –12, u + v = –8 with
u and v having opposite signs. This is impossible.
52. Factoring Trinomials I
Example B.
a. Factor x2 + 4x – 12
We need u and v having opposite signs such that uv = –12,
u + v = +4. Since -12 = (-1)(12) = (-2)(6) = (-3)(4)…
They must be –2 and 6 hence x2 + 4x – 12 = (x – 2)(x + 6).
b. Factor x2 – 8x – 12
We need u and v such that uv = –12, u + v = –8 with
u and v having opposite signs. This is impossible.
Hence x2 – 8x – 12 is prime.