Trinomials are polynomials of the form ax^2 + bx + c, where a, b, and c are numbers. To factor a trinomial, we write it as the product of two binomials (x + u)(x + v) where uv = c and u + v = b. For example, to factor x^2 + 5x + 6, we set uv = 6 and u + v = 5. The only possible values are u = 2 and v = 3, so x^2 + 5x + 6 = (x + 2)(x + 3). Similarly, to factor x^2 - 5x + 6, we set uv = 6 and u + v = -5,

1. Factoring Trinomials I

2. Factoring Trinomials I
Trinomials (three-term) are polynomials of the form
ax2 + bx + c where a, b, and c are numbers.

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.

14. Factoring Trinomials I
Example A.
a. Factor x2 + 5x + 6

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)

34. Factoring Trinomials I
Observations About Signs

35. Factoring Trinomials I
Observations About Signs
Given that x2 + bx + c = (x + u)(x + v) so that uv = c,
we observe the following.

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)

43. Factoring Trinomials I
Example B.
a. Factor x2 + 4x – 12

44. Factoring Trinomials I
Example B.
a. Factor x2 + 4x – 12
We need u and v having opposite signs such that uv = –12,

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.

53. Factoring Trinomials I
Exercise. A. Factor. If it’s prime, state so.
1. x2 – x – 2 2. x2 + x – 2 3. x2 – x – 6 4. x2 + x – 6
5. x2 – x + 2 6. x2 + 2x – 3 7. x2 + 2x – 8 8. x2 – 3x – 4
9. x2 + 5x + 6 10. x2 + 5x – 6
13. x2 – x – 20
11. x2 – 5x – 6
12. x2 – 5x + 6
17. x2 – 10x – 24
14. x2 – 8x – 20
15. x2 – 9x – 20 16. x2 – 9x + 20
18. x2 – 10x + 24 19. x2 – 11x + 24 20. x2 – 11x – 24
21. x2 – 12x – 36 22. x2 – 12x + 36 23. x2 – 13x – 36
24. x2 – 13x + 36
B. Factor. Factor out the GCF, the “–”, and arrange the
terms in order first if necessary.
29. 3x2 – 30x – 7227. –x2 – 5x + 14 28. 2x3 – 18x2 + 40x
30. –2x3 + 20x2 – 24x
25. x2 – 36 26. x2 + 36
31. –2x4 + 18x2
32. –3x – 24x3 + 22x2 33. 5x4 + 10x5 – 5x3

54. Factoring Trinomials I
35. –3x3 – 30x2 – 48x34. –yx2 + 4yx + 5y
36. –2x3 + 20x2 – 24x
40. 4x2 – 44xy + 96y2
37. –x2 + 11xy + 24y2
38. x4 – 6x3 + 36x2 39. –x2 + 9xy + 36y2
C. Factor. Factor out the GCF, the “–”, and arrange the
terms in order first.
D. Factor. If not possible, state so.
41. x2 + 1 42. x2 + 4 43. x2 + 9 43. 4x2 + 25
44. What can you conclude from 41–43?