9 factoring trinomials

Factoring Trinomials I

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

Factoring Trinomials ITrinomials (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


(#x + #)(#x + #) ax2 + bx + cHence, to factor a trinomial, we write the trinomial as a product of two binomials, if possible,


(#x + #)(#x + #) ax2 + bx + cHence, 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.


To factor x2 + bx + c, we note hatif (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:




To factor x2 + bx + c, we note hatif (x + u)(x + v) = x2 + ux + vx + uv





To factor x2 + bx + c, we note hatif (x + u)(x + v) = x2 + ux + vx + uv = x2 + (u + v)x + uv





To factor x2 + bx + c, we note hatif (x + u)(x + v) = x2 + ux + vx + uv = x2 + (u + v)x + uv= x2 + bx + c,





To factor x2 + bx + c, we note hatif (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,





To factor x2 + bx + c, we note hatif (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.





To factor x2 + bx + c, we note hatif (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).




Factoring Trinomials IExample A. a. Factor x2 + 5x + 6

Factoring Trinomials IExample A. a. Factor x2 + 5x + 6We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v where uv = 6

Factoring Trinomials IExample A. a. Factor x2 + 5x + 6We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v where uv = 6 and u + v = 5.

Factoring Trinomials IExample A. a. Factor x2 + 5x + 6We 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)

Factoring Trinomials IExample A. a. Factor x2 + 5x + 6We 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

Factoring Trinomials IExample A. a. Factor x2 + 5x + 6We 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,

Factoring Trinomials IExample A. a. Factor x2 + 5x + 6We 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)


b. Factor x2 – 5x + 6

Example A. a. Factor x2 + 5x + 6We 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)


b. Factor x2 – 5x + 6We want (x + u)(x + v) = x2 – 5x + 6,



b. Factor x2 – 5x + 6We want (x + u)(x + v) = x2 – 5x + 6, so we need u and v where uv = 6



b. Factor x2 – 5x + 6We want (x + u)(x + v) = x2 – 5x + 6, so we need u and v where uv = 6 and u + v = –5.



b. Factor x2 – 5x + 6We 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)



b. Factor x2 – 5x + 6We 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,



b. Factor x2 – 5x + 6We 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).



c. Factor x2 + 5x – 6




c. Factor x2 + 5x – 6 We want (x + u)(x + v) = x2 + 5x – 6, so we need uv = –6




c. Factor x2 + 5x – 6 We want (x + u)(x + v) = x2 + 5x – 6, so we need uv = –6 and u + v = 5.




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)




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,




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).



Factoring Trinomials IObservations About Signs


Given that x2 + bx + c = (x + u)(x + v) so that uv = c, we observe the following.


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.


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)


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)

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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)

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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.

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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.

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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)

Factoring Trinomials IExample B.

a. Factor x2 + 4x – 12


a. Factor x2 + 4x – 12 We need u and v having opposite signs such that uv = –12,


a. Factor x2 + 4x – 12 We need u and v having opposite signs such that uv = –12, u + v = +4.


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)…


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


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



b. Factor x2 – 8x – 12 We need u and v such that uv = –12, u + v = –8 with u and v having opposite signs.



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.



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.

Factoring Trinomials IExercise. 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


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 + 2544. What can you conclude from 41–43?

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9 factoring trinomials