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- 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 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. To factor x2 + bx + c, we note hat if (x + u)(x + v)
- 8. 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. To factor x2 + bx + c, we note hat if (x + u)(x + v) = x2 + ux + vx + uv
- 9. 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. To factor x2 + bx + c, we note hat if (x + u)(x + v) = x2 + ux + vx + uv = x2 + (u + v)x + uv
- 10. 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. 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,
- 11. 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. 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,
- 12. 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. 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.
- 13. 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. 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).
- 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 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) b. Factor x2 – 5x + 6
- 22. 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) b. Factor x2 – 5x + 6 We want (x + u)(x + v) = x2 – 5x + 6,
- 23. 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) b. Factor x2 – 5x + 6 We want (x + u)(x + v) = x2 – 5x + 6, so we need u and v where uv = 6
- 24. 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) 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.
- 25. 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) 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)
- 26. 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) 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,
- 27. 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) 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).
- 28. 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) 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). c. Factor x2 + 5x – 6
- 29. 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) 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). c. Factor x2 + 5x – 6 We want (x + u)(x + v) = x2 + 5x – 6, so we need uv = –6
- 30. 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) 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). c. Factor x2 + 5x – 6 We want (x + u)(x + v) = x2 + 5x – 6, so we need uv = –6 and u + v = 5.
- 31. 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) 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). 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)
- 32. 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) 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). 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,
- 33. 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) 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). 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).
- 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 14. x2 – 8x – 20 15. x2 – 9x – 20 16. x2 – 9x + 20 17. x2 – 10x – 24 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 25. x2 – 36 26. x2 + 36 B. Factor. Factor out the GCF, the “–”, and arrange the terms in order first if necessary. 27. –x2 – 5x + 14 28. 2x3 – 18x2 + 40x 29. 3x2 – 30x – 72 30. –2x3 + 20x2 – 24x 31. –2x4 + 18x2 32. –3x – 24x3 + 22x2 33. 5x4 + 10x5 – 5x3
- 54. Factoring Trinomials I C. Factor. Factor out the GCF, the “–”, and arrange the terms in order first. 34. –yx2 + 4yx + 5y 35. –3x3 – 30x2 – 48x 36. –2x3 + 20x2 – 24x 40. 4x2 – 44xy + 96y2 37. –x2 + 11xy + 24y2 38. x4 – 6x3 + 36x2 39. –x2 + 9xy + 36y2 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?

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