Question

Factorise P(x)= (x+3)(12x^2-3)+(2x-1)(x^2-9) And Q(x)= 4x^4-x^2

Answer 1

Answer:

a2 answers to exercises

ii. yes. suppose a is negative, so write a = -b where b is positive. let c be a

positive number such that c" = b. then (-c)" = a because (-1)" = -1 since

n is odd.

ii, §1, p. 24

3. x negative, y positive 4. x negative, y negative

ii, §3, p. 33

5. y = -~x + i 6. y = -ix + s 7. x = j2

9 9j3 8. y = --;:;- x + 4 - --;:;- 9. y = 4x - 3 10. y = -2x + 2

",3+3 ",3+3

ii. y= -!x+3+ j'!- 12. y=j3x+s+j3 19. -! 20.-8

21. 2 + j2 22.!<3 + j3) 23. y = (x - n>(j22_ n) + 1

24. y=(x-j2)c~-~)+2 25. y= -(x+ l)(j23+ 1)+2

26. y = (x + 1)(3 + j2) + j2 29. (a) x = -4, y = -7 (b) x = ~, y = i

(c) x = -!, y = 1 (d) x = - 6, y = - 5

ii, §4, p. 35

1.j97 2.j2 3.js2 4.ji3 5.!j5 6.(4,-3) 7. sands 8.(-2,5)

9. sand 7

ii, §8, p. 51

5. (x - 2)2 + (y + 1)2 = 25 6. x 2 + (y - 1)2 = 9 7. (x + \)2 + y2 = 3

8. y + ¥ = 2(x + w 9. y - 1 = (x + 2)2 10. y + 4 = (x - \)2

ii. (x + 1)2 + (y - 2)2 = 2 12. (x - 2)2 + (y - \)2 = 2

13. x + ¥- = 2(y + !)2 14. x-i = (y + 2)2

iii, §1, p. 61

1. 4 2. -2 3. 2 4. i s. -! 6. 0 7. 4 8. 6 9. 3 10. 12 ii. 2

12. 3 13. a

Answer 2

Answer:

Q is x^2((4x+1) (4x-1)) and P is (2x-1)(x-3) (x+3) + 3(x+3)(4x+1)(4x-1)

Step-by-step explanation:

You can further factor the equations by simplifying certain areas, i.e. taking 3 out of (12x^2-3) to make it factorable, or taking x^2 out of the second one to make it factorable