Step 2: Find the three terms in the second bracket
We can replacexandyin the factorised form of the expression for the difference of two cubes with 2 tand 5 p.
Doing so we get the second bracket:
(
8 t^3 + 125p^3)
= (2t+ 5p)[
(2t)^2 (2t) (5p) + (5p)^2]
= (2t+ 5p)(
4 t^2 10 tp+ 25p^2)
Step 3: Expand the brackets to check that the expression has been correctly factorised
(2t+ 5p)(
4 t^2 10 tp+ 25p^2)
= 2t(
4 t^2 10 tp+ 25p^2)
+ 5p(
4 t^2 10 tp+ 25p^2)
= 8t^3 20 pt^2 + 50p^2 t+ 20pt^2 50 p^2 t+ 125p^3
= 8t^3 + 125p^3Exercise 1 – 9:
Factorise:
1.w^3 8 2. g^3 + 64 3. h^3 + 1
4.x^3 + 8 5. 27 m^3 6. 2 x^3 2 y^3
- 3 k^3 + 81q^3 8. 64 t^3 1 9. 64 x^2 1
- 125 x^3 + 1 11. 25 x^3 + 1 12. z 125 z^4
- 8 m^6 +n^9 14. 216 n^3 k^3 15. 125 s^3 +d^3
- 8 k^3 +r^3 17. 8 j^3 k^3 l^3 b^3 18. 27 x^3 y^3 +w^3
- 128 m^3 + 2f^3 20.p^15
1
8
y^12 21.27
t^3 s^31
64 q^3 h^3 23. 72 g^3 +1
3
v^3 24. 1 (x y)^325.h^4 (8g^6 +h^3 ) (8g^6 +h^3 ) 26. x(125w^3 h^3 ) +y(125w^3 h^3 )
27.x^2 (27p^3 +w^3 ) 5 x(27p^3 +w^3 ) 6(27p^3 +w^3 )
For more exercises, visit http://www.everythingmaths.co.za and click on ’Practise Maths’.
1.2DMW 2.2DMX 3.2DMY 4.2DMZ 5.2DN2 6.2DN3 7.2DN4
8.2DN5 9.2DN6 10.2DN7 11.2DN8 12.2DN9 13.2DNB 14.2DNC
15.2DND 16.2DNF 17.2DNG 18.2DNH 19.2DNJ 20.2DNK 21.2DNM
22.2DNN 23.2DNP 24.2DNQ 25.2DNR 26.2DNS 27.2DNThttp://www.everythingmaths.co.za m.everythingmaths.co.za30 1.7. Factorisation