Everything Maths Grade 10

10. 4 x^2 (4x 3 y)^2 11. 16 a^2 (3b+ 4c)^2 12. (b4)^2 9(b5)^2


11. 4(a3)^2 49(4a5) 14. 16 k^2 4 15.a^2 b^2 c^2 1


12. 
    

1

9

a^2  4 b^2 17.

1

2

x^2  2 18.y^2  8

19. y^2 13 20.a^2 (a 2 ab 15 b^2 ) 9 b^2 (a^2 2 ab 15 b^2 )

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1.2DJM 2.2DJN 3.2DJP 4.2DJQ 5.2DJR 6.2DJS 7.2DJT 8.2DJV

9.2DJW 10.2DJX 11.2DJY 12.2DJZ 13.2DK2 14.2DK3 15.2DK4 16.2DK5

17.2DK6 18.2DK7 19.2DK8 20.2DK9

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Factorising by grouping in pairs EMAK

The taking out of common factors is the starting point in all factorisation problems. We know that the factors
of 3 x+ 3are 3 and(x+ 1). Similarly, the factors of 2 x^2 + 2xare 2 xand(x+ 1). Therefore, if we have an
expression:
2 x^2 + 2x+ 3x+ 3

there is no common factor to all four terms, but we can factorise as follows:
(
2 x^2 + 2x

)

+ (3x+ 3) = 2x(x+ 1) + 3 (x+ 1)

We can see that there is another common factor(x+ 1). Therefore, we can write:

(x+ 1) (2x+ 3)

We get this by taking out the(x+ 1)and seeing what is left over. We have 2 xfrom the first group and +3
from the second group. This is called factorising by grouping.

Worked example 12: Factorising by grouping in pairs

QUESTION

Find the factors of 7 x+ 14y+bx+ 2by.

SOLUTION

Step 1: There are no factors common to all terms

Step 2: Group terms with common factors together

7 is a common factor of the first two terms andbis a common factor of the second two terms. We see that the

ratio of the coefficients7 : 14is the same asb: 2b.

7 x+ 14y+bx+ 2by= (7x+ 14y) + (bx+ 2by)

= 7 (x+ 2y) +b(x+ 2y)

Chapter 1. Algebraic expressions 23