Q3 of 59 Page 32

Simplify the following:

(i) (2xy – xy)(3xy – 5)


(ii) (3xy2 + 1)(4xy – 6xy2)


(iii) (3x2 + 2x)(2x2 + 3)


(iv) (2m3 + 3m)(5m – 1).

By using the distributive law

(i) (2xy – xy)(3xy – 5)


= 2xy (3xy – 5) –xy (3xy –5)


= 6x2y2 –10xy – 3x2y2 +5xy


= 3x2y2 – 5xy (after adding and subtracting the like terns)


(ii) (3xy2 + 1)(4xy – 6xy2)


By using the distributive law


3xy2 ((4xy – 6xy2) +1((4xy – 6xy2)


= 3xy2 × 4xy – 3xy2 × 6xy2 + 4xy – 6xy2


= 12x2y3 – 18 x2y4 + 4xy –6xy2


(iii) (3x2 + 2x)(2x2 + 3)


By using the distributive law


3x2 (2x2 + 3) +2x (2x2 + 3)


= 3x2 × 2x2 + 3x2 × 3 + 2x × 2x2 + 2x × 3


= 6x4 + 9x2 + 4x3 +6x


(iv) (2m3 + 3m)(5m – 1).


By using the distributive law


2m3 (5m –1) +3m (5m –1)


= 2m3 × 5m – 1× 2m3 + 3m × 5m – 1× 3m


= 10m4 – 2m3 + 15m2 – 3m


More from this chapter

All 59 →
1

Complete the following table of products of two monomials:

 2

Find the products:

(i) (5x + 8)3x


(ii) (–3pq) (–15p3q2 – q3)


(iii) (3a3 – 3b3)


(iv) –x2(x – 15).

 1

Find the product:

(i) (a + 3)(a + 5)


(ii) (3t + 1)(3t + 4)


(iii) (a – 8)(a + 2)


(iv) (a – 6)(a – 2)

 2

Evaluate using suitable identities:

(i) 53 × 55 (ii) 102 × 106


(iii) 34 × 36 (iv) 103 × 96