Question

Multiply the binomials.
(i) (2x + 5) and (4x – 3)
(ii) (y – 8) and (3y – 4)
(iii) (2.5l – 0.5m) and (2.5l + 0.5m)
(iv) (a + 3b) and (x + 5)
(v) (2pq + 3q2) and (3pq – 2q2)
(vi) ($\frac{3}{4}$ a2 + 3b2) and 4(a2 – $\frac{2}{3}$ b2)

Answer 1

(i) (2x + 5) (4x – 3)
= 2x $\times$ 4x – 2x $\times$ 3 + 5 $\times$ 4x – 5$\times$ 3
= 8x² – 6x + 20x -15
= 8x² + 14x -15

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(ii) (y – 8) (3y – 4)
= y $\times$ 3y – 4y – 8 $\times$ 3y + 32
= 3y2 – 4y – 24y + 32
= 3y2 – 28y + 32

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(iii) (2.5l – 0.5m) (2.5l + 0.5m)
= 2.5l $\times$ 2.5 l + 2.5l $\times$ 0.5m – 0.5m $\times$ 2.5l – 0.5m $\times$ 0.5m
= 6.25l2 + 1.25 lm – 1.25 lm – 0.25 m2
= 6.25l2 – 0.25 m2

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**(iv) **(a + 3b) (x + 5)
= ax + 5a + 3bx + 15b

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(v) (2pq + 3q2) (3pq – 2q2)
= 2pq $\times$ 3pq – 2pq $\times$ 2q2 + 3q2 $\times$ 3pq – 3q2 $\times$ 2q2
= 6p2q2 – 4pq3 + 9pq3 – 6q4
= 6p2q2 + 5pq3 – 6q4

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(vi) ($\frac{3}{4}$ a² + 3b² ) and 4( a² – $\frac{2}{3}$ b²)

=($\frac{3}{4}$ a² + 3b²) $\times$ 4(a² – $\frac{2}{3}$ b²)

=($\frac{3}{4}$ a² + 3b²) $\times$ (4a² – $\frac{8}{3}$ b²)

=$\frac{3}{4}$ a² $\times$ (4a² – $\frac{8}{3}$ b²) + 3b²$\times$(4a² – $\frac{8}{3}$ b²)

=$\frac{3}{4}$ a² $\times$ 4a² -$\frac{3}{4}$ a²$\times$ $\frac{8}{3}$ b² + 3b² $\times$ 4a² – 3b² $\times$ $\frac{8}{3}$ b²

=3a4– 2a² b² + 12 a² b² – 8b4

= 3a4 + 10a² b² – 8b4