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) $(\frac{3}{4} a^{2} + 3 b^{2})$ and $4 (a^{2} - \frac{2}{3} b^{2})$

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Solution

(i)(2x + 5) ×(4x − 3) = 2x ×(4x − 3) + 5 ×(4x − 3)
=8x² − 6x + 20x − 15
= 8x² + 14x −15(By adding liketerms)
(ii)(y −8) × (3y − 4) =y × (3y −4) − 8 × (3y −4)
=3y² − 4y − 24y + 32
=3y² − 28y + 32 (By adding liketerms)
(iii)(2.5l −0.5m) × (2.5l + 0.5m)= 2.5l × (2.5l+ 0.5m) − 0.5m (2.5l + 0.5m)
=6.25l² + 1.25lm − 1.25lm −0.25m²
=6.25l² − 0.25m²
(iv)(a + 3b)× (x + 5) = a× (x + 5) + 3b× (x + 5)
= ax + 5a + 3bx + 15b
(v)(2pq + 3q²)× (3pq − 2q²)= 2pq × (3pq −2q²) + 3q² ×(3pq − 2q²)
=6p²q² − 4pq³+ 9pq³ − 6q⁴
=6p²q² + 5pq³ −6q⁴
(vi) Given $(\frac{3}{4} a^{2} + 3 b^{2})$ and $4 (a^{2} - \frac{2}{3} b^{2})$
$\Rightarrow (\frac{3}{4} a^{2} + 3 b^{2}) \times 4 (a^{2} - \frac{2}{3} b^{2}) = \frac{3}{4} a^{2} [4 (a^{2} - \frac{2}{3} b^{2}) + 3 b^{2} [4 (a^{2} - \frac{2}{3} b^{2}]$
$= 3 a^{2} (a^{2} - \frac{2}{3} b^{2}) + 12 b^{2} (a^{2} - \frac{2}{3} b^{2})$
$= 3 a^{4} - 2 a^{2} b^{2} + 12 a^{2} b^{2} - 8 b^{4}$
$= 3 a^{4} + 10 a^{2} b^{2} - 8 b^{4}$

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