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) $(2pq+3q^2)$ and $(3pq-2q^2)$

(vi) $\left(\begin{array}{c}\frac{3}{4}{a}^2+3b^2\end{array}\right)$ and $\left(\begin{array}{c}{a}^2-\frac{2}{3}{b}^2\end{array}\right)$

Answer 1

Multiplying the given polynomials one by one:

$\left(i\right)$

$(2x+5)(4x-3)$

$=2x(4x-3)+5(4x-3)$
$=8x^2-6x+20x-15$
$=8x^2+14x-15$

Hence, $(2x+5)(4x-3)=8x^2+14x-15$.

$\left(ii\right)$

$(y-8)(3y-4)$

$=y(3y-4)-8(3y-4)$

$=3y^2-4y-24y+32$

$=3y^2-28y+32$

Hence, $(y-8)(3y-4)=3y^2-28y+32$

$\left(iii\right)$

$(2.5l-0.5m)(2.5l+0.5m)$

We know that, $(a+b)(a-b)=a^2-b^2$

So, $(2.5l-0.5m)(2.5l+0.5m)=(2.5l{)}^2-(0.5m{)}^2$

$=6.25l^2-0.25m^2$

Hence, $(2.5l-0.5m)(2.5l+0.5m)=6.25l^2-0.25m^2$

$\left(iv\right)$

$(a+3b)(x+5)$

$=a(x+5)+3b(x+5)$

$=ax+5a+3bx+15b$
$=ax+3bx+5a+15b$

Hence, $(a+3b)(x+5)=ax+3bx+5a+15b$

$\left(v\right)$

$(2pq+3q^2)(3pq-2q^2)$

$=2pq(3pq-2q^2)+3q^2(3pq-2q^2)$
$=6p^2{q}^2-4p{q}^3+9p{q}^3-6q^4$
$=6p^2{q}^2+5p{q}^3-6q^4$

Hence, $(2pq+3q^2)(3pq-2q^2)=6p^2{q}^2+5p{q}^3-6q^4$

$\left(vi\right)$

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

$=\frac{3}{4}{a}^2\left(\begin{array}{c}{a}^2-\frac{2}{3}{b}^2\end{array}\right)+3b^2\left(\begin{array}{c}{a}^2-\frac{2}{3}{b}^2\end{array}\right)$

$=\frac{3}{4}{a}^4-\frac{1}{2}{a}^2{b}^2+3a^2{b}^2-2b^4$

$=\frac{3}{4}{a}^4-2b^4+\frac{5}{2}{a}^2{b}^2$

Hence, $(\frac{3}{4}{a}^2+3b^2)(a^2-\frac{2}{3}{b}^2)=\frac{3}{4}{a}^4-2b^4+\frac{5}{2}{a}^2{b}^2$