Write the following squares of binomials as trinomials:

Solution:

We will use the identities $(a+b)^{2}=a^{2}+2 a b+b^{2}$ and $(a-b)^{2}=a^{2}-2 a b+b^{2}$ to convert the squares of binomials as trinomials.

(i) $(x+2)^{2}$

$=x^{2}+2 \times x \times 2+b$

$=x^{2}+4 x+b^{2}$

(ii) $(8 a+3 b)^{2}$

$=(8 a)^{2}+2(8 a)(3 b)+(6 b)^{2}$

$=64 a^{2}+48 a b+36 b^{2}$

(iii) $(2 m+1)^{2}$

$=(2 m)^{2}+2(2 m)(1)+1^{2}$

$=4 m^{2}+4 m+1$

$($ iv $)\left(9 a+\frac{1}{6}\right)^{2}$

$=(9 a)^{2}+2(9 a)\left(\frac{1}{6}\right)+\left(\frac{1}{6}\right)^{2}$

$=81 a^{2}+3 a+\frac{1}{36}$

$(\mathrm{v})\left(x+\frac{x^{2}}{2}\right)^{2}$

$=x^{2}+2 x\left(\frac{x^{2}}{2}\right)+\left(\frac{x^{2}}{2}\right)^{2}$

$=x^{2}+x^{3}+\frac{x^{4}}{4}$

$(\mathrm{vi})\left(\frac{x}{4}-\frac{y}{3}\right)^{2}$

$=\left(\frac{x}{4}\right)^{2}-2\left(\frac{x}{4}\right)\left(\frac{y}{3}\right)+\left(\frac{y}{3}\right)^{2}$

$=\frac{x^{2}}{16}-\frac{1}{6} x y+\frac{y^{2}}{9}$

(viii) $\left(\frac{x}{y}-\frac{y}{x}\right)^{2}$

$=\left(\frac{x}{y}\right)^{2}-2\left(\frac{x}{y}\right)\left(\frac{y}{x}\right)+\left(\frac{y}{x}\right)^{2}$

$=\frac{x^{2}}{y^{2}}-2+\frac{y^{2}}{x^{2}}$

$(\mathrm{ix})\left(\frac{3 a}{2}-\frac{5 b}{4}\right)^{2}$

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

$=\frac{9 a^{2}}{4}-\frac{15 a b}{4}+\frac{25 b^{2}}{16}$

$(\mathrm{x})\left(a^{2} b-b c^{2}\right)^{2}$

$=\left(a^{2} b\right)^{2}-2\left(a^{2} b\right)\left(b c^{2}\right)+\left(b c^{2}\right)^{2}$

$=a^{4} b^{2}-2 a^{2} b^{2} c^{2}+b^{2} c^{4}$

$(\mathrm{xi})\left(\frac{2 a}{3 b}+\frac{2 b}{3 a}\right)^{2}$

$=\left(\frac{2 a}{3 b}\right)^{2}+2\left(\frac{2 a}{3 b}\right)\left(\frac{2 b}{3 a}\right)+\left(\frac{2 b}{3 a}\right)^{2}$

$=\frac{4 a^{2}}{9 b^{2}}+\frac{8}{9}+\frac{4 b^{2}}{9 a^{2}}$

$($ xii $)\left(x^{2}-a y\right)^{2}$

$=\left(x^{2}\right)^{2}-2 x^{2}(a y)+(a y)^{2}$

$=x^{4}-2 x^{2} a y+a^{2} y^{2}$