Factorise : $\frac{25}{4} x^{2}-\frac{y^{2}}{9}$
we have $\frac{25}{4} x^{2}-\frac{y^{2}}{9}=\left(\frac{5}{2} x\right)^{2}-\left(\frac{y}{3}\right)^{2}$
Now comparing it with Identity $III$, we get
$\frac{25}{4} x^{2}-\frac{y^{2}}{9}=\left(\frac{5}{2} x\right)^{2}-\left(\frac{y}{3}\right)^{2}$
$=\left(\frac{5}{2} x+\frac{y}{3}\right)\left(\frac{5}{2} x-\frac{y}{3}\right)$
Write the following cubes in the expanded form : $(5 p-3 q)^{3}$
Check whether $-2$ and $2$ are zeroes of the polynomial $x + 2$.
Factorise of the following : $8 a^{3}-b^{3}-12 a^{2} b+6 a b^{2}$
Expand each of the following, using suitable identities : $(-2 x+3 y+2 z)^{2}$
Determine which of the following polynomials has $(x + 1)$ a factor : $x^{4}+x^{3}+x^{2}+x+1$.