$\displaystyle \begin{array}{{>{\displaystyle}l}} a) 4x^{2} -12x+9\\ =( 2x)^{2} -2.2.3x+3^{2}\\ =( 2x-3)^{2}\\ b) 4x^{2} +4x+1\ \\ =( 2x)^{2} +2.2+1\ \\ =( 2x+1)^{2}\\ c) 1+12x+36x^{2}\\ =1+2.6x+( 6x)^{2}\\ =( 1+6x)^{2}\\ d) 9x^{2} -24xy+16y^{2}\\ =( 3x)^{2} -2.4.3xy+( 4y)^{2}\\ =( 3x-4y)^{2}\\ e)\frac{x^{2}}{4} +2xy+4y^{2}\\ =\frac{x^{2} +8xy+16y^{2}}{4}\\ =\frac{x^{2} +2.4xy+( 4y)^{2}}{4}\\ =\frac{( x+4y)^{2}}{4} \ \\ e)\frac{x^{2}}{4} +2xy+4\\ =\frac{x^{2} +8xy+16}{4}\\ =\frac{( x+4)^{2}}{4} \ \\ f) -x^{2} +10x-25\\ -\left( x^{2} -10x+25\right) =-( x-5)^{2}\\ g) -16a^{4} b^{6} -24a^{5} b^{5} -9a^{6} b^{4}\\ =-\left[\left( 4a^{2} b^{3}\right)^{2} +2.4a^{2} b^{3} .3a^{3} +\left( 3a^{3} b^{2}\right)^{2}\right]\\ =-\left( 4a^{2} b^{3} +3a^{3} b^{2}\right)^{2} \ \\ h) 25x^{2} -20xy+4y^{2}\\ =( 5x)^{2} -2.5.2+( 2y)^{2}\\ =( 5x-2y)^{2}\\ i) 25x^{4} -10x^{2} y+y^{2}\\ =\left( 5x^{2}\right)^{2} -2.5x^{2} y+y^{2}\\ =( 5x-y)^{2}\\ k)( 3x-1)^{2} -16\\ =( 3x-1+4)( 3x-1-4)\\ =( 3x+3)( 3x-5) \ \\ l)( 5x-4)^{2} -49x^{2}\\ =( 5x-4+7x)( 5x-4-7x)\\ =( 12x-4)( -2x-4) \ \\ =-8( x-1)( x-2) \ \\ m)( 2x+5)^{2} -( x-9)^{2}\\ =( 2x+5-x+9)( 2x+5+x-9)\\ =( x+14)( 3x-4) \ \\ n)( 3x+1)^{2} -4( x-2)^{2}\\ =( 3x+1-2x+4)( 3x+1+2x-4)\\ =( x+5)( 5x-3) \ \\ o) 9( 2x+3)^{2} -4( x+1)^{2}\\ =( 6x+9-2x-2)( 6x+9+2x+2)\\ =( 4x-7)( 8x+11) \ \\ p) 4b^{2} c^{2} -\left( b^{2} +c^{2} -a^{2}\right)^{2}\\ =\left( 2bc-b^{2} -c^{2} +a^{2}\right)\left( 2bc+b^{2} +c^{2} -a^{2}\right)\\ =\left[ a^{2} -\left( b^{2} -2bc+c^{2}\right)\right]\left[( b+c)^{2} -a^{2}\right]\\ =( a-b+c)( a+b-c)( b+c-a)( b+c+a) \ \\ \end{array}$
