a)  ${\displaystyle x^{2}+14x+49=x^{2}+2\times 7x+7^{2}=(x+7)^{2}}$

b)  ${\displaystyle 9x^{2}-6x+1=(3x)^{2}-2\times 3x+1^{2}=(3x+1)^{2}}$

c)  ${\displaystyle 2x^{2}+x{\sqrt {24}}+3=2x^{2}+2x{\sqrt {6}}+3=(x{\sqrt {2}})^{2}+2\times x{\sqrt {2}}\times {\sqrt {3}}+({\sqrt {3}})^{2}=(x{\sqrt {2}}+{\sqrt {3}})^{2}}$

d)  ${\displaystyle {\frac {1}{4}}-3x+9x^{2}=\left({\frac {1}{2}}\right)^{2}-2\times {\frac {1}{2}}3x+(3x)^{2}=\left({\frac {1}{2}}-3x\right)^{2}}$

e)  ${\displaystyle x^{2}-25=x^{2}-5^{2}=(x+5)(x-5)}$

f)  ${\displaystyle 9x^{2}-16=(3x)^{2}-4^{2}=(3x+4)(3x-4)}$

g)  ${\displaystyle 3x^{2}-1=(x{\sqrt {3}})^{2}-1^{2}=(x{\sqrt {3}}+1)(x{\sqrt {3}}-1)}$

h)  ${\displaystyle (x-3)^{2}-16=(x-3)^{2}-4^{2}=(x-3+4)(x-3-4)=(x+1)(x-7)}$

i)  ${\displaystyle (x-3)^{2}-2=(x-3)^{2}-({\sqrt {2}})^{2}=(x-3+{\sqrt {2}})(x-3-{\sqrt {2}})}$

j)  ${\displaystyle 4(x-3)^{2}-25(1-3x)^{2}=[2(x-3)]^{2}-[5(1-3x)]^{2}=[2(x-3)+5(1-3x)][2(x-3)-5(1-3x)]=(2x-6+5-15x)(2x-6-5+15x)=(-13x-1)(17x-11)}$

a)  ${\displaystyle (3x+1)(-4x+5)-2(3x+1)=(3x+1)[(-4x+5)-2]=(3x+1)(-4x+3)}$

b)  ${\displaystyle (x-5)(x+3)-(2x+7)(x+3)=(x+3)[(x-5)-(2x+7)]=(x+3)(-x-12)}$

c)  ${\displaystyle (x+2)(5x-1)-(5x+2)(2x+4)=(x+2)(5x-1)-2(5x+2)(x+2)=(x+2)[(5x-1)-2(5x+2)]=(x+2)(5x-1-10x-4)=(x+2)(-5x-5)=-5(x+2)(x+1)}$

d)  ${\displaystyle (2x+3)^{2}-(4x+6)(-4x+2)=(2x+3)^{2}-2(2x+3)(-4x+2)=(2x+3)[(2x+3)-2(-4x+2)]=(2x+3)(2x+3+8x-4)=(2x+3)(10x-1)}$

${\displaystyle {\begin{aligned}a)\quad 4x^{2}-1-(2x-1)(3x+5)&=(2x)^{2}-1-(2x-1)(3x+5)\\&=(2x+1)(2x-1)-(2x-1)(3x+5)\\&=(2x-1)[(2x+1)-(3x+5)]\\&=(2x-1)(-x-4)\\&=-(2x-1)(x+4)\end{aligned}}}$

${\displaystyle {\begin{aligned}b)\quad 4x^{2}+4x+1-(2x+1)(3x+5)&=(2x)^{2}+2\times 2x+1-(2x+1)(3x+5)\\&=(2x+1)^{2}-(2x+1)(3x+5)\\&=(2x+1)[(2x+1)-(3x+5)]\\&=(2x+1)(-x-4)\\&=-(2x+1)(x+4)\end{aligned}}}$

${\displaystyle {\begin{aligned}c)\quad ab(a+b)+bc(a^{2}+2ab+b^{2})+b^{2}(a^{2}-b^{2})&=ab(a+b)+bc(a+b)^{2}+b^{2}(a+b)(a-b)\\&=b(a+b)[a+c(a+b)+b(a-b)]\\&=b(a+b)(a+ac+bc+ab-b^{2})\end{aligned}}}$

${\displaystyle {\begin{aligned}d)\quad 3x(x-1)-6(x^{2}-2x+1)+9x(x^{2}-1)&=3x(x-1)-6(x-1)^{2}+9x(x+1)(x-1)\\&=3(x-1)[x-2(x-1)+3x(x+1)]\\&=3(x-1)(3x^{2}+2x+2)\end{aligned}}}$

${\displaystyle {\begin{aligned}e)\quad 2xy(y^{2}-9)-4x(3-y)+8x(y^{2}-6y+9)&=2xy(y+3)(y-3)+4x(y-3)+8x(y-3)^{2}\\&=2x(y-3)[y(y+3)+2+4(y-3)]\\&=2x(y-3)[y^{2}+3y+2+4y-12]\\&=2x(y-3)(y^{2}+7y-10)\end{aligned}}}$

${\displaystyle {\begin{aligned}a)\quad (x+3y)^{2}-(x-5y)^{2}&=(x+3y+x-5y)(x+3y-x+5y)\\&=(2x-2y)(8y)\\&=16y(x-y)\end{aligned}}}$

${\displaystyle {\begin{aligned}b)\quad (x^{2}+2xy+y^{2})-(m^{2}-2mn+n^{2})&=(x+y)^{2}-(m-n)^{2}\\&=(x+y+m-n)(x+y-m+n)\end{aligned}}}$

${\displaystyle {\begin{aligned}c)\quad (5+2{\sqrt {5a}}+a)-(3-2{\sqrt {3b}}+b)&=(({\sqrt {5}})^{2}+2{\sqrt {5}}{\sqrt {a}}+({\sqrt {a}})^{2})-(({\sqrt {3}})^{2}-2{\sqrt {3}}{\sqrt {b}}+({\sqrt {b}})^{2})\\&=({\sqrt {5}}+{\sqrt {a}})^{2}-({\sqrt {3}}-{\sqrt {b}})^{2}\\&=({\sqrt {5}}+{\sqrt {a}}+{\sqrt {3}}-{\sqrt {b}})({\sqrt {5}}+{\sqrt {a}}-{\sqrt {3}}+{\sqrt {b}})\end{aligned}}}$

${\displaystyle {\begin{aligned}d)\quad a^{2}-b^{2}-c^{2}-2bc&=a^{2}-\left(b^{2}+c^{2}+2bc\right)\\&=a^{2}-\left(b+c\right)^{2}\\&=(a+b+c)(a-b-c)\end{aligned}}}$

${\displaystyle {\begin{aligned}a)\quad abc^{2}-abd^{2}+ce-de&=ab(c^{2}-d^{2})+e(c-d)\\&=ab(c+d)(c-d)+e(c-d)\\&=(c-d)[ab(c+d)+e]\\&=(c-d)(abc+abd+e)\end{aligned}}}$

${\displaystyle {\begin{aligned}b)\quad a^{2}+2ab(a-b)-b^{2}&=a^{2}-b^{2}+2ab(a-b)\\&=(a+b)(a-b)+2ab(a-b)\\&=(a-b)(a+b+2ab)\end{aligned}}}$

${\displaystyle {\begin{aligned}c)\quad x^{2}+2+{\frac {1}{x^{2}}}&=x^{2}+2x{\frac {1}{x}}+\left({\frac {1}{x}}\right)^{2}\\&=\left(x+{\frac {1}{x}}\right)^{2}\end{aligned}}}$

${\displaystyle {\begin{aligned}d)\quad ax+ay+bx+by+(a-b)^{2}+4ab&=a(x+y)+b(x+y)+a^{2}-2ab+b^{2}+4ab\\&=(a+b)(x+y)+a^{2}+2ab+b^{2}\\&=(a+b)(x+y)+(a+b)^{2}\\&=(a+b)(x+y+a+b)\end{aligned}}}$

${\displaystyle {\begin{aligned}a)\quad (x+2{\sqrt {3}}-5)^{2}-(x-3{\sqrt {3}}+2)^{2}&=(x+2{\sqrt {3}}-5+x-3{\sqrt {3}}+2)(x+2{\sqrt {3}}-5-x+3{\sqrt {3}}-2)\\&=(2x-{\sqrt {3}}-3)(5{\sqrt {3}}-7)\end{aligned}}}$

${\displaystyle {\begin{aligned}b)\quad x^{2}+2x(x-1)+(x-1)^{2}&=[x+(x-1)]^{2}\\&=(2x-1)^{2}\end{aligned}}}$

${\displaystyle {\begin{aligned}c)\quad m^{2}+{\frac {1}{m^{2}}}+1&=m^{2}+2+{\frac {1}{m^{2}}}-1\\&=m^{2}+2m.{\frac {1}{m}}+\left({\frac {1}{m}}\right)^{2}-1^{2}\\&=(m+{\frac {1}{m}})^{2}-1^{2}\\&=(m+{\frac {1}{m}}+1)(m+{\frac {1}{m}}-1)\end{aligned}}}$

${\displaystyle {\begin{aligned}d)\quad 4x^{2}+4x(x+3)+(x+3)^{2}+6(x+1)&=(2x)^{2}+2\times 2x(x+3)+(x+3)^{2}+6(x+1)\\&=(2x+x+3)^{2}+6(x+1)\\&=(3x+3)^{2}+6(x+1)\\&=9(x+1)^{2}+6(x+1)\\&=(x+1)[9(x+1)+6]\\&=(x+1)(9x+15)\\&=3(x+1)(3x+5)\end{aligned}}}$

${\displaystyle {\begin{aligned}a)\quad x^{2}-8x+15&=x^{2}-8x+16-1\\&=x^{2}-8x+4^{2}-1^{2}\\&=(x-4)^{2}-1^{2}\\&=(x-4+1)(x-4-1)\\&=(x-3)(x-5)\end{aligned}}}$

${\displaystyle {\begin{aligned}b)\quad x^{2}+2x-35&=x^{2}+2x+1^{2}-6^{2}\\&=(x+1)^{2}-6^{2}\\&=(x+1+6)(x+1-6)\\&=(x+7)(x-5)\end{aligned}}}$

${\displaystyle {\begin{aligned}c)\quad x^{2}+x-6&=\left(x+{\frac {1}{2}}\right)^{2}-{\frac {1}{4}}-6\\&=\left(x+{\frac {1}{2}}\right)^{2}-{\frac {25}{4}}\\&=\left(x+{\frac {1}{2}}\right)^{2}-\left({\frac {5}{2}}\right)^{2}\\&=\left(x+{\frac {1}{2}}+{\frac {5}{2}}\right)\left(x+{\frac {1}{2}}-{\frac {5}{2}}\right)\\&=(x+3)(x-2)\end{aligned}}}$

${\displaystyle {\begin{aligned}d)\quad x^{2}-3x-40&=\left(x-{\frac {3}{2}}\right)^{2}-{\frac {9}{4}}-40\\&=\left(x-{\frac {3}{2}}\right)^{2}-\left({\frac {13}{2}}\right)^{2}\\&=\left(x-{\frac {3}{2}}+{\frac {13}{2}}\right)\left(x-{\frac {3}{2}}-{\frac {13}{2}}\right)\\&=(x+5)(x-8)\\\end{aligned}}}$

${\displaystyle {\begin{aligned}e)\quad x^{2}+6x+7&=(x+3)^{2}-2\\&=(x+3)^{2}-({\sqrt {2}})^{2}\\&=(x+3+{\sqrt {2}})(x+3-{\sqrt {2}})\end{aligned}}}$

${\displaystyle {\begin{aligned}a)\quad 3(x^{2}+2x-35)-(x+2)(x-5)&=3[(x+1)^{2}-6^{2}]-(x+2)(x-5)\\&=3[(x+1+6)(x+1-6)]-(x+2)(x-5)\\&=3(x+7)(x-5)-(x+2)(x-5)\\&=(x-5)[3(x+7)-(x+2)]\\&=(x-5)(2x+19)\end{aligned}}}$

${\displaystyle {\begin{aligned}b)\quad (3x-1)^{2}-(2x+3)^{2}-(x-4)^{2}&=[(3x-1)+(2x+3)][(3x-1)-(2x+3)]-(x-4)^{2}\\&=(3x-1+2x+3)(3x-1-2x-3)-(x-4)^{2}\\&=(5x+2)(x-4)-(x-4)^{2}\\&=(x-4)[(5x+2)-(x-4)]\\&=(x-4)[4x+6]\\&=2(x-4)(2x+3)\end{aligned}}}$