نویسنده : امیر انصاری

1. $$
   4(5b-3)^2+10(5b-3)-6
   $$
   
2. $$
   16(x^2+1)^2-4(2x)^2
   $$
   
3. $$
   -\frac{1}{4}(2x)^2+25(2y^3)^2
   $$
   

پاسخ

1. $$
   4(5b-3)^2+10(5b-3)-6\\
   =4\color{red}{r}^2+10\color{red}{r}-6\\
   =2(2r^2+5r-3)\\
   =2(2r-1)(r+3)\\
   =2(2(\color{red}{5b-3})-1)(\color{red}{5b-3}+3)\\
   =2(10b-7)(5b)\\
   =(10b)(10b-7)
   $$
   
2. $$
   16(x^2+1)^2-4(2x)^2\\
   =\bigl( 4(x^2+1)-2(2x) \bigr) \bigl( 4(x^2+1)+2(2x) \bigr)\\
   =(4x^2+4-4x)(4x^2+4+4x)\\
   =16(x^2-x+1)(x^2+x+1)
   $$
   
3. $$
   -\frac{1}{4}(2x)^2+25(2y^3)^2\\
   =-\frac{1}{4} \bigl( (2x)^2-100(2y^3)^2 \bigr)\\
   =-\frac{1}{4} \bigl( 2x+10(2y^3) \bigr) \bigl( 2x+10(2y^3) \bigr)\\
   =-\frac{1}{4} (2x-20y^3)(2x+20y^3)\\
   =-(x-10y^3)(x+10y^3) \text{ or } (10y^3-x)(10y^3+x)
   $$