نویسنده : امیر انصاری

1. $$
   3(2z+3)^2-9(2z+3)-30
   $$
   
2. $$
   16(m^2-4)^2-4(3n)^2
   $$
   
3. $$
   \frac{1}{9}y^2-\frac{1}{3}yx+\frac{1}{4}x^2
   $$
   
4. $$
   -28(w+\frac{2}{3})^2+7(3w-\frac{1}{3})^2
   $$
   

پاسخ

1. $$
   3(2z+3)^2-9(2z+3)-30\\
   =3\color{red}{r}^2-9\color{red}{r}-30\\
   =3(r^2-3r-10)\\
   =3(r-5)(r+2)\\
   =3(\color{red}{2z+3}-5)(\color{red}{2z+3}+2)\\
   =3(2z-2)(2z+5)\\
   =6(z-1)(2z+5)
   $$
   
2. $$
   16(m^2-4)^2-4(3n)^2\\
   =4 \bigl( 4(m^2-4)^2 - (3n)^2 \bigr)\\
   =4 \bigl( 2(m^2-4)-3n \bigr)\bigl( 2(m^2-4)+3n \bigr)\\
   =4(2m^2-8-3n)(2m^2-8+3n)
   $$
   
3. $$
   \frac{1}{9}y^2-\frac{1}{3}yx+\frac{1}{4}x^2\\
   =\frac{1}{36}(4y^2-12yx+9x^2)\\
   =\frac{1}{36}(2y-3x)(2y-3x)\\
   =\frac{1}{36}(2y-3x)^2
   $$
   
4. $$
   -28(w+\frac{2}{3})^2+7(3w-\frac{1}{3})^2\\
   =-7\bigl( 4(w+\frac{2}{3})^2-(3w-\frac{1}{3})^2 \bigr)\\
   =-7 \bigl( 2(w+\frac{2}{3})-(3w-\frac{1}{3}) \bigr) \bigl( 2(w+\frac{2}{3})+(3w-\frac{1}{3}) \bigr)\\
   =-7(2w+\frac{4}{3}-3w+\frac{1}{3})(2w+\frac{4}{3}+3w-\frac{1}{3})\\
   =-7(-w+\frac{5}{3})(5w+1)\\
   =7(w-\frac{5}{3})(5w+1)
   $$