Question

Ex 3 va rog!
Dau coroana si 20 de puncte!

Answer 1

   
[tex]Ex. ~3) \\ \\ a)~121x^2-144y^2= \\ =11^2x^2-12^2y^2=\boxed{(11x+12y)(11x-12y)}\\\\ b)~(a+1)^2-4= \\ =(a+1)^2-2^2=\boxed{(a+1+2)(a+1-2)}=\boxed{(a+3)(a-1)} \\ \\ c)~9-(2b+3)^2 = 3^2-(2b+3)^2 = \\ =\boxed{[3 + (2b+3)][3 - (2b+3)]} = \\ = (3+2b+3)(3-2b-3)=\boxed{(6-2b)\cdot(-2b)} \\ \\ d)~16c^2 -(3c+5d)^2 =4^2c^2 -(3c+5d)^2= \\ =\boxed{[4c +(3c+5d)][4c -(3c+5d)]}= \\ =(4c +3c+5d)(4c -3c-5d)= \boxed{(7c+5d)(c-5d)} [/tex]


[tex]e)~9(5m-4p)^2-64m^2 = 3^2(5m-4p)^2-8^2m^2 = \\ =\boxed{[3(5m-4p)+8m ][3(5m-4p)-8m]}= \\ =(15m-12p+8m ][15m-12p-8m)=\boxed{(23m-12p)(7m-12p)} \\ \\ f)~(n+3q)^2-4(q-n)^2 = (n+3q)^2-2^2(q-n)^2 = \\ =\boxed{[(n+3q)+2(q-n)][(n+3q)-2(q-n)]}= \\ =(n+3q+2q-2n)(n+3q-2q+2n)=\boxed{(-n+5q)(3n+q)} \\ \\ g)~a^2 + 6a + 9 = a^2 + 2\cdot 3\cdot a + 3^2 = \boxed{(a+3)^2} [/tex]


[tex]h)~-30ab^2+25b^4+9a^2 = 25b^4 -30ab^2+9a^2 = \\ = (5b^2)^2 +2\cdot 5b^2 \cdot (-3a)+ (-3a)^2 =\boxed{(5b^2-3a)^2} \\ \\ i)~-6a^2b^3+1+9a^4b^6 = 9a^4b^6 -6a^2b^3+1 = \\ = (3a^2b^3)^2 +2\cdot(3a^2b^3) \cdot (-1) +(-1)^2 =\boxed{(3a^2b^3-1)^2}[/tex]