Przydatne będą wzory skróconego mnożenia:

${\left(a+b\right)}^2=a^2+2ab+b^2$  

${\left(a-b\right)}^2=a^2-2ab+b^2$  

$\left(a-b\right)\left(a+b\right)=a^2-b^2$  

$a)$  

$f\left(x\right)={\left(3x-2\right)}^2=9x^2-12x+4$  

$\Delta ={\left(-12\right)}^2-4\cdot 9\cdot 4=144-144=0$  

$b)$  

$f\left(x\right)=\left(1-4x\right)\left(4x+1\right)=\left(1-4x\right)\left(1+4x\right)=1-16x^2=-16x^2+1$  

$\Delta =0^2-4\cdot \left(-16\right)\cdot 1=0+64=64$  

$c)$  

$f\left(x\right)=\left(x-3\right)\left(4+2x\right)=4x+2x^2-12-6x=2x^2-2x-12$  

$\Delta ={\left(-2\right)}^2-4\cdot 2\cdot \left(-12\right)=$   $4+96=100$  

$d)$  

$f\left(x\right)={\left(1-4x\right)}^2-{\left(2x+1\right)}^2=\left(1-8x+16x^2\right)-\left(4x^2+4x+1\right)=$  

$=1-8x+16x^2-4x^2-4x-1=$   $12x^2-12x$  

$\Delta ={\left(-12\right)}^2-4\cdot 12\cdot 0=144$  

$e)$  

$f\left(x\right)=-4\left(x-1\right)\left(2x+1\right)=-4\left(2x^2+x-2x-1\right)=$  

$=-4\left(2x^2-x-1\right)=-8x^2+4x+4$  

$\Delta =4^2-4\cdot \left(-8\right)\cdot 4=$   $16+128=144$  

$f)$  

$f\left(x\right)=\frac{x^2-3\left(x-1\right)}{3}=\frac{x^2-3x+3}{3}=\frac{1}{3}{x}^2-x+1$  

$\Delta ={\left(-1\right)}^2-4\cdot \frac{1}{3}\cdot 1=1-\frac{4}{3}=-\frac{1}{3}$