Lösung 2.1:2c

We obtain the answer by using the squaring rule, (a+b)2=a2+2ab+b2 on the quadratic term and expanding the other bracketed terms:

(3x+4)2−(3x−2)(3x−8)

=(3x)2+23x4+42−(3x3x−3x8−23x+28)=(9x2+24x+16)−(9x2−24x−6x+16)=(9x2+24x+16)−(9x2−30x+16)=(9x2+24x+16)−9x2+30x−16=9x2−9x2+24x+30x+16−16=0+54x+0=54x