Question

se alguém puder me ajudar ficarei muito grato!

Answer 1

a) x⁴-y⁴ =(x²+y²)(x²-y²)

=(x²+y²)*(x-y)*(x+y)

b)ax²-4ax =ax*(x-4)

c) 4x³-12x²+9x =x*(4x²-12x+9)

****P(x)=ax²+bx+c =a*(x-x')*(x-x'')  ....a≠0 e x' e x'' são as raízes

****4x²-12x+9=0

****x'=x''=[12+√(12²-144)]/8 =12/8=3/2

****4x²-12x+9=4 *(x-3/2)*(x-3/2) =(2x-3)*(2x-3)

4x³-12x²+9x =x * (2x-3)*(2x-3)

d) 3x²-x-4

3x²-x-4=0

x'=[1+√(1+48)]/6 =(1+7)/6 =8/6=4/3

x''=[1-√(1+48)]/6 =(1-7)/6 =8/6=-1

3x²-x-4 = 3*(x+1)*(x-4/3) =(x+1)*(3x-4)

3x²-x-4 = (x+1)*(3x-4)

e) a²x³+a²x²-4x-4

=a²x²(x+1) -4(x+1)

=(x+1)*(a²x²-4)

=(x+1)*(ax-2)*(ax+2)