Question

Va rog să mă ajute cineva cu ex 2

Answer 1

Răspuns:

Explicație pas cu pas:

a) x³+x²+x+1=x²·(x+1)+(x+1)·1=(x+1)·(x²+1)

b) x³+x²-4x-4=x²·(x+1)-4·(x+1)=(x+1)·(x²-4)=(x+1)·(x²-2²)=(x+1)(x-2)(x+2)

c) x⁶-x⁴-16x²+16=x⁴·(x²-1)-16·(x²-1)=(x²-1)(x⁴-16)=(x²-1²)((x²)²-4²)=(x-1)x+1)(x²-4)(x²+4)=(x-1)(x+1)(x²-2²)(x²+4)=(x-1)(x+1)(x-2)(x+2)(x²+4)

d) x⁵+x³-x²-1=x³(x²+1)-1·(x²+1)=(x²+1)(x³-1)=(x²+1)(x³-1³)=(x²+1)(x-1)(x²+x+1)

e) x³+x²-25x-25=x²(x+1)-25(x+1)=(x+1)(x²-25)=(x+1)(x²-5²)=(x+1)(x-5)(x+5)

f) x³+3x²-x-3=x²(x+3)-1·(x+3)=(x+3)(x²-1)=(x+3)(x²-1²)=(x+3)(x-1)(x+1)

g) x⁴-9x³+x-9=x³·(x-9)+1·(x-9)=(x-9)(x³+1)=(x-9)(x³+1³)=(x-9)(x+1)(x²-x+1)

h) x³+4x²-16x-64=x²(x+4)-16(x+4)=(x+4)(x²-16)=(x+4)(x²-4²)=(x+4)(x-4)(x+4)= (x-4)(x+4)²