Question

descompuneti in factori:
a) 6x+6z-cx-cz
b)y+ay+t+at
c)ay-by+ay²-by²
d)x³+x²-5x-5
e)a³+a²-25(a+1)
f)x²+4x+3
g)x²-7x+6
h)x²+9x-10
i)x²+12x-28
j)3x²-x-2

Answer 1

a) 6x+6z-cx-cz=6(x+z)-c(x+z)=(x+z)(6-c)
b)y+ay+t+at=y(1+a)+t(1+a)=(1+a)(y+t)
c)ay-by+ay²-by²=y(a-b)+y²(a-b)=(a-b)(y+y²)=(a-b)(1+y)y
d)x³+x²-5x-5=x²(x+1)-5(x+1)=(x+1)(x²-5)=(x+1)(x-√5)(x+√5)
e)a³+a²-25(a+1)=a²(a+1)-25(a+1)=(a+1)(a²-25)=(a+1)(a+5)(a-5)
f)x²+4x+3=x²+3x+x+3=x(x+3)+(x+3)=(x+3)(x+1)
g)x²-7x+6=x²-6x-x+6=x(x-6)-(x-6)=(x-6)(x-1)
h)x²+9x-10=x²+10x-x-10=x(x+10)-(x+10)=(x+10)(x-1)
i)x²+12x-28=x²+14x-2x-28=x(x+14)-2(x+14)=(x+14)(x-2)
j)3x²-x-2 =3x²-3x+2x-2=3x(x-1)+2(x-1)=(x-1)(3x+2)

Answer 2

a)6x+6z-cx-cz = (6x+6z)-(cx+cz) = 6(x+z) - c(x+z) = (x+z)(6-c)
b)y+ay+t+at = (y+ay)+(t+at) = y(1+a) + t(1+a) = (1+a)(y+t)
c)ay-by+ay²-by² = (ay-by)+(ay²-by²) = y(a-b) + y²(a-b) = (a-b)(y-y²)
d)x³+x²-5x-5 = (x³+x²)-(5x+5) = x²(x+1) - (5(x+1) = (x+1)(x²-5)
e)a³+a²-25(a+1) = (a³+a²)-25(a+1) = a²(a+1)-25(a+1) = (a+1)(a²-25)
f)x²+4x+3=x²+3x+x+3=x(x+3)+(x+3)=(x+3)(x+1)
g)x²-7x+6=x²-6x-x+6=x(x-6)-(x-6)=(x-6)(x-1)
h)x²+9x-10=x²+10x-x-10=x(x+10)-(x+10)=(x+10)(x-1)
i)x²+12x-28=x²+14x-2x-28=x(x+14)-2(x+14)=(x+14)(x-2)
j)3x²-x-2 =3x²-3x+2x-2=3x(x-1)+2(x-1)=(x-1)(3x+2)