`a)x^3+x^2-4x-4=x^2(x+1)-4(x+1)=(x+1)(x^2-4)=(x+1)(x-2)(x+2)`
`b)x^2-2x-15=x^2-5x+3x-15=x(x-5)+3(x-5)=(x-5)(x+3)`
`c)x^4-8x=x(x^3-8)=x(x-2)(x^2+2x+4)`
\(a,x^3+x^2-4x-4=x^2\left(x+1\right)-4\left(x+1\right)=\left(x+1\right)\left(x^2-4\right)=\left(x+1\right)\left(x+2\right)\left(x-2\right)\)\(b,x^2-2x-15=\left(x^2-2x+1\right)-16=\left(x-1\right)^2-16=\left(x-1-4\right)\left(x-1+4\right)=\left(x-5\right)\left(x+3\right)\)\(c,x^4-8x=x\left(x^3-8\right)=x\left(x-2\right)\left(x^2+2x+4\right)\)
a,
= (x3+x2)+(−4x−4)=\(x^2\left(x+1\right)-4\left(x+1\right)\)
= \(\left(x+1\right)\left(x^2-4\right)\)=\(\left(x+1\right)\left(x+2\right)\left(x-2\right)\)
b,
\(\left(x^2+3x\right)+\left(-5x-15\right)\) =\(x\left(x+3\right)-5\left(x+3\right)=\left(x+3\right)\left(x-5\right)\)
c,
= \(x\left(x^3-8\right)=x\left(x-2\right)\left(x^2+2x+2^2\right)\)
= \(x\left(x-2\right)\left(x^2+2x+4\right)\)