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a) x^4 + 2^3-x -2
=x^4 - x^3 + 3x^3 - 3x^2 + 3x^2 - 3x + 2x-2
=x^3.(x-1) + 3x^2.(x-1) + 3x.(x-1)+2.(x-1)
=(x-1).( x^3+ 3x^2 + 3x+2)
=(X+1).(X^3 + 2X^2 + X^2 +2X +X+2)
=(X+1).(X+2).(X^2 +X + 1)
\(f\left(x\right)=3x^3-7x^2+4x-4=3x^3-6x^2-x^2+2x+2x-4=3x^2\left(x-2\right)-x\left(x-2\right)+2\left(x-2\right)=\left(x-2\right)\left(3x^2-x+2\right)\)
Vì \(f\left(x\right)\) chứa đa thức \(x-2\) nên \(f\left(x\right)\) chia hết cho \(x-2\) (đpcm)
f(x) = x4 + 6x3 +11x2 + 6x
\(=x^4+x^3+5x^3+5x^2+6x^2+6x\)
\(=\left(x^4+x^3\right)+\left(5x^3+5x^2\right)+\left(6x^2+6x\right)\)
\(=x^3\left(x+1\right)+5x^2\left(x+1\right)+6x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3+5x^2+6x\right)\)
\(=x\left(x+1\right)\left(x^2+5x+6\right)\)
\(=x\left(x+1\right)\left[x^2+2x+3x+6\right]\)
\(=x\left(x+1\right)\left[\left(x^2+2x\right)+\left(3x+6\right)\right]\)
\(=x\left(x+1\right)\left[x\left(x+2\right)+3\left(x+2\right)\right]\)
\(=x\left(x+1\right)\left(x+2\right)\left(x+3\right)\)
b)Ta có
\(f\left(x\right)+1=x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)
\(=\left[x\left(x+3\right)\right].\left[\left(x+1\right)\left(x+2\right)\right]+1\)
\(=\left(x^2+3x\right).\left(x^2 +3x+2\right)+1\)
\(=\left(x^2+3x+1-1\right).\left(x^2+3x+1+1\right)+1\)
\(=\left[\left(x^2+3x+1\right)-1\right].\left[\left(x^2+3x+1\right)+1\right]+1\)
\(=\left(x^2+3x+1\right)^2-1+1=\left(x^2+3x+1\right)^2\)
Vậy với mọi x nguyên thì f(x) + 1 luôn có giá trị là 1 số chính phương
a. x3+x2+2x2+2x
= (x3+x2)+(2x2+2x)
= x2(x+1)+2x(x+1)
= (x2+2x)(x+1)
= x(x+2)(x+1)