Question

[Alice]

Phân tích đa thức thành nhân tử

1. x⁵ + x + 1

2. x⁵ + x⁴ +1

3. 2a²b+4ab²-a²c+ac²-4b²c+2bc-4abc

4. y(x-2z)²+ 8xyz +x(y-2z)²-2z(x+y)²

5. (x²+x)²-2(x²+x)-15

Answer 1

[hattori heiji]

1. x⁵ + x + 1

=x⁵-x²+x²+x+1

=(x⁵-x²)+(x²+x+1)

=x²(x³-1)+(x²+x+1)

=x²(x-1)(x²+x+1)+(x²+x+1)

=(x²+x+1)[x²(x-1)+1]

=(x²+x+1)(x³-x2+1)

2. x⁵ + x⁴ +1

=x⁵+x⁴+x³-x³+1

=(x⁵+x⁴+x³)-(x³-1)

=x³(x²+x+1)-(x-1)(x²+x+1)

=(x²+x+1)[x³-(x-1)]

=(x²+x+1)(x³-x+1)