Question

Phân tích các đa thức sau thành nhân tử

a) \(x^2-y^2-4x+4\)

b) \(x^2+5x-6\)

c) \(x^4+4\)

d) (x+2)(x+3)(x+4)(x+5) -48

Answer 1

b, \(x^2+5x-6=x^2+6x-x-6=x\left(x+6\right)-\left(x+6\right)\)

=(x-1)(x+6)

Answer 2

a, \(x^2-y^2-4x+4\)

=\(\left(x^2-4x+4\right)-y^2\)

=\(\left(x-2\right)^2-y^2\)

=\(\left(x-2-y\right)\left(x-2+y\right)\)

Answer 3

\(x^2-y^2-4x+4=\left(x^2-4x+4\right)-y^2=\left(x-2\right)^2-y^2=\left(x-y-2\right)\left(x+y-2\right)\) \(x^2+5x-6=x^2+6x-x-6=x\left(x+6\right)-\left(x+6\right)=\left(x-1\right)\left(x+6\right)\)

\(x^4+4=x^4+4x^2+4-4x^2=\left(x^2+2\right)^2-\left(2x\right)^2=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-48=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-48=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-48=\left(x^2+7x+10\right)^2+2\left(x^2+7x+10\right)-48=\left(x^2+7x+10\right)^2+8\left(x^2+7x+10\right)-6\left(x^2+7x+10\right)-48=\left(x^2+7x+10\right)\left(x^2+7x+18\right)-6\left(x^2+7x+18\right)=\left(x^2+7x+4\right)\left(x^2+7x+18\right)\)

Answer 4

d) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-48\) (1)

=\(\left(x^2+7x+10\right)\left(x^2+7x+12\right)-48\)

Đặt \(x^2+7x+11=y\)

(1) = \(\left(y-1\right)\left(y+1\right)-48\)

=y²-1-48=y²-49=(y-7)(y+7)=\(\left(x^2+7x+11-7\right)\left(x^2+7x+11+7\right)\)

=\(\left(x^2+7x+4\right)\left(x^2+7x+18\right)\)