Question

Phân tích đa thức thành nhân tử

\(\left(x^2+x+4\right)^2+8x\left(x^2+x+4\right)+15x^2\)

\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)

Answer 1

a) \(\left(x^2+x+4\right)^2+8x\left(x^2+x+4\right)+15x^2\) (*)

Đặt \(x^2+x+4=a\) => (*) trở thành:

\(a^2+8ax+15x^2\) = \(\left(a^2+3ax\right)+\left(5ax+15x^2\right)\)

= \(a\left(a+3x\right)+5x\left(a+3x\right)\)

= \(\left(a+3x\right)\left(a+5x\right)\) (1)

Thay \(a=x^2+x+4\) vào (1) ta được:

\(\left(x^2+4x+4\right)\left(x^2+6x+4\right)\) = \(\left(x+2\right)^2\left(x^2+6x+4\right)\)

Answer 2

a) \(\left(x^2+x+4\right)^2+8x\left(x^2+x+4\right)+15x^2\)

\(=\left(x^2+x+4\right)^2+5x\left(x^2+x+4\right)+3x\left(x^2+x+4\right)+15x^2\)

\(=\left(x^2+x+4\right)\left(x^2+x+4+5x\right)+3x\left(x^2+x+4+5x\right)\)

\(=\left(\left(x^2+x+4\right)+3x\right)\left(x^2+x+4+5x\right)\)

\(=\left(x^2+x+4+3x\right)\left(x^2+6x+4\right)\)

\(=\left(x^2+4x+4\right)\left(x^2+6x\right)+4\)

\(=\left(x+2\right)^2\cdot\left(x^2+6x+4\right)\)

Answer 3

b) Đặt D = \(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)

<=> D = \(\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)

Đặt \(x^2+10x+16=d\) => D trở thành:

D = \(d\left(d+8\right)+16\) = \(d^2+8d+16\) = \(\left(d+4\right)^2\)

Thay d = \(x^2+10+16\) ta được:

D = \(\left(x^2+10x+20\right)^2\)