\(\left(x^2+8x+7\right)\cdot\left(x^2+8x+15\right)+15=\left(x^2+8x+11\right)^2-16+15\)
\(\left(x^2+8x+11\right)^2-1=\left(x^2+8x+10\right)\cdot\left(x^2+8x+12\right)=\left(x^2+8x+10\right)\cdot\left(x+2\right)\cdot\left(x+6\right)\)
( x2 + 8x + 7 )( x2 + 8x + 15 ) + 15 (1)
Đặt t = x2 + 8x + 7
(1) <=> t( t + 8 ) + 15
= t2 + 8t + 15
= t2 + 3t + 5t + 15
= t( t + 3 ) + 5( t + 3 )
= ( t + 3 )( t + 5 )
= ( x2 + 8x + 7 + 3 )( x2 + 8x + 7 + 5 )
= ( x2 + 8x + 10 )( x2 + 8x + 12 )
= ( x2 + 8x + 10 )( x2 + 2x + 6x + 12 )
= ( x2 + 8x + 10 )[ x( x + 2 ) + 6( x + 2 ) ]
= ( x2 + 8x + 10 )( x + 2 )( x + 6 )
Đặt \(x^2+8x+11=y\)
Từ đó: \(\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
\(=\left(x^2+8x+11-4\right)\left(x^2+8x+11+4\right)+15\)
\(=\left(y-4\right)\left(y+4\right)+15\)
\(=y^2-16+15\)
\(=y^2-1\)
\(=\left(y-1\right)\left(y+1\right)\)
\(=\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)
\(=\left(x+2\right)\left(x+6\right)\left(x^2+8x+10\right)\)
\(\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
Đặt \(t=x^2+8x\)
\(\left(t+8\right)t+15\)
\(t^2+8t+15\)
\(=t^2+3t+5t+15\)
\(=t\left(t+3\right)+5\left(t+3\right)\)
\(=\left(t+3\right)\left(t+5\right)\)
\(=\left(x^2+8x+3\right)\left(x^2+8x+5\right)\)
