\(\text{a) }\left(x^2+x\right)^2-2\left(x^2+x\right)-15\\ \text{Đặt }\left(x^2+x\right)^2=y\\ \text{Ta được: }\left(x^2+x\right)^2-2\left(x^2+x\right)-15\\ \\ =y^2-2y-15\\ \\ =y^2+3y-5y-15\\ =\left(y^2+3y\right)-\left(5y+15\right)\\ \\ =y\left(y+3\right)-5\left(y+3\right)\\ \\ =\left(y+3\right)\left(y-5\right)\\ Thay\text{ }y=x^2+x\text{ vào đa thức }\\ Ta\text{ lại được: }\left(y+3\right)\left(y-5\right)=\left(x^2+x+3\right)\left(x^2+x-5\right)\)
\(\text{b) }\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\\ \left(x^2+3x+1\right)\left(x^2+3x+1+1\right)-6=\\ \text{Đặt }\left(x^2+3x+1\right)=y\\ \text{Ta được: }\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\\ =y\left(y+1\right)-6\\ \\ =y^2+y-6\\ \\ =y^2+3y-2y-6\\ \\ =\left(y^2+3y\right)-\left(2y+6\right)\\ \\ =y\left(y+3\right)-2\left(y+3\right)\\ \\ =\left(y-2\right)\left(y+3\right)\\ Thay\text{ }y=\left(x^2+3x+1\right)\text{ vào đa thức }\\ \text{Ta lại được: }\left(x^2+3x+1-2\right)\left(x^2+3x+1+3\right)\\ \\ =\left(x^2+3x-1\right)\left(x^2+3x+4\right)\)
\(\text{c) }\left(x^2+8x+7\right)\left(x+3\right)\left(x+5\right)\\ =\left(x^2+x+7x+7\right)\left(x+3\right)\left(x+5\right)\\ =\left[\left(x^2+x\right)+\left(7x+7\right)\right]\left(x+3\right)\left(x+5\right)\\ =\left[x\left(x+1\right)+7\left(x+1\right)\right]\left(x+3\right)\left(x+5\right)\\ \\=\left(x+1\right)\left(x+7\right)\left(x+3\right)\left(x+5\right)\)
( x2 + 8x + 7 ) ( x + 3 )( x +5)
= ( x2 + 2.4x + 42 - 9 )( x + 3)( x + 5)
= [( x + 4)2 - 32] .( x + 3)(x + 5)
= ( x +4 +3)( x + 4 - 3)( x + 3)( x + 5)
= ( x + 7)( x + 1)( x + 3)( x + 5)
Đặt : x + 4 = y , ta có :
( y + 3)( y - 3)( y - 1)(y + 1)
=( y2 - 32)( y2 - 1)
= y4 - y2 - 9y2 + 9
= y2( y2 - 1) - 9( y2 - 1)
= ( y + 3)( y - 3)( y - 1)( y + 1)
