Phân tích đa thức thành nhân tử:
x4 – 2x3 – 2x2 – 2x – 3
Bạn nào giúp mình nhanh nhất mình chọn ﹋o﹋
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x^4+2x^3+2x^2+2x+1\\ =\left(x^4+x^3\right)+\left(x^3+x^2\right)+\left(x^2+x\right)+\left(x+1\right)\\ =x^3\left(x+1\right)+x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\\ =\left(x^3+x^2+x+1\right)\left(x+1\right)\\ =\left[\left(x^3+x^2\right)+\left(x+1\right)\right]\left(x+1\right)\\ =\left[x^2\left(x+1\right)+\left(x+1\right)\right]\left(x+1\right)\\ =\left(x^2+1\right)\left(x+1\right)^2\)
x 4 - 2 x 3 - 2 x 2 - 2 x - 3 = ( x 4 − 1 ) − ( 2 x 3 + 2 x 2 ) − ( 2 x + 2 ) = ( x 2 + 1 ) ( x 2 − 1 ) − 2 x 2 ( x + 1 ) − 2 ( x + 1 ) = ( x 2 + 1 ) ( x − 1 ) ( x + 1 ) − 2 x 2 ( x + 1 ) − 2 ( x + 1 ) = ( x + 1 ) ( x 2 + 1 ) ( x − 1 ) − 2 x 2 – 2 = ( x + 1 ) ( x 2 + 1 ) ( x − 1 ) − 2 ( x 2 + 1 ) = ( x + 1 ) ( x 2 + 1 ) ( x – 1 − 2 ) = ( x + 1 ) ( x 2 + 1 ) ( x − 3 )
x^4 - 2x^3 - 2x^2 - 2x - 3
= x^4 - 1 - 2x^3 - 2x^2 - 2x -2
= ( x - 1 ) ( x + 1 ) ( x^2 + 1 ) - 2x^2 ( x + 1 ) - 2 ( x + 1 )
= ( x + 1 ) [ ( x - 1 ) ( x^2 + 1 ) - 2x^2 - 2 ]
= ( x + 1 ) [ ( x - 1 ) ( x^2 + 1 - 2 ( x^2 - 1 ) ]
= ( x + 1 ) [ ( x - 1 ) ( x^2 + 1 ) - 2 ( x - 1 ) ( x + 1 ) ]
= ( x + 1 ) ( x - 1 ) [ ( x^2 + 1 ) - 2 ( x +1 )
= ( x + 1 ) ( x - 1 ) ( x^2 +1 - 2x - 2 )
= ( x + 1 ) ( x - 1 ) ( x^2 - 2x - 1 )
1/(x+2)2 -(3x-1)2=(x+2+3x-1)(x+2-3x+1)=4x(-2x+3)=-8x2+12x
2/(x4+x2)(-2x3-2x)=x2(x2+1)-2x(x2+1)=(x2+1)(x2-2x)
\(x^4+8x=x\left(x^3+8\right)=x\left(x+2\right)\left(x^2-2x+4\right)\)
Ta có : \(x^4-5x^2+4\)
\(=x^4-x^2-4x^2+4\)
\(=x^2\left(x^2-1\right)-4\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2-4\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
Ta có: \(x^4-5x^2+4\)
\(=x^4-x^2-4x^2+4\)
\(=x^2\left(x^2-1\right)-4\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2-4\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
a) \(x^2-2x-4y^2-4y=\left(x^2-4y^2\right)-\left(2x+4y\right)=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)=\left(x+2y\right)\left(x-2y-2\right)\)
b) \(x^3+2x^2+2x+1=\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)=\left(x+1\right)\left(x^2-x+1+2x\right)=\left(x+1\right)\left(x^2+x+1\right)\)
c) \(x^3-4x^2+12x-27=x^3-3x^2-x^2+3x+9x-27=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)=\left(x-3\right)\left(x^2-x+9\right)\)
d) \(a^6-a^4+2a^3+2a^2=a^2\left(a^4-a^2+2a+2\right)=a^2\left[a^2\left(a-1\right)\left(a+1\right)+2\left(a+1\right)\right]=a^2\left(a+1\right)\left(a^3-a^2+2\right)=a^2\left(a+1\right)\left[a^3+a^2-2a^2+2\right]=a^2\left(a+1\right)\left[a^2\left(a+1\right)-2\left(a-1\right)\left(a+1\right)\right]=a^2\left(a+1\right)^2\left(a^2-2a+2\right)\)
a) Ta có: \(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
b) Ta có: \(x^3+2x^2+2x+1\)
\(=\left(x^3+1\right)+2x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+x+1\right)\)
x^4 - 2x^3 - 2x^2 - 2x - 3
=x^4 + x^3 - 3x^3 - 3x^2 + x^2 + x - 3x - 3
=x^3(x+1) - 3x^2(x+1) + x(x+1 ) - 3(x+1)
=(x+1)(x^3 - 3x^2 + x - 3)
=(x+1)[x^2 (x - 3) + x - 3]
=(x+1)(x - 3)(x^2 + 1)