Phân tích đa thức thành nhân tử bằng cách phối hợp nhiều phương pháp
3x3+3x2-36x
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a: \(3x^3-75x\)
\(=3x\left(x^2-25\right)\)
\(=3x\left(x-5\right)\left(x+5\right)\)
b: \(x^4y^2-12x^3y^2+48x^2y^2-64xy^2\)
\(=xy^2\left(x^3-12x^2+48x-64\right)\)
\(=xy^2\cdot\left(x-4\right)^3\)
a) \(x^4-4x^2-4x-1=\left(x^4-1\right)-4x\left(x+1\right)=\left(x^2+1\right)\left(x-1\right)\left(x+1\right)-4x\left(x+1\right)=\left(x+1\right)\left[\left(x^2+1\right)\left(x-1\right)-4x\right]=\left(x+1\right)\left(x^3-x^2+x-1-4x\right)=\left(x+1\right)\left(x^3-x^2-3x-1\right)\)
b) \(10x^4y^2-10x^3y^2-10x^2y^2+10xy^2=10xy^2\left(x^3-x^2-x+1\right)=10xy^2\left(x-1\right)^2\left(x+1\right)\)
a: \(x^4-4x^2-4x-1\)
\(=\left(x^4-1\right)-4x\left(x+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)-4x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3+x-x^2-1-4x\right)\)
\(=\left(x+1\right)\left(x^3-x^2-3x-1\right)\)
b: \(10x^4y^2-10x^3y^2-10x^2y^2+10xy^2\)
\(=10xy^2\left(x^3-x^2-x+1\right)\)
\(=10xy^2\cdot\left[\left(x+1\right)\left(x^2-x+1\right)-x\left(x+1\right)\right]\)
\(=10xy^2\cdot\left(x+1\right)\left(x-1\right)^2\)
a) \(5x^2y-20xy+20y=5y\left(x^2-4x+4\right)=5y\left(x-2\right)^2\)
b) \(3x^3+6x^2+3x=3x\left(x^2+2x+1\right)=3x\left(x+1\right)^2\)
c) \(3x^2y-12y=3y\left(x^2-4\right)=3y\left(x-2\right)\left(x+2\right)\)
d) \(7x^3-28x^2+28x=7x\left(x^2-4x+4\right)=7x\left(x-2\right)^2\)
a: \(5x^2y-20xy+20y\)
\(=4y\left(x^2-4x+4\right)\)
\(=4x\left(x-2\right)^2\)
b: \(3x^3+6x^2+3x\)
\(=3x\left(x^2+2x+1\right)\)
\(=3x\left(x+1\right)^2\)
c: \(3x^2y-12y\)
\(=3y\left(x^2-4\right)\)
\(=3y\left(x-2\right)\left(x+2\right)\)
d: \(7x^3-28x^2+28x\)
\(=7x\left(x^2-4x+4\right)\)
\(=7x\left(x-2\right)^2\)
\(1,x^2+5x-6=x^2-x+6x-6=x\left(x-1\right)+6\left(x-1\right)=\left(x-1\right)\left(x+6\right)\)
\(3,7x-6x^2-2=-6x^2+7x-2=-6x^2+3x+4x-2=3x\left(-2x+1\right)+2\left(2x-1\right)\)
\(=3x\left(1-2x\right)-2\left(1-2x\right)=\left(1-2x\right)\left(3x-2\right)\)
\(2,5x^2+5xy-x-y=5x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(5x-1\right)\)
Đa thức 3x2 – 8x +1 có các hạng tử là: 3x2 ; -8x ; 1
Ta có: 2x . 3x2 = (2.3). (x.x2) = 6x3
2x. (-8x) = [2.(-8) ]. (x.x) = -16x2
2x. 1 = 2x
Vậy 2x.(3x2 – 8x + 1) = 6x3 -16x2 + 2x
2x - 2y - x² + 2xy - y²
= (2x - 2y) - (x² - 2xy + y²)
= 2(x - y) - (x - y)²
= (x - y)(2 - x + y)
\(x^3-x^2-x+1\)
\(=x^2\left(x-1\right)-\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-1\right)\)
\(3x^3+3x^2-36x\)
\(=3x\left(x^2+x-12\right)\)
\(=3x\left(x+4\right)\cdot\left(x-3\right)\)