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a) \(4x^2-16+\left(3x+12\right)\left(4-2x\right)\)
\(=\left(2x-4\right)\left(2x+4\right)-3\left(x+4\right)\left(2x-4\right)\)
\(=\left(2x-4\right)\left(2x+4-3x-12\right)\)
\(=-\left(2x-4\right)\left(x+8\right)\)
b) \(x^3+x^2y-15x-15y\)
\(=x^2\left(x+y\right)-15\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-15\right)\)
c) \(3\left(x+8\right)-x^2-8x\)
\(=3\left(x+8\right)-x\left(x+8\right)\)
\(=\left(x+8\right)\left(3-x\right)\)
d) \(x^3-3x^2+1-3x\)
\(=x^3+1-3x^2-3x\)
\(=\left(x+1\right)\left(x^2-x+1\right)-3x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1-3x\right)\)
\(=\left(x+1\right)\left(x^2-4x+1\right)\)
d) \(5x^2-5y^2-20x+20y\)
\(=5\left(x^2-y^2\right)-20\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y\right)-20\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y-4\right)\)
a) x3 + 4x2 – 2x – 8
= (x3 + 4x2) - (2x + 8)
= x2(x + 4) - 2(x + 4)
= (x + 4)(x2 - 2)
= (x + 4)(x + √2)(x - √2)
a: \(=x\left(x^2+4x+4-z^2\right)\)
\(=x\left(x+2-z\right)\left(x+2+z\right)\)
a: \(x^2-y^2-x-y\)
\(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-1\right)\)
f: \(x^3-5x^2-5x+1\)
\(=\left(x+1\right)\left(x^2-x+1\right)-5x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-6x+1\right)\)
c: \(x^4+x^3-4x^2+x+1\)
\(=x^4-x^3+2x^3-2x^2-2x^2+2x-x+1\)
\(=\left(x-1\right)\left(x^3+2x^2-2x-1\right)\)
\(=\left(x-1\right)\left[\left(x-1\right)\left(x^2+x+1\right)+2x\left(x-1\right)\right]\)
\(=\left(x-1\right)^2\cdot\left(x^2+3x+1\right)\)
x3 – 4x2 – 12x + 27
(Nhóm để xuất hiện nhân tử chung)
= (x3 + 27) – (4x2 + 12x)
= (x3 + 33) – (4x2 + 12x)
(nhóm 1 là HĐT, nhóm 2 có 4x là nhân tử chung)
= (x + 3)(x2 – 3x + 9) – 4x(x + 3)
= (x + 3)(x2 – 3x + 9 – 4x)
= (x + 3)(x2 – 7x + 9)
a) x² - 9
= x² - 3²
= (x - 3)(x + 3)
b) 4x² - 1
= (2x)² - 1²
= (2x - 1)(2x + 1)
c) x⁴ - 16
= (x²)² - 4²
= (x² - 4)(x² + 4)
= (x² - 2²)(x² + 4)
= (x - 2)(x + 2)(x + 4)
d) x² - 4x + 4
= x² - 2.x.2 + 2²
= (x - 2)²
e) x³ - 8
= x³ - 2³
= (x - 2)(x² + 2x + 4)
f) x³ + 3x² + 3x + 1
= x³ + 3.x².1 + 3.x.1² + 1³
= (x + 1)³
1)\(4x^2-4xy+y^2-8x+4y=\left(4x^2-4xy+y^2\right)-\left(8x-4y\right)=\left(2x-y\right)^2-4\left(2x-y\right)=\left(2x-y\right)\left(2x-y-4\right)\)
2) \(2x^3-3x^2+3x-1=x^2\left(2x-1\right)-x\left(2x-1\right)+\left(2x-1\right)=\left(2x-1\right)\left(x^2-x+1\right)\)