Phân tích đa thức thành nhân tử
\(3x-\sqrt{x}-20\)
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Ta có: \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)
\(=\left(12x^2+8x+3x+2\right)\left(12x^2+12x-x-1\right)-4\)
\(=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)
\(=\left(12x^2+11x\right)^2+\left(12x^2+11x\right)-6\)
\(=\left(12x^2+11x+3\right)\left(12x^2+11x-2\right)\)
\(x^3-3x^2-4x+12\)
\(=x^2\left(x-3\right)-\left(4x-12\right)\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x^2-4\right)\left(x-3\right)\)
\(=\left(x+2\right)\left(x-2\right)\left(x-3\right)\)
\(x^3-3x^2-4x+12\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x^2-4\right)\left(x-3\right)\)
\(=\left(x+2\right)\left(x-2\right)\left(x-3\right)\)
x2-2xy+y2+3x-3y-10
= (x-y)2+3(x-y)-10
= [(x-y)2+5(x-y)]-[2(x-y)+10]
= (x-y)(x-y+5)-2(x-y+5)
= (x-y+5)(x-y-2)
Ta có: \(x^2-2xy+y^2+3x-3y-10\)
\(=\left(x-y\right)^2+3\left(x-y\right)-10\)
\(=\left(x-y+5\right)\left(x-y-2\right)\)
\(\left(x^2-3x\right)^2-14x^2+42x+40\\ =\left(x^2-3x-7\right)^2-9\\ =\left(x^2-3x-10\right)\left(x^2-3x-4\right)\)
a) x2+x-2
= x2-x+2x-2
= x(x-1)+2(x-1)
= (x+2)(x-1)
b) 2x2+5x+3
= 2x2+2x+3x+3
= 2x(x+1)+3(x+1)
= (2x+3)(x+1)
c) 3x2+5x-2
= 3x2+6x-1x-2
= 3x(x+2)-1(x+2)
= (3x-1)(x+2)
Ta có: (x2+6x-5)(x2+6x+3)-20
= [(x2+6x-1)-4][(x2+6x-1)+4]-20
= (x2+6x-1)2-16-20
= (x2+6x-1)2-36
= (x2+6x-7)(x2+6x-5)
= (x+7)(x-1)(x2+6x-5)
\(\left(x^2+6x-5\right)\left(x^2+6x+3\right)\\ =\left(x^2+6x-1\right)^2-16-20\\ =\left(x^2+6x-1\right)^2-36\\ =\left(x^2+6x-1-6\right)\left(x^2+6x-1+6\right)\\ =\left(x^2+6x-7\right)\left(x^2+6x+5\right)\\ =\left(x-1\right)\left(x+7\right)\left(x+1\right)\left(x+5\right)\)
1: \(x^2-2x-24=\left(x-6\right)\left(x+4\right)\)
2: \(x^2-8x+15=\left(x-3\right)\left(x-5\right)\)
3: \(x^2-9x+14=\left(x-2\right)\left(x-7\right)\)