Question

Phân tích đa thức thành nhân tử

a, \(x^2-7x+12\)

b, \(x^2-5x-14\)

c, \(4x^2-3x-1\)

d, \(x^2-2xy+y^2+3x-3y-10\)

Answer 1

a, \(x^2-7x+12=x^2-4x-3x+12\)

\(=x\left(x-4\right)-3\left(x-4\right)=\left(x-3\right)\left(x-4\right)\)

b, \(x^2-5x-14=x^2+2x-7x-14\)

\(=x\left(x+2\right)-7\left(x+2\right)=\left(x-7\right)\left(x+2\right)\)

c, \(4x^2-3x-1=4x^2-4x+x-1\)

\(=4x\left(x-1\right)+\left(x-1\right)=\left(4x+1\right)\left(x-1\right)\)

Answer 2

a)\(x^2-7x+12\)

\(=x^2-3x-4x+12\\ =\left(x^2-3x\right)-\left(4x-12\right)\\ =x\left(x-3\right)-4\left(x-3\right)\\ =\left(x-3\right)\left(x-4\right)\)

b)\(x^2-5x-14\)

\(=x^2-7x+2x-14\\ =\left(x^2-7x\right)+\left(2x-14\right)\\ =x\left(x-7\right)+2\left(x-7\right)\\ =\left(x-7\right)\left(x+2\right)\)

c)\(4x^2-3x-1\)

\(=4x^2-2x+x-1\\ =\left(4x^2-2x\right)+\left(x-1\right)\\ =2x\left(x-1\right)+\left(x-1\right)\\ =\left(x-1\right)\left(2x+1\right)\)

d) \(x^2-2xy+y^2+3x-3y-10\)

\(=\left(x^2-2xy+y^2\right)+\left(3x-3y\right)-10\\ =\left(x-y\right)^2+3\left(x-y\right)-10\\ =\left(x-y\right)^2+2.1,5\left(x-y\right)+1.5^2-10-1.5^2\\ =\left(\left(x-y\right)+1,5\right)^2-12,25\\ =\left(x-y+1,5-3,5\right)\left(x-y+1,5+3,5\right)\\ =\left(x-y-2\right)\left(x-y+5\right)\)

Answer 3

a,\(x^2-3x-4x+12=\left(x^2-3x\right)-\left(4x-12\right)=x\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x-4\right)\\ \\ \\ \\ \\ \\ b,x^2+2x-7x-14=\left(x^2+2x\right)-\left(7x+14\right)=x\left(x+2\right)-7\left(x+2\right)=\left(x-7\right)\left(x+2\right)\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ c,4x^2+4x-x-1=\left(4x^2+4x\right)-\left(x+1\right)=4x\left(x+1\right)-\left(x+1\right)=\left(x+1\right)\left(4x-1\right)\)

Answer 4

a) x² - 7x + 12

= x² - 2. \(\dfrac{7}{2}\) .x + ^{\(\left(\dfrac{7}{2}\right)^2-\dfrac{1}{4}\)}

^{= \(\left(x+\dfrac{7}{2}\right)^2-\left(\dfrac{1}{2}\right)^2\)}

= \(\left(x+\dfrac{7}{2}+\dfrac{1}{2}\right).\left(x+\dfrac{7}{2}-\dfrac{1}{2}\right)\)

= \(\left(x+2\right).\left(x+3\right)\)

b) \(x^2-5x-14=x^2+2x-7x-14\)

\(=x.\left(x+2\right)-7.\left(x+2\right)\)

\(=\left(x+2\right).\left(x-7\right)\)

c) \(4.x^2-3x-1=4.x^2+x-4x-1\)

\(=x.\left(4x+1\right)-\left(4x+1\right)\)

= (4x + 1) . ( x - 1)

d) \(x^2-2xy+y^2+3x-3y-10\)=

\(y^2-xy+2y-xy+x^2-2x-5y+5x-10\)

= \(y\left(y-x+2\right)-x\left(y-x+2\right)-5\left(y-x+2\right)\) = \(\left(y-x+2\right).\left(y-x-5\right)\)