Question

Bài 1: Phân tích các đa thức sau thành nhân tử

    a) x² - 7x + 12;

  b) x² - 5x – 14;

  c) 4x² - 3x – 1;

  d) 2x² - 3x + 1

  e) x⁴ + 64

Answer 1

a) x² - 7x + 12 

= x² - 4x - 3x + 12 

= x(x - 4) - 3(x - 4) 

 =(x - 3)(x - 4) 

b) x² - 5x - 14 

= x² + 2x - 7x - 14

= x(x + 2) - 7(x + 2) 

= (x - 7)(x + 2)

c) 4x² - 3x - 1 

= 4x² - 4x + x - 1

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

= (4x + 1)(x - 1) 

d) 2x² - 3x + 1

= 2x² - 2x - x + 1

= 2x(x - 1) - (x - 1)

= (2x - 1)(x - 1) 

e) x⁴ + 64 

= x⁴ + 16x² + 64 - 16x²

= (x² + 8)² - (4x)²

= (x² - 4x + 8)(x² + 4x + 8) 

Answer 2

a) = x² - 3x - 4x + 12 = x( x - 3 ) - 4( x - 3 ) = ( x - 3 )( x - 4 )

b) = x² - 7x + 2x - 14 = x( x - 7 ) + 2( x - 7 ) = ( x - 7 )( x + 2 )

c) = 4x² - 4x + x - 1 = 4x( x - 1 ) + ( x - 1 ) = ( x - 1 )( 4x + 1 )

d) = 2x² - 2x - x + 1 = 2x( x - 1 ) - ( x - 1 ) = ( x - 1 )( 2x - 1 )

d) = x⁴ + 16x² + 64 - 16x² = ( x² - 8 )² - (4x)² = ( x² - 4x - 8 )( x² - 4x + 8 )

Answer 3

Trả lời:

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

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

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

d, \(2x^2-3x+1=2x^2-2x-x+1=2x\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(2x-1\right)\)

e, \(x^4+64=\left(x^2\right)^2+16x^2+64-16x^2=\left(x^2+8\right)-\left(4x\right)^2=\left(x^2+8-4x\right)\left(x^2+8+4x\right)\)