实数范围内因式分解：(1) 3x^2-5x-4 (2) -2x^2+3x+1 (3) (x^2-8x)+5(x^2+8x)+6

题目

实数范围内因式分解：(1) 3x^2-5x-4 (2) -2x^2+3x+1 (3) (x^2-8x)+5(x^2+8x)+6

答案

配方后用平方差公式.
(1) 3x^2-5x-4
=3(X^2-5/3X-4/3)
=3［(X-5/6)^2-4/3-25/36］
=3［(X-5/6)^2-73/36］
=3(X-5/6+√73/6)(X-5/6-√73/6］

(2) -2x^2+3x+1
=-2(X^2-3/2X-1/2)
=-2［(X-3/4)^2-17/16］
=-2(X-3/4+√17/4)(X-3/4-√17/4)

(3) (x^2-8x)+5(x^2+8x)+6
=6X^2+32X+6
=6(X^2+16/3X+1)
=6［(X^2+8/3)^2+1-64/9］
=6［(X+8/3)^2-55/9］
=6(X+8/3+√55/3)(X-8/3-√55/3)