Question

帮忙在实数范围内分解因式 2x^2-3xy-3y^2 急急急急急！！！

要详细过程！！好的话再加分！！

Answer 1

先求出2x^2-3xy-3y^2 =0的两个根为x=-（3+√33)y/4或x==-（3-√33)y/4
2x^2-3xy-3y^2=2[x+（3+√33)y/4][x+（3-√33)y/4]

Answer 2

2x² - 3xy - 3y²
= 2[x² - (3/2)xy] - 3y²
= 2[x² - (3/2)xy + (3y/4)² - (3y/4)²] - 3y²
= 2(x - 3y/4)² - 18y²/16 - 3y²
= 2(x - 3y/4)² - 33y²/8
= [√2(x - 3y/4)]² - [(√33y)/(2√2)]²
= [√2(x - 3y/4) + (√33y)/(2√2)][√2(x - 3y/4) - (√33y)/(2√2)]

Answer 3

用一元二次方程的求根公式算出两根x1 x2
结果就是（x-x1）（x-x2）