Factorization of Quadratic Polynomials(二次多項式的因式分解)

用戶37009

2007-12-23 6:43 am

Suppose I have a quadratic polynomial Ax^2+Bx+C, where A, B and C are integral constants and x is a variable.
How can I know whether this polynomial can be factorized to become
(Dx+E)(Fx+G)?
(where D, E, F and G are integral constants)

假設一個二次多項式Ax^2+Bx+C中, A、B及C均為整數, 而x則為變量.
怎樣才能知道這個多項式可否被因式分解為(Dx+E)(Fx+G)?
(D、E、F及G均為整數)

更新1:

我想問一問henrytsang168168: 依你的方法, 以下兩個二次多項式可否被因式分解為(Dx+E)(Fx+G)?要加解釋. (D、E、F及G均為整數) (a) 2,124x^2+2,311x-4,774 (b) 980y^2-5,509y+7,411

更新2:

werwer 的講法好似有點問題. 例如:x^2+2x+1 Discriminant(判別式)=(2)^2-4(1)(1)=0 咁係咪呢個二次多項式唔可以被因式分解為(Dx+E)(Fx+G)? 但我唸你知,x^2+2x+1=(x+1)(x+1)

回答 (2)

用戶218573

2007-12-23 7:12 am

✔ 最佳答案

Ax^2+Bx+C=(Dx+E)(Fx+G)
Ax^2+Bx+C=DFx^2+DGx+EFx+EG
Ax^2+Bx+C=DFx^2+(DG+EF)x+EG
A=DF
B=DG+EF
C=EG
可以

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Octavius

2007-12-23 7:27 am

If and only if
(the discriminant is non-zero)
and
(the discriminant is a perfect square or a ratio of two perfect squares)

which guarantees that
- there are real roots
- the square root of the discriminant is rational, hence the roots to the quadratic polynomial are rational.

參考: 自己，http://en.wikipedia.org/wiki/Square_root

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