Factorization of Quadratic Polynomials(二次多項式的因式分解)－wszspj2的部落格

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

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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,... 顯示更多 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: 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)

最佳解答:

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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其他解答:

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.481517FD598DAD6B