Consider two quadratic polynomials x^2-3x+2 and x^2+2x-3 , they can be factorized into (x-1)(x-2) and (x-1)(x+3) respectively. Note that the coefficient of x and the constant term are interchanged between these two quadratic polynomials. In other words,these two quadratic polynomials are in the form of x^2+ax+b and x^2+bx+a respectively.
Can you find out three(or more)pairs of quadratic polynomials in the above foemat which can be factorized?
1條中2數學!!<急20點>
2009-11-01 7:06 pm
回答 (3)
2009-11-01 7:15 pm
✔ 最佳答案
e.g.x^2 + 5x + 6 = 0 x^2 + 6x + 5 = 0
(x + 2)(x + 3) = 0 (x + 5)(x + 1) = 0
x^2 + 12x + 11 = 0 x^2 + 11x + 12 = 0
(x + 11)(x + 1) = 0 (x + 6)(x + 5) = 0
x^2 - 2x + 1 = 0 x^2 + x - 2 = 0
(x - 1)(x - 1) = 0 (x + 2)(x - 1) = 0
2009-11-02 9:26 pm
x^2 + 5x + 6 = 0 x^2 + 6x + 5 = 0
(x + 2)(x + 3) = 0 (x + 5)(x + 1) = 0
x^2 + 12x + 11 = 0 x^2 + 11x + 12 = 0
(x + 11)(x + 1) = 0 (x + 6)(x + 5) = 0
x^2 - 2x + 1 = 0 x^2 + x - 2 = 0
(x - 1)(x - 1) = 0 (x + 2)(x - 1) = 0
(x + 2)(x + 3) = 0 (x + 5)(x + 1) = 0
x^2 + 12x + 11 = 0 x^2 + 11x + 12 = 0
(x + 11)(x + 1) = 0 (x + 6)(x + 5) = 0
x^2 - 2x + 1 = 0 x^2 + x - 2 = 0
(x - 1)(x - 1) = 0 (x + 2)(x - 1) = 0
2009-11-01 11:21 pm
x^2 + 5x + 6 = 0 x^2 + 6x + 5 = 0
(x + 2)(x + 3) = 0 (x + 5)(x + 1) = 0
x^2 + 12x + 11 = 0 x^2 + 11x + 12 = 0
(x + 11)(x + 1) = 0 (x + 6)(x + 5) = 0
x^2 - 2x + 1 = 0 x^2 + x - 2 = 0
(x - 1)(x - 1) = 0 (x + 2)(x - 1) = 0
(x + 2)(x + 3) = 0 (x + 5)(x + 1) = 0
x^2 + 12x + 11 = 0 x^2 + 11x + 12 = 0
(x + 11)(x + 1) = 0 (x + 6)(x + 5) = 0
x^2 - 2x + 1 = 0 x^2 + x - 2 = 0
(x - 1)(x - 1) = 0 (x + 2)(x - 1) = 0
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