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因式分解問題....簡單~~
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最佳解答:
= (p^2 - q^2 ) - 16(p^2 - q^2)x^4 = (p^2 - q^2)( 1 - 16x^4) = (p +q)(p -q)[ 1 - (2x)^4] = (p+q)(p-q)[ 1 - (2x)^2][1 + 4x^2] = (p+q)(p-q)(1+ 2x)( 1 - 2x)((1 + 4x^2). 2008-09-13 10:10:50 補充： Q2 = abx^2 + a^2x + b^2x + ab = ax(bx + a) + b(bx + a) = (ax + b)(bx + a). Q3 = (x^2 + 5x + 4)(x^2 + 5x + 6) - 120. Let x^2 + 5x = y, we get (y + 4)(y + 6) - 120 = y^2 + 10y - 96 = (y +16)(y -6) = (x^2 + 5x + 16)(x^2 + 5x - 6) = (x^2 + 5x + 16)(x+6)(x -1).
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因式分解問題....簡單~~
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p^ - q^ + 16 ( q^ - p^ ) x* ^為2次方 *為4次方 更新: 重有2題! abx^ + (a^+b^)x + ab (x+1)(x+2)(x+3)(x+4)-120最佳解答:
= (p^2 - q^2 ) - 16(p^2 - q^2)x^4 = (p^2 - q^2)( 1 - 16x^4) = (p +q)(p -q)[ 1 - (2x)^4] = (p+q)(p-q)[ 1 - (2x)^2][1 + 4x^2] = (p+q)(p-q)(1+ 2x)( 1 - 2x)((1 + 4x^2). 2008-09-13 10:10:50 補充： Q2 = abx^2 + a^2x + b^2x + ab = ax(bx + a) + b(bx + a) = (ax + b)(bx + a). Q3 = (x^2 + 5x + 4)(x^2 + 5x + 6) - 120. Let x^2 + 5x = y, we get (y + 4)(y + 6) - 120 = y^2 + 10y - 96 = (y +16)(y -6) = (x^2 + 5x + 16)(x^2 + 5x - 6) = (x^2 + 5x + 16)(x+6)(x -1).
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