分解因式(x2+3x-4)2-(3x2-4x+2)2+(2x2-7x+6)2
2个回答
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原式=[(x+4)(x-1)]^2+[(2x^2-7x+6)+(3x^2-4x+2)][(2x^2-7x+6)-(3x^2-4x+2)]
=[(x+4)(x-1)]^2-(5x^2-11x+8)(x^2+3x-4)
=[(x+4)(x-1)]^2-(5x^2-11x+8)(x+4)(x-1)
=(x+4)(x-1)[(x^2+3x-4)-(5x^2-11x+8)]
=(x+4)(x-1)(-4x^2+14x-12)
=-2(x+4)(x-1)(2x^2-7x+6)
=-2(x+4)(x-1)(2x-3)(x-2)
=[(x+4)(x-1)]^2-(5x^2-11x+8)(x^2+3x-4)
=[(x+4)(x-1)]^2-(5x^2-11x+8)(x+4)(x-1)
=(x+4)(x-1)[(x^2+3x-4)-(5x^2-11x+8)]
=(x+4)(x-1)(-4x^2+14x-12)
=-2(x+4)(x-1)(2x^2-7x+6)
=-2(x+4)(x-1)(2x-3)(x-2)
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