F.2 Math
2017/06/21 02:56
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標題:
F.2 Math
發問:
Factorize x3+27 Hence, Factorize x3+x2+4x+30
Q1. 3x+27 =3(x+9) =3(x+33) Q2. 3x+2x+4x+30 =9x+30 =3(3x+10) **Actually, we put the constant (e.g. 0, 1, 2...) before the algebra (e.g. a, x, y...), and mostly in alphabetical order (a, b ,c...) as it is more easy for us to calculate. So, in the above question, x3+27 --> 3x+27 x3+x2+4x+30 --> 3x+2x+4x+30 Hope I can help you ^_^
其他解答:
其實你應該打錯,你應該係想 Factorize x 3次方+27 係咪? 我已假設你打錯 a^3+b^3 =(a+b)(a^2-ab+b^2) x^3+27 =x^3+3^3 =(x+3)(x^2-3x+3^2) =(x+3)(x^2-3x+9) 如果無,就係: 3x+27 =3(x+9) 我又假設你打錯,因為你又要hence ,但結果又唔同,因為如果係 x 3次方+x 2 次方+4x +30,計出黎好怪... It should be ... x^3+27+x^2+4x+3 =(x+3)(x^(2)-3x+9)+(x+3)(x+1) =(x+3)(x^2-3x+9+x+1) =(x+3)(x^2-2x+10) 又或者你無錯,咁就係: 3x+2x+4x+30 = x(3+2+4)+30 =x(9)+30 =9x+30 =3(x+10) 希望幫到你la! 2014-11-27 20:57:59 補充： oh sorry ,I made a mistake! 3x+2x+4x+30 = x(3+2+4)+30 =x(9)+30 =9x+30 =3(x+10) Should be 3x+2x+4x+30 = x(3+2+4)+30 =x(9)+30 =9x+30 =3(3x+10) 2014-11-27 20:59:37 補充： oh sorry ,I made a mistake! 3x+2x+4x+30 = x(3+2+4)+30 =x(9)+30 =9x+30 =3(x+10) Should be 3x+2x+4x+30 = x(3+2+4)+30 =x(9)+30 =9x+30 =3(3x+10)|||||a) x^3+27 =x^3+(3)^3 =(x+3)(x^(2)-3x+9) b)x^3+x^2+4x+30 =x^3+27+x^2+4x+3 =(x+3)(x^(2)-3x+9)+(x+3)(x+1) =(x+3)(x^2-3x+9+x+1) =(x+3)(x^2-2x+10)|||||a3+b3=(a+b)(a2-ab+b2) x3+27 =x3+33 =(x+3)(x2-3x+32) =(x+3)(x2-3x+9) x3+x2+4x+30 =(x3+27)+(x2+4x+3) =(x+3)(x2-3x+9)+(x+3)(x+1) =(x+3)(x2-3x+9+x+1) =(x+3)(x2-2x+10)
F.2 Math
發問:
Factorize x3+27 Hence, Factorize x3+x2+4x+30
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最佳解答:Q1. 3x+27 =3(x+9) =3(x+33) Q2. 3x+2x+4x+30 =9x+30 =3(3x+10) **Actually, we put the constant (e.g. 0, 1, 2...) before the algebra (e.g. a, x, y...), and mostly in alphabetical order (a, b ,c...) as it is more easy for us to calculate. So, in the above question, x3+27 --> 3x+27 x3+x2+4x+30 --> 3x+2x+4x+30 Hope I can help you ^_^
其他解答:
其實你應該打錯,你應該係想 Factorize x 3次方+27 係咪? 我已假設你打錯 a^3+b^3 =(a+b)(a^2-ab+b^2) x^3+27 =x^3+3^3 =(x+3)(x^2-3x+3^2) =(x+3)(x^2-3x+9) 如果無,就係: 3x+27 =3(x+9) 我又假設你打錯,因為你又要hence ,但結果又唔同,因為如果係 x 3次方+x 2 次方+4x +30,計出黎好怪... It should be ... x^3+27+x^2+4x+3 =(x+3)(x^(2)-3x+9)+(x+3)(x+1) =(x+3)(x^2-3x+9+x+1) =(x+3)(x^2-2x+10) 又或者你無錯,咁就係: 3x+2x+4x+30 = x(3+2+4)+30 =x(9)+30 =9x+30 =3(x+10) 希望幫到你la! 2014-11-27 20:57:59 補充： oh sorry ,I made a mistake! 3x+2x+4x+30 = x(3+2+4)+30 =x(9)+30 =9x+30 =3(x+10) Should be 3x+2x+4x+30 = x(3+2+4)+30 =x(9)+30 =9x+30 =3(3x+10) 2014-11-27 20:59:37 補充： oh sorry ,I made a mistake! 3x+2x+4x+30 = x(3+2+4)+30 =x(9)+30 =9x+30 =3(x+10) Should be 3x+2x+4x+30 = x(3+2+4)+30 =x(9)+30 =9x+30 =3(3x+10)|||||a) x^3+27 =x^3+(3)^3 =(x+3)(x^(2)-3x+9) b)x^3+x^2+4x+30 =x^3+27+x^2+4x+3 =(x+3)(x^(2)-3x+9)+(x+3)(x+1) =(x+3)(x^2-3x+9+x+1) =(x+3)(x^2-2x+10)|||||a3+b3=(a+b)(a2-ab+b2) x3+27 =x3+33 =(x+3)(x2-3x+32) =(x+3)(x2-3x+9) x3+x2+4x+30 =(x3+27)+(x2+4x+3) =(x+3)(x2-3x+9)+(x+3)(x+1) =(x+3)(x2-3x+9+x+1) =(x+3)(x2-2x+10)
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