F.2 Math

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)