factorization... @ 部落格

標題:

factorization...

發問:

1. 2y^3 + 1/4 2. t^2 + (t+3)^2 - 17 3. It is given that a^2 + b^2=39 and ab=18. a. find the values of (a+b)^2 and (a-b)^2 b. hence find the value of a^4 - 2a^2b^2 + b^4 4. a. factorize x^3 -8 b. use the result in (a) to factorize x^3 + x -10 更新: (p^2 + q^2)^2 - 4p^2q^2 36x^2 - 12x - 169y^2 + 26y

最佳解答:

1. 2y^3 + 1/4 =2(y^3 + 1/8) =2(y^3 + 1/2^3) =2(y+ 1/2)[y^2 -(1/2)*y+(1/2)^2)] , 因為(a^3 + b^3)=[a^2 -ab+b^2)] =2(y+ 1/2)[y^2 -(1/2)*y+(1/4))] 2. t^2 + (t+3)^2 - 17 =t^2 + (t^2+6t+9) - 17 =2t^2 +6t - 8 =2(t^2 +3t - 4) =2(t+4)(t-1) 3. It is given that a^2 + b^2=39 and ab=18. a. find the values of (a+b)^2 and (a-b)^2 (a+b)^2 =a^2 + b^2 +2 ab =(39) +2 (18) =75 (a-b)^2 =a^2 + b^2 -2 ab =(39) -2 (18) =3 b. hence find the value of a^4 - 2a^2b^2 + b^4 a^4 - 2a^2b^2 + b^4 =(a^2 - b^2)^2 =[(a-b)(a+b)]^2 =(a-b)^2(a+b)^2 =(75)(3) =225 4. a. factorize x^3 -8 =x^3 -2^3 =(x-2)(x^2+2x+2^2) ,因為(a^3 - b^3)=[a^2 +ab+b^2)] =(x-2)(x^2+2x+4) b. use the result in (a) to factorize x^3 + x -10 x^3 + x -10 =(x^3 -8)+( x -2) =(x-2)(x^2+2x+4)+( x -2) ,from (a) =(x-2)(x^2+2x+4+1) =(x-2)(x^2+2x+5)

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