S. 3 Factorization 有少少唔明

2010-10-08 6:10 am
1b & 2b 有少少唔明

1a) Factorize c^2-8c-20
﹦(c-10)(c+2)

1b) Hence factorize y^2(y-3)^2-8y(y-3)-20
﹦(1b 欠詳解)
﹦(c-10)(c+2)

2a) Factorize the following expressions.
i) d^2+2d-8
﹦(d-2)(d+4)

ii) d^2+5d-14
﹦(d-2)(d+7)

2b) Hence factorize (﹣3d^2-6d+24)+(d^3+5d^2-14d)
﹦(2d 欠詳解)
﹦(d-2)^2(d+6)

回答 (1)

2010-10-08 7:37 am
✔ 最佳答案
1a) Factorize c^2-8c-20
﹦(c-10)(c+2)1b) Hence factorize y^2(y-3)^2-8y(y-3)-20Sub c = y(y - 3) to a) : y^2(y-3)^2-8y(y-3)-20= (y(y -3 ) - 10) (y(y - 3) + 2)= (y^2 - 3y - 10) (y^2 - 3y + 2)= (y - 5)(y + 2) (y - 1)(y - 2) 2a) Factorize the following expressions.
i) d^2+2d-8
﹦(d-2)(d+4)ii) d^2+5d-14
﹦(d-2)(d+7)2b) Hence factorize (﹣3d^2-6d+24)+(d^3+5d^2-14d)﹦- 3 (d^2 + 2d - 8) + d(d^2 + 5d - 14) by part a) := - 3 (d - 2)(d + 4) + d(d - 2)(d + 7)= (d - 2)( - 3(d + 4) + d(d + 7))= (d - 2)(d^2 + 4d - 12)= (d - 2)(d - 2)(d + 6)﹦(d - 2)^2 (d + 6)


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