*** Steps ***
Factorize the following quadratic polynomials using cross method.
(a) 2c^2 + 8c - 24
(b) -d^2 - 10d - 21
(c) P^2 - (q - 2q + 1 )
(d) (a+1)^2 + 2(a+1) + 1
中2Maths..急..Factorize
2009-08-19 3:15 am
回答 (2)
2009-08-19 3:57 am
✔ 最佳答案
(a)2c^2 + 8c - 24
=( 2c + 12 )( C - 2 )
2c \ / 12
c / \ -2
---------------
12c - 4c = 8c
(b)
-d^2 - 10d - 21
=( d - 7 )( 3 - d )
d \ / -7
-d / \ 3
---------------
7d + 3d = 10d
(c)( I'm not sure!!! )
P^2 - ( q - 2q + 1 )
=P^2 - q + 2q - 1
=P^2 + q - 1
=( P^2 + q - 1 )( 1 + 0 )
P^2 \ / q - 1
1 / \ 0
-------------------
q - 1 + 0 = q - 1
(d)
( a + 1 )^2 + 2( a + 1 ) + 1
=( a + 2 )^2
a+1 \ / 1
a+1 / \ 1
---------------
a + 1 + a + 1 = 2a + 2 = 2( a + 1 )
參考: myself
2009-08-19 3:43 am
(a) 2c^2 + 8c - 24
=2( c^2 + 4c - 12)
=2(c-2) (C+6)
(b) -d^2 - 10d - 21
=-(d^2 + 10d + 21)
=-(d+3)(d+7)
(c) P^2 - (q - 2q + 1 )
= P^2 + q- 1
(d) (a+1)^2 + 2(a+1) + 1
= (a+1+1)^2
=(a+2)^2
=2( c^2 + 4c - 12)
=2(c-2) (C+6)
(b) -d^2 - 10d - 21
=-(d^2 + 10d + 21)
=-(d+3)(d+7)
(c) P^2 - (q - 2q + 1 )
= P^2 + q- 1
(d) (a+1)^2 + 2(a+1) + 1
= (a+1+1)^2
=(a+2)^2
收錄日期: 2021-05-01 17:46:10
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