a) (a-1)
b) (a-3)
c) (5a+1)
d) (5a-1)
e) (5a ^2+14a-3)
2007-04-16 09:53:32 · 8 answers · asked by jonathan w 1 in Science & Mathematics ➔ Mathematics
d) (5a -1)
using factor theorem of if f(a) = 0 then a is a factor
in the case of d. a = 1/5
totally factorised the eqn. is 4(5a -1)(a + 3)
2007-04-16 09:59:48 · answer #1 · answered by James H 1 · 1⤊ 0⤋
20a^2 + 56a -12 (div by 4)
= 5a^2 + 14a -3
= (5a - 1) ( a + 3)
Answer : (5a - 1) => d)
2007-04-16 17:10:10 · answer #2 · answered by frank 7 · 1⤊ 0⤋
4[5a² + 14a - 3] =
4(5a - 1)(a + 3)
2007-04-16 17:04:11 · answer #3 · answered by Darlene 4 · 1⤊ 0⤋
d 4(5a-1)(a+3)
2007-04-16 17:04:37 · answer #4 · answered by dwinbaycity 5 · 1⤊ 0⤋
4(5a-1)(a+3)
so D is correct although E is also true.
2007-04-16 17:02:08 · answer #5 · answered by ooorah 6 · 2⤊ 0⤋
4(5a^2+14a-3)
(5a-1)(a+3)
The answer is d.
2007-04-16 17:46:42 · answer #6 · answered by Anonymous · 0⤊ 0⤋
Let´s find the zeros of the expression
a=((-56+-sqrt(4096))/40 =(-56+-64)/40
so a=-3 and a= 1/5
So one factor is 5a-1
2007-04-16 17:03:53 · answer #7 · answered by santmann2002 7 · 1⤊ 0⤋
(e) is the original expression divided by 4 so it is a factor.
2007-04-16 16:58:31 · answer #8 · answered by peateargryfin 5 · 1⤊ 0⤋