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2006-09-24 16:08:37 · 12 answers · asked by jeff f 1 in Science & Mathematics Mathematics

12 answers

first factor a 2 out of everything;
2(a^2 - 5a+4) Then factor
2(a -4)(a-1)

2006-09-24 16:15:21 · answer #1 · answered by teacher2006 3 · 1 0

2a^2 - 10a + 8 = 2(a^2 - 5a + 4) = 2(a - 4)(a - 1)

2006-09-24 16:36:17 · answer #2 · answered by Sherman81 6 · 0 0

As a first step, notice that all the numbers are even and so have a factor of 2: remove this to give 2(a^2-5a+4).

Now a^2-5a+4 factors to (a-1)(a-4)

So the final answer is 2(a-1)(a-4)

2006-09-24 16:21:58 · answer #3 · answered by Andrew B 1 · 0 0

2a^2-10a+8
(2a-2)(a-4)
a=1, a=4

2006-09-24 16:23:25 · answer #4 · answered by ronw 4 · 0 0

2a² - 10a + 8

2(a² - 5a + 4)

2(a - 4)(a - 1)

2006-09-25 00:22:14 · answer #5 · answered by SAMUEL D 7 · 0 0

2aa-10a+8 =
= 2aa - 2a - 8a + 8 =
= 2a(a - 1) - 8(a - 1) =
= (2a - 8)(a - 1) =
= 2(a - 4)(a - 1)

[]'s

2006-09-24 16:44:13 · answer #6 · answered by Eric Campos Bastos Guedes 3 · 0 0

2a^2-10a+8

2a - 2 2a
a - 4 8a

(2a-2)(a-4)

2006-09-24 17:14:50 · answer #7 · answered by ayie 2 · 0 0

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2016-12-12 14:29:30 · answer #8 · answered by ? 4 · 0 0

first divide by 2. then answer is
2(a-4)(a-1)

2006-09-25 01:13:20 · answer #9 · answered by sweetfloss8 2 · 0 0

(2a -2)
(a-4)

2006-09-24 17:58:57 · answer #10 · answered by Big mama 4 · 0 0

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