Factorisation problem? please help
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Factorisation problem? please help

[From: ] [author: ] [Date: 12-04-28] [Hit: ]
..Then my calculator came up with 4(w^2 -w +1.25) but Im not sure if thats right, and its a homework sheet in VCE so I dont have the answer either........
When I was trying to solve this problem, none of my methods worked:
4w^2 - 4w + 5

Because of the 4 in front, I tried the normal (4w .....) (4w.....) all divided by four, and so times the 5 by four,
like this now: 4w^2 -4w +20
but I just couldn't figure out what 2 numbers add to give -4 and multiply to give +20...
Then my calculator came up with 4(w^2 -w +1.25) but I'm not sure if that's right, and it's a homework sheet in VCE so I don't have the answer either...

any advice?

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4w² - 4w + 5
4[w² - w] + 5                       ← Now, complete the square for [w² - w]
4[w² - w + ¼ - ¼] + 5
4[(w - ½)² - ¼] + 5              ← Now, distribute 4 through [(w - ½)² - ¼]
4(w - ½)² - 1 + 5
4(w - ½)² + 4                       ← Now, factor out 4
4[(w - ½)² + 1]
4[(w - ½)² - i²]                       ← set up to factor the difference of squares
4[(w - ½) + i][(w - ½) - i]
4(w - ½ + i)(w - ½ - i)            ← ANSWER


𝐄𝐃𝐈𝐓
If you don't want the fraction ½ in your answer, you could do the following:
4(w - ½ + i)(w - ½ - i)
2(w - ½ + i)2(w - ½ - i)
(2w - 1 + 2i)(2w - 1 - 2i)            ← ANSWER

𝐄𝐃𝐈𝐓
i is the "imaginary number" or the "imaginary unit" and is defined as i = √ ̅-̅1̅ ̅
so that    i² = -1



Have a good one!
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Have you tried using the quadratic formula?

This is excellent for use with trinomials that you cannot figure out or that are not factorable.

This website will explain it:
http://www.purplemath.com/modules/quadfo…

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4w^2 - 4w + 5 is prime!
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