polynomial question (or part of) (1 Viewer)

[abcd9146]

FITZ 3u 27d 15a
α+β+γ=-3
αβ+αγ+βγ=-2
αβγ=-1

i needed to simplify this:
1. α²(β+γ)+β²(α+γ)+γ²(α+β)
2. α²β+α²γ+β²γ+β²α+γ²α+γ²β
3. α²β+αβ²+α²γ+αγ²+β²γ+βγ²
4. αβ(α+β)+αγ(α+γ)+βγ(β+γ)
5. αβ(α+β)+αβγ+αγ(α+γ)+αβγ+βγ(β+γ)+αβγ-3αβγ
6. (α+β+γ)(αβ+αγ+βγ)-3αβγ

my teacher did this, and i wanna know how she factorised from line 4 to line 6, any ideas?

 

[insert-username]

FITZ 3u 27d 15a
α+β+γ=-3
αβ+αγ+βγ=-2
αβγ=-1

i needed to simplify this:
1. α²(β+γ)+β²(α+γ)+γ²(α+β)
2. α²β+α²γ+β²γ+β²α+γ²α+γ²β
3. α²β+αβ²+α²γ+αγ²+β²γ+βγ²
4. αβ(α+β)+αγ(α+γ)+βγ(β+γ)
5. αβ(α+β)+αβγ+αγ(α+γ)+αβγ+βγ(β+γ)+αβγ-3αβγ
6. (α+β+γ)(αβ+αγ+βγ)-3αβγ

αβγ = -1. From line 4 to 5, all your teacher has done is 'added' -1, then subtracted '-3' (i.e. + αβγ + αβγ + αβγ - 3αβγ). From that line, she's taken αβ out of the first two terms, αγ out of the next two, and βγ out of the next two. Then she's factorised that using the distributive law, so:

αβ(α+β)+αβγ+αγ(α+γ)+αβγ+βγ(β+γ)+αβγ-3αβγ

= αβ[(α+β)+γ] + αγ[(α+γ)+β] + βγ[(β+γ)+α]-3αβγ

= αβ(α+β+γ) + αγ(α+β+γ) + βγ(α+β+γ)-3αβγ

= (α+β+γ)(αβ+αγ+βγ)-3αβγ

Hope that helps,


I_F