Asked by Chujung
                Factor x3 + 9x2 + 27x + 27 
Factor 8a3 + 60a2b + 150 ab2+ 125b3
I tried doing them, but I just don't get how; perfect squares didn't work and neither did decomposition... someone please help me.
            
            
        Factor 8a3 + 60a2b + 150 ab2+ 125b3
I tried doing them, but I just don't get how; perfect squares didn't work and neither did decomposition... someone please help me.
Answers
                    Answered by
            Chujung
            
    Factor the following:
a)x4 + 11x2 + 30
b)x4 – 7x2y + 12y2
c)x6 - 3x3 -54
d)3(x – 5)2 + 27(x – 5) – 66
    
a)x4 + 11x2 + 30
b)x4 – 7x2y + 12y2
c)x6 - 3x3 -54
d)3(x – 5)2 + 27(x – 5) – 66
                    Answered by
            drwls
            
    <<Factor x^3 + 9x^2 + 27x + 27 >>
There is no simple general formula for solving cubic equations or factoring cubic poynomials. First look for one root of
x^3 + 9x^2 + 27x + 27 = 0
by graphing or trial and error.
One root is x = -3. That means that x+3 is one of the factors of
x^3 + 9x^2 + 27x + 27. Now use long division to get the quadratic factor. Divide x^3 + 9x^2 + 27x + 27 by (x+3), which will give you (x^2 + 6x + 9). That can be factored to (x+3)^2
The factored form is thus
(x+3)^3
    
There is no simple general formula for solving cubic equations or factoring cubic poynomials. First look for one root of
x^3 + 9x^2 + 27x + 27 = 0
by graphing or trial and error.
One root is x = -3. That means that x+3 is one of the factors of
x^3 + 9x^2 + 27x + 27. Now use long division to get the quadratic factor. Divide x^3 + 9x^2 + 27x + 27 by (x+3), which will give you (x^2 + 6x + 9). That can be factored to (x+3)^2
The factored form is thus
(x+3)^3
                    Answered by
            Chujung
            
    oh, okay, thanks!
    
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