Woodencrawdad wrote:Can someone explain how you simplify #1 to #2? Can't seem to see how they arrived at the simplification in #2.
1) x(x-1)^2 + 5(x-1)^2 = 0
2) (x+5)(x-1)^2 = 0
Hi Woodencrawdad!
This factorization might be a bit difficult to see right away, so I would recommend using a "substitution" trick that I sometimes find handy! Notice that both terms in #1 have an (x-1)^2 term. Let's use a simple substitution variable to put in this place, so why don't we say that H = (x-1)^2.
Now #1 becomes:
x(H) + 5(H) = 0
Since both terms have an H, we can factor it out!
H(x+5) = 0
or
(x+5)(H) = 0
And now, we just replace our H with what it was originally
(x+5)(x-1)^2 = 0
And then, since this product =0, then one piece must =0 for it to be true and we can solve as we would any other quadratic:
x+5=0, or (x-1)^2=0
x = -5, or x-1=0 --> x=1
So the solutions for this equation are x=-5, 1

Whit
Whitney Garner
GMAT/GRE/EA Instructor & Anxiety/Accommodations Coach
www.whitneygarner.com
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Math is a lot like love - a simple idea that can easily get complicated
