# The square root of (x^2+8x+15) (x^2+x-6) (x^2+3x-10)

1
by umeshbala

2014-08-02T23:03:49+05:30

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Square root of [ (x^2+8x+15) (x^2+x-6) (x^2+3x-10) ]
we can get an idea that in such cases that, the polynomials (quadratic expressions) given above may be factorized ..
x²+8x+15  = (x+a)(x+b) => a+b = 8    a*b = 15  easy to find a = 5  b = 3  as factors of 15 are easy
x²+x-6  = (x+c)(x+d)    => c+d = 1  c*d = -6  factors of -6 are 3 & 2  since sum is +1,  c=+3  d= - 2
x² + 3x - 10 = (x+e)(x+f)  =>  e+f = 3  ef = -10  factors of 10 5, 2. since sum is 3, e= +5,  f= -2

NOW squaretoot  is equal to
square root [   (x+3) (x+5)  (x+3)  (x-2)  (x+5) (x-2)  ]
square root [  (x+3)² (x+5)²  (x-2)² ]
=  (x+3) (x+5) (x-2)

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If factorizaion is difficult, then do as :
(x^2+8x+15) = 0      Δ = 8² - 4*15 = 4    => x = (-8 +- 2 ) /2 = -5 or -3
so  (x^2+8x+15) = (x - -5)  (x - -3)  = (x+5)(x+3)
(x^2+x-6) =0          Δ = 1 +24 = 25    => x = -1 +- 5  / 2  = -3 or 2
so factors  (x - -3)(x - 2)
similarly the other expression