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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

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 ]

=========================================================

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