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2014-07-24T19:59:40+05:30

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To solve    y = x³ + 13 x² + 32 x + 20  = 0
You could solve it by writing (x+a)(x+b)(x+c) =0  and solve for a, b,c.  OR, you can do as follows:

From the coefficients of the polynomial it is easy to see that at x = -1, y =0
  as -1 + 13 - 32 + 20 = 0    ==> x = -1  =>  (x+1) is a factor of y.
write now  y as (x + 1) (x² + a x + b) 
  It is easy to see that 1 * b = 20  => b = 20
   now coefficient of x2 will be    a + 1  = 13    ==>  a =12
   solve x² + 12 x + 20 = 0    =>  x = -2 or -10
the zeros are :  -1, -2, -10

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2014-07-24T19:59:59+05:30
The zeros of polynomial x3 + 13x2 +32x +20 are
x = -1 , -2 , -10
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