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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

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

The zeros of polynomial x3 + 13x2 +32x +20 are

x = -1 , -2 , -10

x = -1 , -2 , -10