Step-by-step explanation:

Let p(x) = x³+13x²+32x+20

By trial, we find that

p( - 1) = ( - 1 {)}^{3} + 13( - 1 {)}^{2} + 32( - 1) + 20 = - 1 + 13 - 32 + 20 = 0

By factor theorem, x-(-1) i, e(x+1) is a factor of p(x)

Now,

{x}^{3} + 13x + 32x + 20 = {x}^{2}(x + 1) + 12x(x + 1) + 20(x + 1) = (x + 1)( {x}^{2} + 12x + 20) = (x + 1)( {x}^{2} + 2x + 10x + 20) = (x + 1)(x(x + 2) + 10(x + 2) = (x + 1)(x + 2)(x + 10)

Answer:

(x+1)(x+2)(x+10)

Step-by-step explanation:

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Let p(x) = x3 + 13x2 + 32x + 20

p(-1) = -1 + 13 - 32 + 20 = -33 + 33 = 0

Therefore (x + 1) is a factor of p(x).

On dividing p(x) by (x + 1) we get

p(x)   (x + 1) = x2 + 12x + 20  

Thus,

x3 + 13x2 + 32x + 20 = (x + 1)(x2 + 12x + 20)  

= (x + 1) (x2 + 10x + 2x + 20)  

= (x + 1)[x(x + 10) + 2(x + 10)]  

= (x + 1) (x +2) (x + 10)

Hence, x3 + 13x2 + 32x + 20 = (x + 1) (x +2) (x + 10).