Using factor theorem, factorize each of the following polynomial:x3 +

1.	Using factor theorem, factorize each of the following polynomial:x3 + 13x2 + 32x + 20
Answer» Let, f (x) = x³ + 13x² + 32x + 20 The factors of the constant term + 20 are \(\pm\) 1, \(\pm\) 2,\(\pm\) 4, \(\pm\) 5, \(\pm\) 10 and 20 Putting x = -1, we have f (-1) = (-1)³ + 13 (-1)² + 32 (-1) + 20 = -1 + 13 – 32 + 20 = 0 So, (x + 1) is a factor of f (x) Let us now divide f (x) = x³ + 13x² + 32x + 20 by (x + 1) to get the other factors of f (x) Using long division method, we get x³ + 13x² + 32x + 20 = (x + 1) (x² + 12x + 20) x² + 2x + 20 = x² + 10x + 2x + 20 = x (x + 10) + 2 (x + 10) = (x + 10) (x + 2) Hence, x³ + 13x² + 32x + 20 = (x + 1) (x + 10) (x + 2)

Using factor theorem, factorize each of the following polynomial:x3 + 13x2 + 32x + 20

Answer»

Let, f (x) = x³ + 13x² + 32x + 20

The factors of the constant term + 20 are \(\pm\) 1, \(\pm\) 2,\(\pm\) 4, \(\pm\) 5, \(\pm\) 10 and 20

Putting x = -1, we have

f (-1) = (-1)³ + 13 (-1)² + 32 (-1) + 20

= -1 + 13 – 32 + 20

= 0

So,

(x + 1) is a factor of f (x)

Let us now divide

f (x) = x³ + 13x² + 32x + 20 by (x + 1) to get the other factors of f (x)

Using long division method, we get

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

x² + 2x + 20 = x² + 10x + 2x + 20

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

= (x + 10) (x + 2)

Hence,

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