Using factor theorem, factorize the polynomials: x3 + 6x2 + 11x + 6

1.	Using factor theorem, factorize the polynomials: x3 + 6x2 + 11x + 6
Answer» Let f(x) = x³ + 6x² + 11x + 6 Step 1: Find the factors of constant term Here constant term = 6 Factors of 6 are ±1, ±2, ±3, ±6 Step 2: Find the factors of f(x) Let x + 1 = 0 ⇒ x = -1 Put the value of x in f(x) f(-1) = (−1)³ + 6(−1)² + 11(−1) + 6 = -1 + 6 -11 + 6 = 12 – 12 = 0 So, (x + 1) is the factor of f(x) Let x + 2 = 0 ⇒ x = -2 Put the value of x in f(x) f(-2) = (−2)³ + 6(−2)² + 11(−2) + 6 = - 8 + 24 – 22 + 6 = 0 So, (x + 2) is the factor of f(x) Let x + 3 = 0 ⇒ x = -3 Put the value of x in f(x) f(-3) = (−3)³ + 6(−3)2 + 11(−3) + 6 = -27 + 54 – 33 + 6 = 0 So, (x + 3) is the factor of f(x) Hence, f(x) = (x + 1)(x + 2)(x + 3).

Answer»

Let f(x) = x³ + 6x² + 11x + 6

Step 1: Find the factors of constant term

Here constant term = 6

Factors of 6 are ±1, ±2, ±3, ±6

Step 2: Find the factors of f(x)

Let x + 1 = 0

⇒ x = -1

Put the value of x in f(x)

f(-1) = (−1)³ + 6(−1)² + 11(−1) + 6

= -1 + 6 -11 + 6

= 12 – 12

= 0

So, (x + 1) is the factor of f(x)

Let x + 2 = 0

⇒ x = -2

Put the value of x in f(x)

f(-2) = (−2)³ + 6(−2)² + 11(−2) + 6

= - 8 + 24 – 22 + 6

= 0

So, (x + 2) is the factor of f(x)

Let x + 3 = 0

⇒ x = -3

Put the value of x in f(x)

f(-3) = (−3)³ + 6(−3)2 + 11(−3) + 6

= -27 + 54 – 33 + 6

= 0

So, (x + 3) is the factor of f(x)

Hence, f(x) = (x + 1)(x + 2)(x + 3).

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