Question

using factor theorem factorise x^3+2x^2-x-2​

Answer 1

Let f(x) = x3 + 2x2 – x – 2 Constant term = -2 Factors of -2 are ±1, ±2 Let x – 1 = 0 ⇒ x = 1 Put the value of x in f(x) f(1) = (1)3 + 2(1)2 – 1 – 2 = 1 + 2 – 1 – 2 = 0 So, (x – 1) is factor of f(x) Let x + 1 = 0 ⇒ x = -1 Put the value of x in f(x) f(-1) = (-1)3 + 2(-1)2 – 1 – 2 = -1 + 2 + 1 – 2 = 0 (x + 1) is a factor of f(x) Let x + 2 = 0 ⇒ x = -2 Put the value of x in f(x) f(-2) = (-2)3 + 2(-2)2 – (-2) – 2 = -8 + 8 + 2 – 2 = 0 (x + 2) is a factor of f(x) Let x – 2 = 0 ⇒ x = 2 Put the value of x in f(x) f(2) = (2)3 + 2(2)2 – 2 – 2 = 8 + 8 – 2 – 2 = 12 ≠ 0 (x – 2) is not a factor of f(x) Hence f(x) = (x + 1)(x - 1)(x + 2)Read more on Sarthaks.com - https://www.sarthaks.com/607724/using-factor-theorem-factorize-the-polynomials-x-3-2x-2-x-2

Answer 2

Answer:

ACTUALLY ITS A BIG PROBLEM HOPE YOU UNDERSTOOD

Step-by-step explanation:

Let f(x) = x3 + 2x2 – x – 2

Constant term = -2

Factors of -2 are ±1, ±2

Let x – 1 = 0 ⇒ x = 1

Put the value of x in f(x)

f(1) = (1)3 + 2(1)2 – 1 – 2 = 1 + 2 – 1 – 2 = 0

So, (x – 1) is factor of f(x)

Let x + 1 = 0

⇒ x = -1

Put the value of x in f(x) f(-1) = (-1)3 + 2(-1)2 – 1 – 2

= -1 + 2 + 1 – 2 = 0 (x + 1) is a factor of f(x)

Let x + 2 = 0 ⇒ x = -2

Put the value of x in f(x) f(-2) = (-2)3 + 2(-2)2 – (-2) – 2

= -8 + 8 + 2 – 2

= 0 (x + 2) is a factor of f(x)

Let x – 2 = 0

⇒ x = 2

Put the value of x in f(x)

f(2) = (2)3 + 2(2)2 – 2 – 2

= 8 + 8 – 2 – 2 = 12 ≠ 0 (x – 2) is not a factor of f(x)

Hence f(x) = (x + 1)(x - 1)(x + 2)