Question

The factors of x^3 - 2 x^2 - x + 2 are
2 points
(a) (x+1)(x-1)(x-2)
(b) (x+1)(x+1)(x+5)
(c) (x+1)(x-1)(x+5)
(d) None of these
​

Answer 1

Solution :

Let p(x) = x³ - 2x² - x + 2

We shall now look for all the factors of 2. These are : ±1, ± 2

By Hit and Trial Method :

→ p(1) = 0

→ p = 0 - 1

→ p = -1

Put the value of p = -1 in p(x)

→ x³ - 2x² - x + 2

→ (-1)³ - 2(-1)² - (-1) + 2

→ -1 -2(1) + 1 + 2

→ -1 - 2 + 1 + 2

→ 0

We find that (x-1) is factor of the polynomial p(x)

[ Refer to the attachment]

We find that (x³ - 2x² - x + 2) ÷ (x - 1) is x² - x - 2

Now,By Splliting the middle term,we've :

→ x² - x - 2

→ x² - 2x + x - 2

→ x(x - 2) + 1(x - 2)

→ (x + 1) (x - 2)

So,

x³ - 2x² - x + 2 = (x - 1) (x + 1) (x - 2)

$\therefore$ Option a) is correct answer.

Answer 2

ANSWER:

$\small \sf{x - 1) \: \: \: \cancel {{x}^{3}} - 2x^{2} - x + 2( {x}^{2} - x + 2 } \\ \small \sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \cancel{{x}^{3} } - {x}^{2} } \\ \small \sf{ \: \: \: \: \: \: \: \: \: \: \underline{ ( -) \: \: \: \: ( +) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }} \\ \sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - \: \: \cancel{{x}^{2}} - x + 2} \\ \sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - \: \: \cancel{{x}^{2} } + x } \\ \sf{ \: \: \: \: \: \: \: \: \: \underline{ \: \: \: \: \: \: \: \: \: \: (+ ) \: \: \: \: (-) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }} \\ \sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \cancel{- 2x + 2}} \\ \sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \cancel {- 2x + 2}} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underline{ ( - ) \: ( + ) \: \: \: \: } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0$

x - 1 = 0

x = 1

p(x) = x³ - 2x² - x + 2

Putting p(x) = p(1)

x² - x + 2 = p(1)

By hit and trial method

p(1) = 0

p = -1

Putting p = -1 in p(x)

→(- 1)³ - 2(-1)² - (-1) + 2

→ - 1 - 2 + 1 + 2

→ 0

Simplifying the quotient

→ x² - x + 2

→ x² - 2x + x + 2

→ x(x - 2) + 1(x - 2)

→ (x - 2)(x + 1)

So,

x³ - 2x² - x + 2 = (x + 1)(x - 2)(x - 2)

Hence, option a is answer