Question

Factorise p(x) = x^3 - 2x^2 - x +2​

Answer 1

Answer:

$()=3−22−+2 \\ \\ </p><p></p><p>p(x) \\ \\ =x^{3}-2x^{2} \\ \\ -x+2p(x) \\ \\ \\ =x3−2x2−x+2</p><p></p><p>$

Answer 2

We can factorise it by many methods , two of them are:-

Method 1 Factorization

${x}^{3} - 2 {x}^{2} - x + 2$

${x}^{2} (x - 2) - 1(x - 2) \: \: (common)$

$( {x}^{2} - {1})(x - 2)$

$by \: using \: {a}^{2} - {b}^{2} = (a + b)(a - b)$

$(x + 1)(x - 1)(x - 2)$

Method 2 Hit and trial method

${x}^{3} - 2 {x}^{2} - x + 2$

Here constant term is 2

Factors of constant term 2 are ± 1, ± 2

Two is having 4 values we will put it in p (x) and will see it becomes zero or not

For x = 1

${x}^{3} - 2 {x}^{2} - x + 2$

$p(1) = {(1)}^{3} - 2 {(1)}^{2} - 1 + 2$

$( 1) - 2(1) - (1) + 2$

$(1) - 2 - 1 + 2$

$3 - 3$

$0$

Therefore by factor therom ( x - 1) is the factor of P (x)

For x= -1

${x}^{3} - 2 {x}^{2} - x + 2$

$p( - 1) = {( - 1)}^{3} - 2 {( - 1)}^{2} - ( - 1) + 2$

$( - 1) - 2(1) + 1 + 2$

$- 1 - 2 + 1 + 2$

$- 3 + 3$

$0$

Therefore by factor therom ( x +1) is the factor of P (x)

For x= 2

${x}^{3} - 2 {x}^{2} - x + 2$

$p(2) = {2}^{3} - 2 {(2)}^{2} - (2) + 2$

$8 - 8 - 2 + 2$

$10 - 10$

$0$

Therefore by factor therom ( x + 2) is the factor of P (x)

Therefore P(x) = ( x-1) ( x+1) (x+2)

Note :-

We need only three factors as degree is three

Hope it helps

• There are also other methods like long method or method of inspection