Question

Factories: (b) x^3+2x^2-x-2​

Answer 1

Answer:

x3−2x2−x+2

=x3−x−2x2+2

Grouping the 1st 2 terms together and the 2nd 2 together:

=x(x2−1)−2(x2−1)

=(x2−1)(x−2)

Using the identity: a2−b2=(a+b)(a−b)

=(x2−12)(x−2)

=(x−1)(x+1)(x−2)

Alternate Method:-

The above method is an easy on to solve this question. For factorizing other cubic polynomials, the following method can be used:

First, by trial and error method, you can find one factor as follows:

x3−2x2−x+2

When replacing 1,

⇒13−2×12−1+2

⇒1

−2−1+2

⇒0

So, we get (x−1) as factor.

Then by long division, divide (x−1) by (x3−2x2−x+2)

You get⇒(x−1)(x2−x−2)

Then you have to factorize it by splitting the middle term method.

⇒(x−1)[x2+x−2x−2]

⇒(x−1)[x(x+1)−2(x+1)]

⇒(x−1)(x+1)(x−2)

Step-by-step explanation:

Answer 2

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\: {x}^{3} + {2x}^{2} - x - 2$

$\rm \: = \: \: \red{ ({x}^{3} + {2x}^{2})} + \purple{( - x - 2)}$

$\rm \: = \: \: (x \times x \times x + 2 \times x \times x) + ( - 1 \times x - 2 \times 1)$

$\rm \: = \: \: {x}^{2} \purple{(x + 2)} - 1 \purple{(x + 2)}$

$\rm \: = \: \: \purple{(x + 2)}( {x}^{2} - 1)$

$\rm \: = \: \: (x + 2) \bigg( {x}^{2} - {1}^{2} \bigg)$

$\rm \: = \: \: (x + 2)(x + 1)(x - 1)$

$\: \: \: \: \: \red{\bigg \{ \because \: {x}^{2} - {y}^{2} = (x + y)(x - y)\bigg \}}$

Hence,

$\red{\boxed{\rm :\longmapsto\: {x}^{3} + {2x}^{2} - x - 2 = (x + 2)(x - 1)(x + 1)}}$

Concept Used :-

1. Factorisation by Regrouping Terms

Sometimes it happens that there is no common term in the expressions then

• We have to make the groups of the terms.

• Then choose the common factor among these groups.

• Find the common factor and it will give the required factors.

More Identities to know:

• (a + b)² = a² + 2ab + b²

• (a - b)² = a² - 2ab + b²

• a² - b² = (a + b)(a - b)

• (a + b)² = (a - b)² + 4ab

• (a - b)² = (a + b)² - 4ab

• (a + b)² + (a - b)² = 2(a² + b²)

• (a + b)³ = a³ + b³ + 3ab(a + b)

• (a - b)³ = a³ - b³ - 3ab(a - b)