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Question: Factorise: (i) \[{x^3} - 2{x^2} - x + 2\] (ii) \[{x^3} - 3{x^2} - 9x - 5\] (iii) \[{x^3} + 13{...

Factorise:
(i) x3−2x2−x+2{x^3} - 2{x^2} - x + 2
(ii) x3−3x2−9x−5{x^3} - 3{x^2} - 9x - 5
(iii) x3+13x2+32x+20{x^3} + 13{x^2} + 32x + 20
(iv) 2y3+y2−2y−12{y^3} + {y^2} - 2y - 1

Explanation

Solution

According to the given question, which is in the form of ax3+bx2+cx+da{x^3} + b{x^2} + cx + d in which we have to separate into two or three parts to take a common and we can use the splitting the middle terms method to solve these parts.

Complete step by step solution:
(i) x3−2x2−x+2{x^3} - 2{x^2} - x + 2
Let’s start by separating the equation into two parts that is x3−2x2{x^3} - 2{x^2}, and −x+2 - x + 2 ⇒(x3−2x2)+(−x+2) \Rightarrow \left( {{x^3} - 2{x^2}} \right) + \left( { - x + 2} \right)
Now, taking common the highest value possible in both of the parts that is x2{x^2} and −1 - 1 .
⇒x2(x−2)−1(x−2)\Rightarrow {x^2}\left( {x - 2} \right) - 1\left( {x - 2} \right)
Here, we will take (x−2)\left( {x - 2} \right) common from the above equation to make pairs and factorise the equation
⇒(x2−1)(x−2)\Rightarrow \left( {{x^2} - 1} \right)\left( {x - 2} \right)
As we know 1 is equal to 12{1^2} . So, we will rewrite 1 as 12{1^2} .
⇒(x2−12)(x−2)\Rightarrow \left( {{x^2} - {1^2}} \right)\left( {x - 2} \right)
For simplifying further we will use formula (a2−b2)=(a−b)(a+b)\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right) .Here, a=xa = x and b=1b = 1 .
So, we get ⇒(x−1)(x+1)(x−2) \Rightarrow \left( {x - 1} \right)\left( {x + 1} \right)\left( {x - 2} \right) .
(ii) x3−3x2−9x−5{x^3} - 3{x^2} - 9x - 5
Let’s start by rewriting −3x2 - 3{x^2} and −9x - 9x as x2−4x2{x^2} - 4{x^2} and −4x−5x - 4x - 5x respectively.
⇒x3+x2−4x2−4x−5x−5\Rightarrow {x^3} + {x^2} - 4{x^2} - 4x - 5x - 5
Here we will separate the equation into three parts that is (x3+x2)\left( {{x^3} + {x^2}} \right) , (\-4x2−4x)\left( { \- 4{x^2} - 4x} \right) , (−5x−5)\left( { - 5x - 5} \right) .
⇒(x3+x2)+(−4x2−4x)+(−5x−5)\Rightarrow \left( {{x^3} + {x^2}} \right) + \left( { - 4{x^2} - 4x} \right) + \left( { - 5x - 5} \right)
Now, taking common the highest value possible in all of the parts that is x2{x^2} , −4x - 4x , −5 - 5
⇒x2(x+1)−4x(x+1)−5(x+1)\Rightarrow {x^2}\left( {x + 1} \right) - 4x\left( {x + 1} \right) - 5\left( {x + 1} \right)
Here, we will take (x+1)\left( {x + 1} \right) common from the above equation to make pairs and factorise the equation
⇒(x+1)(x2−4x−5)\Rightarrow \left( {x + 1} \right)\left( {{x^2} - 4x - 5} \right)
We will rewrite −4x - 4x as −5x+x - 5x + x .
⇒(x+1)(x2−5x+x−5)\Rightarrow \left( {x + 1} \right)\left( {{x^2} - 5x + x - 5} \right)
Now, taking the highest value possible in all the sub parts of the second part of the equation that is xx , 11 .
⇒(x+1)(x(x−5)+1(x−5))\Rightarrow \left( {x + 1} \right)\left( {x\left( {x - 5} \right) + 1\left( {x - 5} \right)} \right)
Taking (x−5)\left( {x - 5} \right) common from the second part of the equation.
So, we get ⇒(x+1)(x+1)(x−5) \Rightarrow \left( {x + 1} \right)\left( {x + 1} \right)\left( {x - 5} \right) .
(iii) x3+13x2+32x+20{x^3} + 13{x^2} + 32x + 20
Let’s start by rewriting 13x213{x^2} and 32x32x as x2+12x2{x^2} + 12{x^2} and 12x+20x12x + 20x respectively.
⇒x3+x2+12x2+12x+20x+20\Rightarrow {x^3} + {x^2} + 12{x^2} + 12x + 20x + 20
Here we will separate the equation into three parts that is (x3+x2)\left( {{x^3} + {x^2}} \right) , (12x2+12x)\left( {12{x^2} + 12x} \right) , (20x+20)\left( {20x + 20} \right) .

\right)$$ Now, taking common the highest value possible in all of the parts that is $${x^2}$$ , $$12x$$ , $$20$$ . $$ \Rightarrow {x^2}\left( {x + 1} \right) + 12x\left( {x + 1} \right) + 20\left( {x + 1} \right)$$ Here, we will take $$\left( {x + 1} \right)$$ common from the above equation to make pairs and factorise the equation $$ \Rightarrow \left( {x + 1} \right)\left( {{x^2} + 12x + 20} \right)$$ We will rewrite $$12x$$ as $$10x + 2x$$ . $$ \Rightarrow \left( {x + 1} \right)\left( {{x^2} + 10x + 2x + 20} \right)$$ Now, taking the highest value possible in all the sub parts of the second part of the equation, that is $$x$$ , $$2$$ . $$ \Rightarrow \left( {x + 1} \right)\left( {x\left( {x + 10} \right) + 2\left( {x + 10} \right)} \right)$$ Taking $$\left( {x + 10} \right)$$ common from the second part of the equation. So, we get $$ \Rightarrow \left( {x + 1} \right)\left( {x + 10} \right)\left( {x + 2} \right)$$ . (iv) $$2{y^3} + {y^2} - 2y - 1$$ Let’s start by separating the equation into two parts that is $$2{y^3} + {y^2}$$, and $$ - 2y - 1$$ . $$ \Rightarrow \left( {2{y^3} + {y^2}} \right)\left( { - 2y - 1} \right)$$ Now, taking common the highest value possible in both of the parts that is $${y^2}$$ and $$ - 1$$ . $$ \Rightarrow {y^2}\left( {2y + 1} \right) - 1\left( {2y + 1} \right)$$ Here, we will take $$\left( {2y + 1} \right)$$ common from the above equation to make pairs and factorise the equation $$ \Rightarrow \left( {{y^2} - 1} \right)\left( {2y + 1} \right)$$ As we know 1 is equal to $${1^2}$$ . So, we will rewrite 1 as $${1^2}$$ . $$ \Rightarrow \left( {{y^2} - {1^2}} \right)\left( {2y + 1} \right)$$ For simplifying further we will use formula $$\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)$$ .Here, $$a = y$$ and $$b = 1$$ . **So, we get $$ \Rightarrow \left( {y - 1} \right)\left( {y + 1} \right)\left( {2y + 1} \right)$$** **Note:** In these types of questions we will divide the equation into parts and solve it further. Sometimes we have to divide it more than one time to factorise it and we can also use the algebraic identities to get the desired results.