Questoin 5. Factorise (i) X3 - 2x2 - X + 2 ( Exercise 2.4 Chapter 2 Polynomial Maths Class 9 NCERT

Questoin 5. Factorise :
(i) x³ - 2x² - x + 2
(ii) x³ - 3x² - 9x - 5
(iii) x³ + 13x² + 32x + 20
(iv) 2y³ + y² - 2y - 1

Solution: (i) x³ - 2x² - x + 2   
Solution  (i) Let take f(x) = x3 - 2x2 - x + 2
The constant term in f(x) is are ±1 and  ±2
Putting x = 1 in f(x), we have
f(1) = (1)³ - 2(1)² -1 + 2
= 1 - 2 - 1 + 2 = 0
According to remainder theorem f(1) = 0 so that  (x - 1) is a factor of x3 - 2x2 - x + 2
Putting x = - 1 in f(x), we have
f(-1) = (-1)³ - 2(-1)² –(-1) + 2
= -1 - 2 + 1 + 2 = 0
According to remainder theorem f(-1) = 0 so that  (x + 1) is a factor of x3 - 2x2 - x + 2
Putting x =  2 in f(x), we have
f(2) = (2)³ - 2(2)² –(2) + 2
= 8 -82  - 2 + 2 = 0
According to remainder theorem f(2) = 0 so that  (x – 2 ) is a factor of x3 - 2x2 - x + 2

Here maximum power of x is 3 so that its can have maximum 3 factors
So our answer is (x-1)(x+1)(x-2)
(ii) x³ - 3x² - 9x - 5
Possible zeros are factors of ± constant term / coefficient of leading term
Here constant term is -5 and coefficient of leading term is 1
So that possible zeros will ±1 and  ±5
Take f(x) = x³ - 3x² - 9x - 5
Plug x  = 1 we get
f(1) = (1)³ – 3(1)² – 9(1) – 5
f(1) = 1 – 3  -  9  – 5
F(1) = -16 ≠ 0
So that (x-1)  is not a factor of x³ - 3x² - 9x - 5
Plug x = -1
f(-1) = (-1)³ – 3(-1)² – 9(-1) – 5
f(1) = -1 – 3  +  9  – 5
F(1) = 0  so it a zero
x-1 = 0
x+1 = 0
So that (x+1) is  a factor of x³ - 3x² - 9x - 5
Divide the expression by x+1 we get

(iii) x³ + 13x² + 32x + 20
Here coefficient of leading term is 1 and constant term is 20
So possible zeros are factors of ± 20/1
So possible zeros are  ±1 ,±2,±4,±5,±10, and ±20
And here all terms are positive so that  zeros cannot positive
Plug x = -1
=>x³ + 13x² + 32x + 20
=> (-1)³ + 13(-1)² + 32(-1) + 20
=> -1+ 13  - 32 + 20
=> 0
So that (x+1) is a factor x³ + 13x² + 32x + 20

Plug x = - 2

=>x³ + 13x² + 32x + 20
=> (-2)³ + 13(-2)² + 32(-2) + 20
=> -8+ 52 - 64 + 20
=> 0
So that (x+2) is a factor x³ + 13x² + 32x + 20

As we have already find two zeros third zeros can  20 / 1*2 = 10
Plug x = 10 we get
Plug x = - 2

=>x³ + 13x² + 32x + 20
=> (-10)³ + 13(-10)² + 32(-10) + 20
=> -1000+ 1300 - 320 + 20
=> 0
So that (x+10) is a factor x³ + 13x² + 32x + 20

As leading term has 3 powers so that there are only 3 roots are possible
Answer (x+1)(x+2)(x+10)

(iv) 2y³ + y² - 2y – 1
Here constant term is -1
Coefficient of leading term is 2
So possible zeros are ±1 ,±1/2
Plug y = 1
=>2y³ + y² - 2y – 1
=>2(1)³ + (1)² – 2(1) – 1
=> 2+ 1 -2-1
=>0
Here   y= -1
Y+1 =0 so that  
(y+1) is factor of 2y³ + y² - 2y – 1

Plug y  = -1
=>2y³ + y² - 2y – 1
=>2(-1)³ + (-1)² – 2(-1) – 1
=> - 2+ 1 + 2-1
=>0
Here   y= 1
y - 1 =0 so that 
(y - 1) is factor of 2y³ + y² - 2y – 1

Plug  y = ½
=>2y³ + y² - 2y – 1
=>2(½)³ + (½)² – 2(½) – 1
=> 2/8+ 1/4 -1-1
=>-3/2 ≠ 0 
(y – ½) is factor of 2y³ + y² - 2y – 1

Plug  y = -½
=>2y³ + y² - 2y – 1
=>2(-½)³ + (-½)² – 2(-½) – 1
=> -2/8+ ¼ + 1-1
=>0
(y + ½) is factor of 2y³ + y² - 2y – 1
Here y has max powers 3 so there are 3 possible factors
And our answer is (y-1)(y+1)(y+ -½)