x3 – 2x2 – x + 2
please solve this
Answers
Answer:
Step-by-step explanation:
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)