Question

please answer all this questions​

Answer 1

1.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)

2.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

p(x)=(x+1)(x+1)(x-5)

3.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)

4.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)

Answer 2

Answer:hi

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)

2.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

p(x)=(x+1)(x+1)(x-5)

3.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)

4.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)

Step-by-step explanation:

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