Factor and Calculate the Roots of the Following Polynomials

Exercise 1

{ x }^{ 3 }+{ x }^{ 2 }

Exercise 2

2{ x }^{ 4 }+4{ x }^{ 2 }

Exercise 3

{ x }^{ 2 }-4

Exercise 4

{ x }^{ 4 }-16

Exercise 5

9+6x+{ x }^{ 2 }

Exercise 6

{ x }^{ 2 }-x-6=0

Exercise 7

{ x }^{ 4 }-10{ x }^{ 2 }+9=0

Exercise 8

{ x }^{ 4 }-2{ x }^{ 2 }-3=0

Exercise 9

{ 2x }^{ 4 }+{ x }^{ 3 }-8{ x }^{ 2 }-x+6

Exercise 10

{ 2x }^{ 3 }-7{ x }^{ 2 }+8{ x }-3

Exercise 11

{ x }^{ 3 }-{ x }^{ 2 }-4

Exercise 12

{ x }^{ 3 }+{ 3x }^{ 2 }-4x-12

Exercise 13

{ 6x }^{ 3 }+{ 7x }^{ 2 }-9x+2

Exercise 14

Factorize the following.

  1. 9{ x }^{ 4 }-4{ x }^{ 2 }=
  2. { x }^{ 2 }(3x+2)(3x-2)
  3. { x }^{ 5 }+20{ x }^{ 3 }+100x=
  4. 3{ x }^{ 5 }-18{ x }^{ 3 }+27x=
  5. 2{ x }^{ 3 }-50x=
  6. 2{ x }^{ 5 }-32x=
  7. 2{ x }^{ 2 }+x-28=

Exercise 15

Factorize the following.

  1. \frac { 2 }{ 5 } { x }^{ 5 }-\frac { 6 }{ 5 } { x }^{ 4 }-\frac { 14 }{ 15 } { x }^{ 2 }=
  2. 2xy-4x-3y+6=
  3. 25{ x }^{ 2 }-1=
  4. 36{ x }^{ 6 }-49=
  5. { 5x }^{ 2 }-10x+5=
  6. { x }^{ 2 }-6x+9=
  7. { x }^{ 2 }-20x+100=
  8. { x }^{ 2 }+10x+25=
  9. { x }^{ 2 }+14x+49=
  10. { x }^{ 3 }-4{ x }^{ 2 }+4x=
  11. 3{ x }^{ 7 }-27x=
  12. { x }^{ 2 }-11x+30=
  13. 3{ x }^{ 2 }-10x+3
  14. 2{ x }^{ 2 }-x-1

 

Solution of exercise 1

{ x }^{ 3 }+{ x }^{ 2 }

{ x }^{ 3 }+{ x }^{ 2 }={ x }^{ 2 }(x+1)

 

Solution of exercise 2

2{ x }^{ 4 }+4{ x }^{ 2 }

2{ x }^{ 4 }+4{ x }^{ 2 }=2{ x }^{ 2 }({ x }^{ 2 }+2)

 

Solution of exercise 3

{ x }^{ 2 }-4

{ x }^{ 2 }-{ 2 }^{ 2 }=(x-2).(x+2)

 

Solution of exercise 4

{ x }^{ 4 }-16

{ ({ x }^{ 2 }) }^{ 2 }-{ 4 }^{ 2 }=({ x }^{ 2 }-4).({ x }^{ 2 }+4)

 

Solution of exercise 5

9+6x+{ x }^{ 2 }

 

{ x }^{ 2 }+6x+9

a{ x }^{ 2 }+bx+c

a=1,\quad b=6,\quad c=9

In order to break the middle term, you need to multiply "a" with "c":

1\times 9=9

Now break 9 into least possible factors:

3\times 3=9

The next step is to check the sign of the c and in this case, it is positive this means that you need to add the factors to come up with 6:

3+3=6

Now your middle term is broken, you can carry forward:

{ x }^{ 2 }+x(3+3)+9

{ x }^{ 2 }+3x+3x+9

x(x+3)+3(x+3)

(x+3)(x+3)

{(x+3) }^{ 2 }

Solution of exercise 6

{ x }^{ 2 }-x-6=0

a{ x }^{ 2 }+bx+c=0

a=1,\quad b=-1,\quad c=-6

x=\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }

x=\quad \frac { -(-1)+\sqrt { { (-1 })^{ 2 }-4(1)(-6) } }{ 2(1) } ,\frac { -(-1)-\sqrt { { { (-1 }) }^{ 2 }-4(1)(-6) } }{ 2(1) }

x=\quad \frac { 1+\sqrt { { 1 }+24 } }{ 2 } ,\frac { 1-\sqrt { { 1 }+24 } }{ 2 }

x=\quad \frac { 1+\sqrt { 25 } }{ 2 } ,\frac { 1-\sqrt { 25 } }{ 2 }

x=\quad \frac { 1+5 }{ 2 } ,\frac { 1-5 }{ 2 }

x=\quad \frac { 6 }{ 2 } ,\frac { -4 }{ 2 }

x=3,\quad x=-2

{ x }^{ 2 }-x-6=(x-3).(x+2)

Solution of exercise 7

{ x }^{ 4 }-10{ x }^{ 2 }+9=0

Assuming { x }^{ 2 }=t:

{ t }^{ 2 }-10t+9=0

a{ t }^{ 2 }+bt+c=0

a=1,\quad b=-10,\quad c=9

t=\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }

t=\quad \frac { -(-10)+\sqrt { { (-10) }^{ 2 }-4(1)(9) } }{ 2(1) } ,\frac { -(-10)-\sqrt { { (-10) }^{ 2 }-4(1)(9) } }{ 2(1) }

t=\quad \frac { 10+\sqrt { { 100 }-36 } }{ 2 } ,\frac { 10-\sqrt { { 100 }-36 } }{ 2 }

t=\quad \frac { 10+\sqrt { 64 } }{ 2 } ,\frac { 1-\sqrt { 64 } }{ 2 }

t=\quad \frac { 10+8 }{ 2 } ,\frac { 10-8 }{ 2 }

t=\quad \frac { 18 }{ 2 } ,\frac { 2 }{ 2 }

t=9,\quad t=1

Putting the value of t:

{ x }^{ 2 }=9,\quad { x }^{ 2 }=1

x=\pm 3,\quad x=\pm 1

{ x }^{ 4 }-10{ x }^{ 2 }+9=(x+1).(x-1).(x+3).(x-3)

Solution of exercise 8

{ x }^{ 4 }-2{ x }^{ 2 }-3=0

Assuming { x }^{ 2 }=t:

{ t }^{ 2 }-2t-3=0

a{ t }^{ 2 }+bt+c=0

a=1,\quad b=-2,\quad c=-3

t=\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }

t=\quad \frac { -(-2)+\sqrt { { (-2) }^{ 2 }-4(1)(-3) } }{ 2(1) } ,\frac { -(-2)-\sqrt { { (-2) }^{ 2 }-4(1)(-3) } }{ 2(1) }

t=\quad \frac { 2+\sqrt { 4+12 } }{ 2 } ,\frac { 2-\sqrt { 4+12 } }{ 2 }

t=\quad \frac { 2+\sqrt { 16 } }{ 2 } ,\frac { 2-\sqrt { 16 } }{ 2 }

t=\quad \frac { 2+4 }{ 2 } ,\frac { 2-4 }{ 2 }

t=\quad \frac { 6 }{ 2 } ,\frac { -2 }{ 2 }

t=3,\quad t=-1

Putting the value of t:

{ x }^{ 2 }=3,\quad { x }^{ 2 }=-1

x=\pm \sqrt { 3 } ,\quad x=\pm \sqrt { -1 } \notin R

 

Solution of exercise 9

2{ x }^{ 4 }+{ x }^{ 3 }-8{ x }^{ 2 }-x+6

P(1)=2{ (1) }^{ 4 }+{ (1) }^{ 3 }-8{ (1) }^{ 2 }-1+6=0

 

 

(x-1).(2{ x }^{ 3 }+3{ x }^{ 2 }-5x-6)

P(1)=2{ (1) }^{ 3 }+{ 3(1) }^{ 2 }-5{ (1) }-6\neq 0

P(-1)=2{ (-1) }^{ 3 }+{ 3(-1) }^{ 2 }-5{ (-1) }-6=-2+3+5-6=0

 

 

(x-1)(x+1)(2{ x }^{ 2 }+x-6)

P(2)=2{ (2) }^{ 2 }+(2)-6\neq 0

P(-2)=2{ (-2) }^{ 2 }+(-2)-6=0

 

 

(x-1).(x+1).(x+2).(2x-3)

2x-3=2(x-\frac { 3 }{ 2 } )

2{ x }^{ 4 }+{ x }^{ 3 }-8{ x }^{ 2 }-x+6=2(x-1).(x+1).(x+2).(x-\frac { 3 }{ 2 } )

x=1,\quad x=-1,\quad x=-2,\quad x=\frac { 3 }{ 2 }

Solution of exercise 10

2{ x }^{ 3 }-7{ x }^{ 2 }+8x-3

P(1)=2{ (1) }^{ 3 }-7{ (1) }^{ 2 }+8(1)-3=2-7+8-3=0

(x-1).(2{ x }^{ 2 }-5x+3)

P(1)=2{ (1) }^{ 2 }-5(1)+3=2-5+3=0

(x-1).(x-1).(2x-3)={ (x-1) }^{ 2 }.2(x-\frac { 3 }{ 2 } )=2{ (x-1) }^{ 2 }(x-\frac { 3 }{ 2 } )

x=\frac { 3 }{ 2 } ,\quad x=1

Solution of exercise 11

{ x }^{ 3 }-{ x }^{ 2 }-4

\{ \quad \pm 1,\quad \pm 2,\quad \pm 4\quad \}

P(1)={ (1) }^{ 3 }-{ (1) }^{ 2 }-4=1-1-4\neq 0

P(-1)={ (-1) }^{ 3 }-{ (-1) }^{ 2 }-4=-1-1-4\neq 0

P(2)={ (2) }^{ 3 }-{ (2) }^{ 2 }-4=8-4-4=0

(x-2).({ x }^{ 2 }+x+2)

{ x }^{ 2 }+x+2=0

a{ x }^{ 2 }+bx+c=0

a=1,\quad b=1,\quad c=2

x=\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }

x=\quad \frac { -(1)+\sqrt { { (1 })^{ 2 }-4(1)(2) } }{ 2(1) } ,\frac { -(1)-\sqrt { { { (1 }) }^{ 2 }-4(1)(2) } }{ 2(1) }

x=\quad \frac { 1+\sqrt { { 1 }-8 } }{ 2 } ,\frac { 1-\sqrt { { 1 }-8 } }{ 2 }

x=\quad \frac { 1+\sqrt { -7 } }{ 2 } \notin R,\frac { 1-\sqrt { -7 } }{ 2 } \notin R

(x-2).({ x }^{ 2 }+x+2)

x = 2

 

Solution of exercise 12

{ x }^{ 3 }+3{ x }^{ 2 }-4x-12

\{ \pm 1,\quad \pm 2,\quad \pm 3,\quad \pm 4,\quad \pm 6\quad ,\pm 12\}

P(1)={ (1) }^{ 3 }+3{ (1) }^{ 2 }-4(1)-12=1+3-4-12\neq 0

P(-1)={ (-1) }^{ 3 }+{ 3(-1) }^{ 2 }-4(-1)-12=-1+3+4-12\neq 0

P(2)={ (2) }^{ 3 }+3{ (2) }^{ 2 }-4(2)-12=8+12-8-12=0

(x-2).({ x }^{ 2 }+5x+6)

{ x }^{ 2 }+5x+6=0

a{ x }^{ 2 }+bx+c=0

a=1,\quad b=5,\quad c=6

x=\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }

x=\quad \frac { -(5)+\sqrt { { (5 })^{ 2 }-4(1)(6) } }{ 2(1) } ,\frac { -(5)-\sqrt { { { (5 }) }^{ 2 }-4(1)(6) } }{ 2(1) }

x=\quad \frac { -5+\sqrt { { 25 }-24 } }{ 2 } ,\frac { -5-\sqrt { { 25 }-24 } }{ 2 }

x=\quad \frac { -5+\sqrt { 1 } }{ 2 } ,\frac { -5-\sqrt { 1 } }{ 2 }

x=\quad \frac { -5+1 }{ 2 } ,\frac { -5-1 }{ 2 }

x=\quad \frac { -4 }{ 2 } ,\frac { -6 }{ 2 }

x=-2,\quad x=-3

(x-2).(x+2).(x+3)

x=2, \quad x=-2, \quad x=-3

 

Solution of exercise 13

{ 6x }^{ 3 }+7{ x }^{ 2 }-9x+2

\{ \pm 1,\quad \pm 2,\}

P(1)={ 6(1) }^{ 3 }+7{ (1) }^{ 2 }-9(1)+2=6+7-9+2\neq 0

P(-1)={ 6(-1) }^{ 3 }+7{ (-1) }^{ 2 }-9(-1)+2=-6+7+9+2\neq 0

P(2)={ 6(2) }^{ 3 }+7{ (2) }^{ 2 }-9(2)+2=48+28-18+2\neq 0

P(-2)={ 6(-2) }^{ 3 }+7{ (-2) }^{ 2 }-9(-2)+2=-48+28-18+2=0

(x+2).({ 6x }^{ 2 }-5x+1)

{ 6x }^{ 2 }-5x+1=0

a{ x }^{ 2 }+bx+c=0

a=6,\quad b=-5,\quad c=1

x=\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }

x=\quad \frac { -(-5)+\sqrt { { (5 })^{ 2 }-4(6)(1) } }{ 2(6) } ,\frac { -(-5)-\sqrt { { { (5 }) }^{ 2 }-4(6)(1) } }{ 2(6) }

x=\quad \frac { 5+\sqrt { { 25 }-24 } }{ 12 } ,\frac { 5-\sqrt { { 25 }-24 } }{ 12 }

x=\quad \frac { 5+\sqrt { 1 } }{ 12 } ,\frac { 5-\sqrt { 1 } }{ 12 }

x=\quad \frac { 5+1 }{ 12 } ,\frac { 5-1 }{ 12 }

x=\quad \frac { 6 }{ 12 } ,\frac { 4 }{ 12 }

x=\frac { 1 }{ 2 } \quad x=\frac { 1 }{ 3 }

6(x+2).(x-\frac { 1 }{ 2 } ).(x-\frac { 1 }{ 3 } )

x=-2,\quad x=\frac { 1 }{ 2 } ,\quad x=\frac { 1 }{ 3 }

 

Solution of exercise 14

9{ x }^{ 4 }-4{ x }^{ 2 }

{ x }^{ 2 }(9{ x }^{ 2 }-4)

 

{ x }^{ 2 }(3x+2)(3x-2)

{ x }^{ 2 }[{ (3x) }^{ 2 }-{ (2) }^{ 2 }]

{ x }^{ 2 }(9{ x }^{ 2 }-4)

9{ x }^{ 4 }-4{ x }^{ 2 }

 

x^{ 5 }+20{ x }^{ 3 }+100x

x(x^{ 4 }+20x^{ 2 }+100)

x{ (x^{ 2 }+10) }^{ 2 }

 

3x^{ 5 }-18{ x }^{ 3 }+27x

3x(x^{ 4 }-6x^{ 2 }+9)

3x{ ({ x }^{ 2 }-3) }^{ 2 }

 

2{ x }^{ 3 }-50x

2x({ x }^{ 2 }-25)

2x({ x }^{ 2 }-{ 5 }^{ 2 })

2x[(x-5)(x+5)]

 

2{ x }^{ 5 }-32x

2x({ x }^{ 4 }-16)

2x[{ ({ x }^{ 2 }) }^{ 2 }-{ 4 }^{ 2 }]

2x[({ x }^{ 2 }-4)({ x }^{ 2 }+4)]

2x[({ x }^{ 2 }-{ 2 }^{ 2 })({ x }^{ 2 }+4)]

2x[({ x }-{ 2 })({ x }+{ 2 })({ x }^{ 2 }+4)]

 

2{ x }^{ 2 }+x-28=0

a{ x }^{ 2 }+bx+c=0

a=2,\quad b=1,\quad c=-28

x=\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }

x=\quad \frac { -(1)+\sqrt { { (1 })^{ 2 }-4(2)(-28) } }{ 2(2) } ,\frac { -(1)-\sqrt { { { (1 }) }^{ 2 }-4(2)(-28) } }{ 2(2) }

x=\quad \frac { -1+\sqrt { { 1 }+224 } }{ 4 } ,\frac { -1-\sqrt { { 1 }+224 } }{ 4 }

x=\quad \frac { -1+\sqrt { 225 } }{ 4 } ,\frac { -1-\sqrt { 225 } }{ 4 }

x=\quad \frac { -1+15 }{ 4 } ,\frac { -1-15 }{ 4 }

x=\quad \frac { 14 }{ 4 } ,\frac { -16 }{ 4 }

x=\frac { 7 }{ 2 } \quad x=-4

2{ x }^{ 2 }+x-28=2(x+4)(x-\frac { 7 }{ 2 } )

 

Solution of exercise 15

\frac { 2 }{ 5 } { x }^{ 5 }-\frac { 6 }{ 5 } { x }^{ 4 }+\frac { 14 }{ 15 } { x }^{ 2 }

\frac { 2 }{ 5 } { x }^{ 2 }({ x }^{ 3 }-3{ x }^{ 2 }+\frac { 7 }{ 3 } )

 

2xy-4x-3y+6

2x(y-2)-3(y-2)

(2x-3)(y-2)

 

25{ x }^{ 2 }-1

{ (5x) }^{ 2 }-{ 1 }^{ 2 }

(5x-1)(5x+1)

 

36{ x }^{ 2 }-49

{ (6x) }^{ 2 }-7^{ 2 }

(6x-7)(6x+7)

 

5{ x }^{ 2 }-10x+5

5({ x }^{ 2 }-2x+1)

5[{ (x) }^{ 2 }-2(x)(1)+{ 1 }^{ 2 }]

5{ (x-1) }^{ 2 }

 

{ x }^{ 2 }-6x+9

{ x }^{ 2 }-2(x)(3)+{ 3 }^{ 2 }

{ (x-3) }^{ 2 }

 

{ x }^{ 2 }-20x+100

{ x }^{ 2 }-2(x)(10)+{ 10 }^{ 2 }

{ (x-10) }^{ 2 }

{ x }^{ 2 }+10x+25{ x }^{ 2 }+2(x)(5)+{ 5 }^{ 2 }

{ (x+5) }^{ 2 }

 

{ x }^{ 2 }+14x+49

{ x }^{ 2 }+2(x)(7)+{ 7 }^{ 2 }

{ (x+7) }^{ 2 }

 

{ x }^{ 3 }-4{ x }^{ 2 }+4x

x({ x }^{ 2 }-4x+4)

x({ x }^{ 2 }-2(x)(2)+{ 2 }^{ 2 })

x{ (x-2) }^{ 2 }

 

3{ x }^{ 7 }-27x

3x({ x }^{ 6 }-9)

3x[{ ({ x }^{ 3 }) }^{ 2 }-{ 3 }^{ 2 }]

3x{ ({ x }^{ 3 }-3) }({ x }^{ 3 }+3)

 

{ x }^{ 2 }-11x+30=0

a{ x }^{ 2 }+bx+c=0

a=1,\quad b=-11,\quad c=30

x=\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }

x=\quad \frac { -(-11)+\sqrt { { (-11 })^{ 2 }-4(1)(30) } }{ 2(1) } ,\frac { -(-11)-\sqrt { { { (-11 }) }^{ 2 }-4(2)(30) } }{ 2(1) }

x=\quad \frac { 11+\sqrt { { 121 }-120 } }{ 2 } ,\frac { 11-\sqrt { { 121 }-120 } }{ 2 }

x=\quad \frac { 11+\sqrt { 1 } }{ 2 } ,\frac { 11-\sqrt { 1 } }{ 2 }

x=\quad \frac { 11+1 }{ 2 } ,\frac { 11-1 }{ 2 }

x=\quad \frac { 12 }{ 2 } ,\frac { 10 }{ 2 }

x=6\quad x=5

{ x }^{ 2 }-11x+30=(x-5)(x-6)

 

3{ x }^{ 2 }-10x+3=0

a{ x }^{ 2 }+bx+c=0

a=3,\quad b=-10,\quad c=3

x=\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }

x=\quad \frac { -(-10)+\sqrt { { (-10 })^{ 2 }-4(3)(3) } }{ 2(3) } ,\frac { -(-10)-\sqrt { { { (-10 }) }^{ 2 }-4(3)(3) } }{ 2(3) }

x=\quad \frac { 10+\sqrt { { 100 }-36 } }{ 6 } ,\frac { 10-\sqrt { { 100 }-36 } }{ 6 }

x=\quad \frac { 10+\sqrt { 64 } }{ 6 } ,\frac { 10-\sqrt { 64 } }{ 6 }

x=\quad \frac { 10+8 }{ 6 } ,\frac { 10-8 }{ 6 }

x=\quad \frac { 18 }{ 6 } ,\frac { 2 }{ 6 }

x=3\quad x=\frac { 1 }{ 3 }

3{ x }^{ 2 }-10x+3=(x-3)(x-\frac { 1 }{ 3 } )

 

2{ x }^{ 2 }-x-1=0

a{ x }^{ 2 }+bx+c=0

a=2,\quad b=-1,\quad c=-1

x=\quad \frac { -(-1)+\sqrt { { (-1 })^{ 2 }-4(2)(-1) } }{ 2(2) } ,\frac { -(-1)-\sqrt { { { (-1 }) }^{ 2 }-4(2)(-1) } }{ 2(2) }

x=\quad \frac { 1+\sqrt { { 1 }+8 } }{ 4 } ,\frac { 1-\sqrt { { 1 }+8 } }{ 4 }

x=\quad \frac { 1+\sqrt { 9 } }{ 4 } ,\frac { 1-\sqrt { 9 } }{ 4 }

x=\quad \frac { 1+3 }{ 4 } ,\frac { 1-3 }{ 4 }

x=\quad \frac { 4 }{ 4 } ,\frac { -2 }{ 4 }

x=1\quad x=-\frac { 1 }{ 2 }

2{ x }^{ 2 }-x-1=(x-1)(x+\frac { 1 }{ 2 } )

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Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.

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