Question

Using factor theorem, factorize the following polynomials:
x4 + x3- 7x2- x + 6.

Answer 1

here we have x4 + x 3 - 7x2 - x + 6  

here we have constant term 6 and cofiicient of x4 is 1 , so 

we can find zeros of this equation for  ± 1  , ± 2 , ± 3 , ± 6

so  we put x  =  1 and check if that satisfied our equation , 

= ( 1 )4 + ( 1 ) 3  - 7 ( 1 )2  - ( 1 ) + 6 

= 1 + 1  - 7 - 1 + 6 

= 0 

So ( x  - 1  )  is a factor of our equation  

now we divide our equation by ( x - 1 )  , and get 
x4 + x3- 7x2- x + 6/x-1=x3 + 2x2  - 5x  - 6 
S0 we get 

x3 + 2x2  - 5x  - 6 
Here we have constant term 6 and cofiicient of x3 is 1 , so 

we can find zeros of this equation for  ± 1  , ± 2 , ± 3 , ± 6


so  we put x =  - 1 and check if that satisfied our equation , 

= ( - 1 )4 + ( - 1 ) 3 - 7 ( - 1 )2 - ( - 1 ) + 6 

= 1 - 1  - 7 + 1 + 6 

= 0 
So ( x + 1  )  is a factor of our equation  

now we divide our equation by ( x + 1 )  , and get
So, we get 

x2 + x - 6 

Now we write it As  : 

x2 + 3x - 2x  - 6 

x ( x + 3 ) - 2 ( x + 3 ) 

( x - 2 ) ( x + 3 ) 

So we have factors  

 x4 + x 3 - 7x2 - x + 6    = ( x  - 1 ) ( x + 1 ) ( x - 2 ) ( x + 3 ) 

Answer 2

Answer:
The factor of the equation $x^4 + x^3- 7x^2- x + 6=0$ are
$(x-1)(x+1)(x-2)(x+3)=0$

Step-by-step explanation:

• The polynomial given to us is
  $x^4 + x^3- 7x^2- x + 6=0$
• By using factor theorm we have to use hit and trial method so let us assume $x - 1 = 0$  be one of its solutions so by putting x=1 in the equation we get
  $1+1-7-1+6 =0\\8-8=0$
• Hence x=1 satisfies the equation so $x - 1 = 0$ is one of the solution
• Now by deviding the equation $x^4 + x^3- 7x^2- x + 6=0$ by $x - 1 = 0$ we can get the other factor of it.
• So on division we get the quotient as
  $x^3 + 2x^2 - 5x - 6 =0$
• Now on factorising the above equation we get its factors as
  $x+1 = 0, x-2=0 , x+3=0$
• Hence the factor of the equation $x^4 + x^3- 7x^2- x + 6=0$ are
  $(x-1)(x+1)(x-2)(x+3)=0$
  #SPJ2