Question

factorise p(x) = x ^ 4 + x ^ 3 - 7x ^ 2 - x + 6 by factor theorem.
plea se do fa st ple ase
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Answer 1

Answer:

hope it will help you!!!!!!

Answer 2

Answer:

Answer:

here we have x4 + x 3 - 7x2 - x + 6  

here we have constant term 6 and coefficient 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 )

hope it helps u ______✽