Question

Using factor theorem Factorise polynomial x^4-x^3-7x^2+x+6

Answer 1

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 )

Answer 2

Here we have

p(x) = x⁴+x³−7x²−x+6

Here we have constant term

6 and coefficient of x⁴ is 1 ,

So, the factors of constant term i.e.

6 are ±1,±2,±3,±6

Case 1: Put x=1and check if that satisfied our equation

$x⁴+x³−7x²−x+6$

$p(1)=(1)⁴+(1)³−7(1)²−(1)+6$

$= 1 + 1 - 7 - 1 + 6$

=0

So

(x−1) is a factor of the given equation p(x).

[Factor theorem, if f(a)=0, then (x−a) is a factor of polynomial f(x) ]

Case 2: Put x=−1

and check if that satisfied our equation ,

$x⁴+x³−7x²−x+6$

$p(1)=(-1)⁴+(-1)³−7(-1)²−(1)+6$

= 1-1-7+1+6

= 0

So (x+1) is a factor of the given equation p(x).

Case 3: Put x=2 and check if that satisfied our equation ,

$x⁴+x³−7x²−x+6$

Find solutions

Case 3: Put x=2 and check if that satisfied our equation ,

$x⁴+x³−7x²−x+6$

$p(2)=(2)⁴+(2)³−7(2)²−(2)+6$

$=16+8−28−2+6=0$

So (x−2) is a factor of the given equation p(x).

Case 4: Put x=−2 and check if that satisfied our equation ,

$x⁴+x³−7x²−x+6$

$p(−2)=(−2)4+(−2)3−7(−2)2−(−2)+6$

$= 16−8−28+2+6$

=−12≠0

So (x+2) is not a factor of the given equation p(x).

Case 5: Put x=−3 and check if that satisfied our equation ,

$x⁴+x³−7x²−x+6$

$p(−2)=(−3)⁴+(−3)³−7(−3)²−(−3)+6$

$=81−27−63+3+6$

=0

So (x+3) is a factor of the given equation p(x).

So we have, the factors of

x4+x3−7x2−x+6 are (x−1)(x+1)(x−2)(x+3)

#SP J2