# Factorise the polynomial x4 + x3 - 7x2 - x + 6 using the factor theorem.

By Ritesh|Updated : November 8th, 2022

By factorising the polynomial x4 + x3 - 7x2 - x + 6 using factor theorem we get (x - 1) (x + 1) (x - 2) (x + 3). It is given that p(x) = x4 + x3 - 7x2 - x + 6. In the above equation, the constant term is 6 and the coefficient of x4 is 1 Factor of constant term 6 are ±1, ±2, ±3, ±6.

Case 1: Substitute x = 1 and check if it satisfies our equation x4 + x3 - 7x2 - x + 6

p(1) = 14 + 13 - 7(1)2 - 1 + 6

= 1 + 1 - 7 - 1 + 6

On simplifying we get

= 0

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

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

Case 2: Substitute x = -1 and check if it satisfies our equation, x4 + x3 - 7x2 - x + 6

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

= 1 - 1 - 7 + 1 + 6

On simplifying we get

= 0

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

Case 3: Substitute x = 2 and check if it satisfies our equation, x4 + x3 - 7x2 - x + 6

p(2) = 24 + 23 - 7(2)2 - 2 + 6

= 16 + 8 - 28 - 2 + 6

On simplifying we get

= 0

(x - 2) is a factor of the equation p(x).

Case 4: Substitute x = -2 and check if it satisfies our equation, x4 + x3 - 7x2 - x + 6

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

= 16 - 8 - 28 + 2 + 6

On simplifying we get

= -12

≠ 0

(x + 2) is a factor of the equation p(x).

Case 5: Substitute x = -3 and check if it satisfies our equation, x4 + x3 - 7x2 - x + 6

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

= 81 - 27 - 63 + 3 + 6

On simplifying we get

= 0

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

So we have, the factors of x4 + x3 - 7x2 - x + 6 are (x - 1) (x + 1) (x - 2) (x + 3)

Summary:

## Factorise the polynomial x4 + x3 - 7x2 - x + 6 using the factor theorem.

By factorising the polynomial x4 + x3 - 7x2 - x + 6 using factor theorem we get (x - 1) (x + 1) (x - 2) (x + 3). In simple terms, the reverse process of expansion of an algebraic expression is known as its factorization.