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Factorise the polynomial x4 + x3 – 7×2 – x + 6 using the factor theorem.
By BYJU'S Exam Prep
Updated on: September 25th, 2023
By factorising the polynomial x4 + x3 – 7×2 – x + 6 using factor theorem we get (x – 1) (x + 1) (x – 2) (x + 3). It is given that p(x) = x4 + x3 – 7×2 – 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 – 7×2 – 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 – 7×2 – 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 – 7×2 – 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 – 7×2 – 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 – 7×2 – 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 – 7×2 – x + 6 are (x – 1) (x + 1) (x – 2) (x + 3)
Summary:
Factorise the polynomial x4 + x3 – 7×2 – x + 6 using the factor theorem.
By factorising the polynomial x4 + x3 – 7×2 – 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.
