Polynomials

Given, p(X) = x² – 7x + 12

Now, we need to write the given polynomial as a product of first degree polynomial and p(x) = 0

⇒ x² – 7x + 12 = 0

⇒ The given equation can be written as follows

⇒ (x – 4)(x – 3) = 0

⇒ Since, we know that x² – 7x + 12 can be written as (x – a)(x – b)

⇒ x² – 7x + 12 = x² – (a + b)x + ab

∴ coefficient on either side of the polynomial is same and a + b = 7 and ab = 12

⇒ we must find two numbers that satisfy a + b and ab we get, 4 and 3 as the numbers

∴ (x – 4)(x – 3) are the product of first degree polynomial

⇒ Substitute x = 4 and x = 3, We get p(x) as 0

⇒ x² – 7x + 12 = 4² – 7(4) + 1 2 = 16 – 28 + 12 = 0

⇒ x² – 7x + 12 = 3² – 7(3) + 12 = 9 – 21 + 12 = 0

Hence, (x – 4)(x – 3) are the first degree factors of the polynomial and 4, 3 are the solutions of the given polynomial x² – 7x + 12

Given, p(X) = x² + 7x + 12

Now, we need to write the given polynomial as a product of first degree polynomial and p(x) = 0

⇒ x² + 7x + 12 = 0

⇒ the given equation can be written as follows

⇒ (x + 4)(x + 3) = 0

⇒ Since, we know that x² + 7x + 12 can be written as (x + a)(x + b)

⇒ x² + 7x + 12 = x² + (a + b)x + ab

∴ coefficient on either side of the polynomial is same and a + b = 7 and ab = 12

⇒ we must find two numbers that satisfy a + b and ab we get, 4 and 3 as the numbers

∴ (x + 4)(x + 3) are the product of first degree polynomial

⇒ P(x) = 0 if (x + 3) is 0 and (x + 4) is 0

∴ x + 4 = 0

⇒ x = – 4

⇒ x² + 7x + 12 = (– 4)² + 7(– 4) + 12 = 16 – 28 + 12 = 0

And x + 3 = 0

⇒ x = – 3

⇒ x² + 7x + 12 = (– 3)² + 7(– 3) + 12 = 9 – 21 + 12 = 0

Hence, (x + 4)(x + 3) are the first degree factors of the polynomial and – 4, – 3 are the solutions of the given polynomial x² + 7x + 12

Given, p(X) = x² – 8x + 12

Now, we need to write the given polynomial as a product of first degree polynomial and p(x) = 0

⇒ x² – 8x + 12 = 0

⇒ The given equation can be written as follows

⇒ (x – 6)(x – 2) = 0

⇒ Since, we know that x² – 8x + 12 can be written as (x – a)(x – b)

⇒ x² – 8x + 12 = x² – (a + b)x + ab

∴ coefficient on either side of the polynomial is same and a + b = 8 and ab = 12

⇒ we must find two numbers that satisfy a + b and ab we get, 6 and 2 as the numbers

∴ (x – 6)(x – 2) are the product of first degree polynomial

⇒ P(x) = 0 if (x – 6) is 0 and (x – 2) is 0

∴ x – 6 = 0

⇒ x = 6

⇒ x² – 8x + 12 = 6² – 8(6) + 12 = 36 – 48 + 12 = 0

And x – 2 = 0

⇒ x = 2

⇒ x² – 8x + 12 = 2² – 8(2) + 12 = 4 – 16 + 12 = 0

Hence, (x – 6)(x – 2) are the first degree factors of the polynomial and 6, 2 are the solutions of the given polynomial x² – 8x + 12

Given, p(X) = x² + 13x + 12

Now, we need to write the given polynomial as a product of first degree polynomial and p(x) = 0

⇒ x² + 13x + 12 = 0

⇒ the given equation can be written as follows

⇒ (x + 12)(x + 1) = 0

⇒ Since, we know that x² + 12x + 13 can be written as (x + a)(x + b)

⇒ x² + 13x + 12 = x² + (a + b)x + ab

∴ coefficient on either side of the polynomial is same and a + b = 7 and ab = 12

⇒ we must find two numbers that satisfy a + b and ab we get, 12 and 1 as the numbers

∴ (x + 12)(x + 1) are the product of first degree polynomial

⇒ P(x) = 0 if (x + 12) is 0 and (x + 1) is 0

∴ x + 1 = 0

⇒ x = – 1

⇒ x² + 13x + 12 = (– 1)² + 13(– 1) + 12 = 1 – 13 + 12 = 0

And x + 12 = 0

⇒ x = – 12

⇒ x² + 13x + 12 = (– 12)² + 13(– 12) + 12 = 144 – 156 + 12 = 0

Hence, (x + 12)(x + 1) are the first degree factors of the polynomial and – 12, – 1 are the solutions of the given polynomial x² + 13x + 12

Given, p(X) = x² – 2x + 1

Now, we need to write the given polynomial as a product of first degree polynomial and p(x) = 0

⇒ x² – 2x + 1 = 0

⇒ The given equation can be written as follows

⇒ (x – 1)(x – 1) = 0

⇒ Since, we know that x² – 2x + 1 can be written as (x – a)(x – b)

⇒ x² – 2x + 1 = x² – (a + b)x + ab

∴ coefficient on either side of the polynomial is same and a + b = 2 and ab = 1

⇒ we must find two numbers that satisfy a + b and ab we get, 1 and 1 as the numbers

∴ (x – 1)(x – 1) are the product of first degree polynomial

⇒ p(x) is 0 if (x – 1) is 0

∴ x – 1 = 0

⇒ x² – 2x + 1 = 1² – 2(1) + 1 = 1 – 2 + 1 = 0

Hence, (x – 1)(x – 1) are the first degree factors of the polynomial and 1 is the solution of the given polynomial x² – 2x + 1

Given, p(x) = x² + x – 1

Now, we need to write the given polynomial as a product of first degree polynomial and p(x) = 0

⇒ The given equation can be written as = x² + x + (– 1) which is of the form x² + (a + b)x + ab

∴ a + b = 1 and ab = – 1

⇒ we know that (a + b)² –(a – b)² = 4ab

∴ (a + b)² – 4ab = (a – b)²

⇒ we know a + b = 1 and ab = – 1

⇒ 1 – 4(– 1) = (a – b)²

⇒ (a – b)² = 5

⇒ a – b =

⇒ Solving the equation both equation a + b and a – b we get as follows

⇒ a + b + a – b = 1 + √5

⇒ a = (1 + √5)

⇒ (a + b) – (a – b) = 1 + √5

⇒ b = (1 – √5)

∴ x² + x + (– 1) has factors (x + (1 + √5))(x + (1 – √5))

⇒ p(x) = 0 if (x + (1 + √5))is 0 and (x + (1 – √5))is 0

⇒ (x + (1 + √5)) = 0

⇒ x = – (1 + √5)

⇒ (– (1 + √5))² + (– (1 + √5)) + (– 1) = 0

And

⇒ (x + (1 – √5)) = 0

⇒ x = – (1 – √5)

⇒ (– (1 – √5))² + (– (1 – √5)) + (– 1) = 0

Hence, (x + (1 + √5))(x + (1 – √5)) are the first degree factors of the polynomial and – (1 + √5) and – (1 – √5) are the solution of the given polynomial x² + x – 1

Given, p(x) = 2x² – 5x + 2

Now, we need to write the given polynomial as a product of first degree polynomial and p(x) = 0

⇒ 2x² – 5x + 2 can be written as 2(x² – x + 1)

⇒ Now, we will find out the factors for x² – x + 1

⇒ x² – x + 1 is in the form x² – (a + b)x + ab and (x – a)(x – b) are the factors

∴ coefficient on either side of the polynomial is same and a + b =

and ab = 1

⇒ we know that (a + b)² –(a – b)² = 4ab

∴ (a + b)² – 4ab = (a – b)²

⇒ – 4(1) = (a – b)²

⇒ – 4 = (a – b)²

⇒ = (a – b)²

⇒ = (a – b)²

⇒ a – b =

⇒ a – b =

⇒ We need to find out the values of a and b by solving (a + b) + (a – b) and (a + b) – (a – b)

⇒ we take a – b =

⇒ (a + b) + (a – b) = +

⇒ 2a =

⇒ a = 2

⇒ (a + b) – (a – b) = –

⇒ 2b =

⇒ b =

∴ we get a = 2 and b =

And if we take a – b = – we get as follows

⇒ (a + b) + (a – b) = + (– )

⇒ 2a =

⇒ a =

⇒ (a + b) – (a – b) = –

⇒ 2b =

⇒ b = 2

∴ a = and b = 2

⇒ So, we have 2(x – )(x – 2)

⇒ (2x – 1)(x – 2)

∴ (2x – 1)(x – 2) are the factors

⇒ p(x) = 0 if (2x – 1) is 0 and (x – 2) is 0

⇒ (2x – 1) = 0

⇒ 2x = 1

⇒ x =

⇒ Substitute, x = in 2x² – 5x + 2 = 2(² – 5( + 2 = – + 2 = 0

⇒ (x – 2) = 0

⇒ x = 2 substitute, 2x² – 5x + 2 = 2(2)² – 5(2) + 2 = 8 – 10 + 2 = 0

Hence, (2x – 1)(x – 2) are the first degree factors of the polynomial and and 2 are the solution of the given polynomial 2x² – 5x + 2

Given, p(x) = 6x² – 7x + 2

Now, we need to write the given polynomial as a product of first degree polynomial and p(x) = 0

⇒ 6x² – 7x + 2 can be written as 6(x² – x + ) = 6(x² – x + )

⇒ Now, we will find out the factors for x² – x +

⇒ x² – x + is in the form x² – (a + b)x + ab and (x – a)(x – b) are the factors

∴ coefficient on either side of the polynomial is same and a + b =

and ab =

⇒ we know that (a + b)² –(a – b)² = 4ab

∴ (a + b)² – 4ab = (a – b)²

⇒ – 4() = (a – b)²

⇒ – = (a – b)²

⇒ = (a – b)²

⇒ = (a – b)²

⇒ a – b =

⇒ a – b =

⇒ We need to find out the values of a and b by solving (a + b) + (a – b) and (a + b) – (a – b)

⇒ we take a – b =

⇒ (a + b) + (a – b) = +

⇒ 2a =

⇒ a = =

⇒ (a + b) – (a – b) = –

⇒ 2b =

⇒ b =

∴ we get a = and b =

And if we take a – b = – we get as follows

⇒ (a + b) + (a – b) = + (– )

⇒ 2a =

⇒ a =

⇒ (a + b) – (a – b) = – (– )

⇒ 2b =

⇒ b =

∴ a = and b =

⇒ So, we have 6(x – )(x – )

⇒ (6x – )( x – )

⇒ (6x – 3)(x – )

∴ (6x – 3)(x – ) are the factors

⇒ p(x) = 0 if (6x – 3) is 0 and (x – ) is 0

⇒ (6x – 3) = 0

⇒ 6x = 3

⇒ x = =

⇒ Substitute, x = in 6x² – 7x + 2 = 6(² – 7( + 2 = – + 2 = 0

⇒ (x – ) = 0

⇒ x = substitute, 6x² – 7x + 2 = 6()² – 7() + 2 = – + 2 = 0

Hence, (6x – 3)(x – ) are the first degree factors of the polynomial and and are the solution of the given polynomial 6x² – 7x + 2

Given, p(1) = 0, p(– 2) = 0

Need to find a polynomial p(x) of second degree

⇒ since we know that p(1) = 0

∴ if x = 1 is substituted in p(x) then it satisfies the equation

⇒ x – 1 = 0 , and x – 1 is one factor of p(x)

And p(– 2) = 0 is given

⇒ if x = – 2 is substituted in p(x) then it satisfies the equation

⇒ x + 2 = 0, and x + 2 is one factor of p(x)

⇒ since, x – 1 and x + 2 are the factors of p(x), it can be written as follows

⇒ p(x) = (x – 1)(x + 2)

⇒ p(x) = x² – x + 2x – 2

⇒ p(x) = x² + x – 2

∴ x² + x – 2 is the second degree polynomial which satisfies p(1) = 0 and p(– 2) = 0.

Given, p(1 + √3) = 0, p(1 – √3) = 0

Need to find a polynomial p(x) of second degree

⇒ since we know that p(1 + √3) = 0

∴ if x = 1 + √3 is substituted in p(x) then it satisfies the equation

⇒ x – (1 + √3) = 0 , and ((x – 1) – √3) is one factor of p(x)

And p(1 – √3) = 0 is given

⇒ if x = 1 – √3 is substituted in p(x) then it satisfies the equation

⇒ x – (1 – √3) = 0, and ((x – 1) + √3) is one factor of p(x)

⇒ since, ((x – 1) – √3) and ((x – 1) + √3) are the factors of p(x), it can be written as follows

⇒ p(x) = ((x – 1) – √3)((x – 1) + √3)

⇒ p(x) = (x – 1 – √3)(x – 1 + √3)

⇒ p(x) = x² – x + √3x –x + 1 – √3 – √3x + √3 – 3

⇒ p(x) = x² – 2x – 2

∴ x² – 2x – 2 is the second degree polynomial which satisfies p(1 + √3) = 0 and p(1 – √3) = 0.

Given, p(1) = 0, p(√2) = 0 and p(– √2) = 0

Need to find the third degree polynomial p(x)

⇒ p(1) = 0 is given which satisfy p(x)

∴ if x = 1 is substituted in p(x) then it satisfies the equation

⇒ x – 1 is one factor of p(x)

⇒ p(√2) = 0 is given

∴ if x = √2 is substituted in p(x) then it satisfies the equation

⇒ x – √2 is another factor

⇒ p(– √2) = 0 is given

∴ if x = – √2 is substituted in p(x) then it satisfies the equation

⇒ x + √2 is third factor of the p(x)

⇒ Since, (x – 1)(x – √2)(x + √2) are the factors of the third degree polynomials

∴ p(x) = (x – 1)(x – √2)(x + √2)

⇒ p(x) = (x² – x – √2x + √2)(x + √2)

⇒ p(x) = (x³ + √2x² – x² – √2x – √2x² – 2x + √2x + 2)

⇒ p(x) = (x³ – x² – 2x + 2)

Hence, x³ – x² – 2x + 2 is the third degree polynomial which satisfies

p(1) = 0, p(√2) = 0 and p(– √2) = 0

Given, a second degree polynomial x² + x + 1

Need to prove the given equation cannot be written as product of first degree polynomial

⇒ we know that a polynomial equation of degree 2, x² + (a + b)x + ab can be written as (x + a)(x + b)

⇒ Here, x² + x + 1 is written as x² + (a + b)x + ab

⇒ coefficient on either side are equal, we get

⇒ a + b = 1 and ab = 1

⇒ We need to find the values of a, b such that it satisfies the given equation to get the factors of first degree polynomial

⇒ Since, a + b = 1 and ab = 1 it is not possible to find out the values of a, b which satisfy the equation x² + x + 1

Hence, x² + x + 1 cannot be splited into factors of first degree polynomial

Given, a pair of polynomial as x – 1, x³ + 4x² – 3x – 6

Need to find out the first polynomial is factor of second and if not a factor need to find the remainder

⇒ To check x – 1 is a factor of x³ + 4x² – 3x – 6 we must substitute x = 1 in the second polynomial, we get as follows

⇒ 1 + 4 – 3 – 6 = – 4 not equal to 0

∴ x – 1 is not a factor of x³ + 4x² – 3x – 6

⇒ To find the remainder by using divide second polynomial by first polynomial

⇒ so, we can subtract a number from the second polynomial to get the remainder

∴ x³ + 4x² – 3x – 6 = (x – 1)q(x) + c

⇒ x³ + 4x² – 3x – 6 –c = (x – 1)q(x)

⇒ c = ((x³ + 4x² – 3x – 6) – (x – 1)) × q(x)

⇒ Now, substitute x = 1 in the above equation we get

⇒ c = (1 + 4 – 3 – 6 – 1 + 1) × q(1)

∴ c = – 4

– 4 is the remainder

Given, a pair of polynomial as x + 1, x³ + 4x² – 3x – 6

Need to find out the first polynomial is factor of second and if not a factor need to find the remainder

⇒ To check x + 1 is a factor of x³ + 4x² – 3x – 6 we must substitute x = – 1 in the second polynomial, we get as follows

⇒ – 1 + 4 + 3 – 6 = 0

∴ x + 1 is a factor

Given, a pair of polynomial as x – 2, x³ + 3x² – 4x – 12

Need to find out the first polynomial is factor of second and if not a factor need to find the remainder

⇒ To check x – 2 is a factor of x³ + 3x² – 4x – 12 we must substitute x = 2 in the second polynomial, we get as follows

⇒ 2³ + 3(2)² – 4(2) – 12 = 8 + 12 – 8 – 12 = 0

∴ x – 2 is a factor

Given, a pair of polynomial as x + 2, x³ + 3x² – 4x – 12

Need to find out the first polynomial is factor of second and if not a factor need to find the remainder

⇒ To check x + 2 is a factor of x³ + 3x² – 4x – 12 we must substitute x = – 2 in the second polynomial, we get as follows

⇒ (– 2)³ + 3(– 2)² – 4(– 2) – 12 = – 8 + 12 + 8 – 12 = 0

∴ x + 2 is a factor

Given, a pair of polynomial as 2x – 1, 3x³ – x² – 8x + 6

Need to find out the first polynomial is factor of second and if not a factor need to find the remainder

⇒ To check 2x – 1 is a factor of 3x³ – x² – 8x + 6 we must substitute x = in the second polynomial, we get as follows

⇒ 3()³ – ()² – 8() + 6 = + 6 = – + 6 = + 2 = 0

∴ 2x – 1 is not a factor

⇒ To find the remainder divide second polynomial by first polynomial

⇒ so, we can subtract a number from the second polynomial to get the remainder

∴ 3x³ – x² – 8x + 6 = (2x – 1)q(x) + c

⇒ 3x³ – x² – 8x + 6 –c = (2x – 1)q(x)

⇒ c = ((3x³ – x² – 8x + 6) – (2x – 1)) × q(x)

⇒ Now, substitute x = in the above equation we get

⇒ c = (3()³ – ()² – 8() + 6 – 2() + 1) × q(1)

= – + 6

∴ c =

is the remainder

Given, a pair of polynomial as 3x – 1, 3x³ – 10x² + 9x – 2

Need to find out the first polynomial is factor of second and if not a factor need to find the remainder

⇒ To check 3x – 1 is a factor of 3x³ – 10x² + 9x – 2 we must substitute x = in the second polynomial, we get as follows

⇒ 3()³ – 10()² + 9() – 2 = – – – 2

= – – 3 – 2

= – 5

= 0

∴ 3x – 1 is not a factor

⇒ To find the remainder divide second polynomial by first polynomial

⇒ so, we can subtract a number from the second polynomial to get the remainder

∴ 3x³ – 10x² + 9x – 2 = (3x – 1)q(x) + c

⇒ 3x³ – 10x² + 9x – 2 –c = (3x – 1)q(x)

⇒ c = ((3x³ – 10x² + 9x – 2) – (3x – 1)) × q(x)

⇒ Now, substitute x = in the above equation we get

⇒

= – – – 2 – 0

∴ c =

is the remainder

Given, x³ – 1, x – 1 as pair of polynomials

Need to find the quotient and remainder

⇒ To find the quotient and remainder the given equation can be written as p(x) = (x – a)q(x) + b

⇒ since, the polynomial is of third degree we can write the q(x) as x² + ax + b

∴ p(x) = (x – a)( x² + ax + b) + c

⇒ x³ – 1 = (x – 1)( x² + ax + b) + c

⇒ x³ – 1 = (x³ –x² + ax² – ax + bx – b) + c

⇒ x³ – 1 = x³ + (a–1)x² + (b – a)x + (c – b)

∴ a – 1 = 0, b – a = 0, c – b = – 1

⇒ a = 1, b = 1

⇒ c = b – 1 = 0

Quotient = x² + ax + b = x² + x + 1

Remainder = 0

Given, x³ – 1, x – 1 as pair of polynomials

Need to find the quotient and remainder

⇒ To find the quotient and remainder the given equation can be written as p(x) = (x – a)q(x) + b

⇒ Since, the polynomial is of third degree we can write the q(x) as x² + ax + b

∴ p(x) = (x – a)( x² + ax + b) + c

⇒ x³ – 1 = (x + 1)( x² + ax + b) + c

⇒ x³ – 1 = (x³ + x² + ax² + ax + bx + b) + c

⇒ x³ – 1 = x³ + (a + 1)x² + (a + b)x + (c + b)

∴ a + 1 = 0, a + b = 0, c + b = – 1

⇒ a = – 1

⇒ a + b = 0

⇒ – 1 + b = 0

⇒ b = 1

⇒ c + b = – 1

⇒ c + 1 = – 1

⇒ c = – 2

Quotient = x² + ax + b = x² – x + 1

Remainder = – 2

Given, x³ – 1, x – 1 as pair of polynomials

Need to find the quotient and remainder

⇒ To find the quotient and remainder the given equation can be written as p(x) = (x – a)q(x) + b

⇒ Since, the polynomial is of third degree we can write the q(x) as x² + ax + b

∴ p(x) = (x – a)( x² + ax + b) + c

⇒ x³ + 1 = (x – 1)(x² + ax + b) + c

⇒ x³ + 1 = (x³ –x² + ax² – ax + bx – b) + c

⇒ x³ + 1 = x³ + (a–1)x² + (b – a)x + (c – b)

∴ a – 1 = 0, b – a = 0, c – b = 1

⇒ a = 1, b = 1

⇒ c – b = 1

⇒ c = 2

Quotient = x² + ax + b = x² + x + 1

Remainder = 2

Given, x³ – 1, x – 1 as pair of polynomials

Need to find the quotient and remainder

⇒ To find the quotient and remainder the given equation can be written as p(x) = (x – a)q(x) + b

⇒ Since, the polynomial is of third degree we can write the q(x) as x² + ax + b

∴ p(x) = (x – a)( x² + ax + b) + c

⇒ x³ + 1 = (x + 1)(x² + ax + b) + c

⇒ x³ + 1 = (x³ + x² + ax² + ax + bx + b) + c

⇒ x³ + 1 = x³ + (a + 1)x² + (b + a)x + (c + b)

∴ a + 1 = 0, b + a = 0, c + b = 1

⇒ a = – 1, b = 1

⇒ c + b = 1

⇒ c = 0

Quotient = x² + ax + b = x² – x + 1

Remainder = 0

Given p(x) = x³ + x² + x

Let the number to be added be “k”.

Then, the new polynomial q(x) = x³ + x² + x + k

Now, (x – 1) is a factor of x³ + x² + x + k.

i.e. x = 1 is the root of the polynomial.

Then, put the polynomial to zero we get,

x³ + x² + x + k = 0

⇒ (1)³ + (1)² + 1 + k = 0

⇒ 1 + 1+1 + k = 0

⇒ k = – 3

Hence, “ – 3” should be added to the polynomial such that (x – 1) is a factor of q(x).

Given p(x) = x³ + x² + x

Let the number to be added be “k”.

Then, the new polynomial q(x) = x³ + x² + x + k

Now, (x+1) is a factor of x³ + x² + x + k.

i.e. x = – 1 is the root of the polynomial.

Then, put the polynomial to zero we get,

x³ + x² + x + k = 0

⇒ ( – 1)³ + ( – 1)² + ( – 1) + k = 0

⇒ – 1 + 1 – 1 + k = 0

⇒ k = – 1

Hence, “ – 1” should be added to the polynomial such that (x – 1) is a factor of q(x).

Given, x – 1, xⁿ – 1 pair of polynomials

Need to find out n such as first polynomial is factor of second

⇒ To check x – 1 is factor of xⁿ – 1 we must get xⁿ – 1 = 0 when substituted x with 1 from the first polynomial

⇒ since, when x is substituted with 1 it will satisfy irrespective of n in the given polynomial

Consider n as 1 then the polynomial equation itself wiil be x – 1 and x – 1 will be the factor

Hence, n is 1

Given, x – 1, xⁿ + 1 pair of polynomials

Need to find out n such as first polynomial is factor of second

⇒ To check x – 1 is factor of xⁿ + 1 we must get xⁿ + 1 = 0 when substituted x with 1 from the first polynomial

⇒ Consider n as 1 then the polynomial equation wiil be x + 1 which is not equal to zero and x – 1 will not be the factor.

⇒ For any n value x – 1 cannot be a factor

Given, x + 1, xⁿ – 1 pair of polynomials

Need to find out n such as first polynomial is factor of second

⇒ To check x + 1 is factor of xⁿ – 1 we must get xⁿ – 1 = 0 when substituted x with – 1 from the first polynomial

⇒ Consider n as 1 then the polynomial equation will be x – 1 which is not equal to zero and x + 1 will not be the factor.

⇒ Then consider n as 2 then x² – 1 be the polynomial equation substitute x = – 1 in the equation

We get, (– 1)² – 1 = 0

∴ x + 1 is the factor of xⁿ – 1

Hence, n = 2

Given, x + 1, xⁿ + 1 pair of polynomials

Need to find out n such as first polynomial is factor of second

⇒ To check x + 1 is factor of xⁿ + 1 we must get xⁿ + 1 = 0 when substituted x with – 1 from the first polynomial

⇒ Consider n as 1 then the polynomial equation will be x + 1 which is equal to zero and x + 1 will is the factor.

We get, (– 1) + 1 = 0

∴ x + 1 is the factor of xⁿ + 1

Hence, n = 1

Given, x² – 1, xⁿ – 1 pair of polynomials

Need to find out n such as first polynomial is factor of second

⇒ To check x² – 1 is factor of xⁿ – 1 we must get xⁿ – 1 = 0 when substituted with x values from the first polynomial

⇒ Here, x² – 1 so, x = √1 = 1

⇒ Consider n as 1 then the polynomial equation will be x¹ – 1 which is equal to zero and x² – 1 is the factor.

We get, 1 – 1 = 0

∴ x² – 1 is the factor of xⁿ – 1

Hence, n = 1

Given, ax³ + bx² + cx + d

Need to show a = – c and b = – d if x² – 1 is a factor

⇒ consider, x² – 1 is a factor of ax³ + bx² + cx + d

⇒ then x = + 1, x = – 1

⇒ substitute x value in the equation ax³ + bx² + cx + d

We get as follows

⇒ for x = 1 we get a(1)³ + b(1)² + c(1) + d

= a + b + c + d ……..eq(1)

⇒ for x = – 1 we get a(– 1)³ + b(– 1)² + c(– 1) + d

= – a + b – c + d ………eq(2)

⇒ Solving the two equations we get

⇒ (a + b + c + d) + (– a + b – c + d) = 0

⇒ b + d = 0

∴ b = – d

And if (a + b + c + d) – (– a + b – c + d) = 0

⇒ a + c = 0

∴ a = – c

Given, 2x³ – 3x² + 5x + 1

Need to find the first degree polynomial if added to the equation gets the multiple of x² – 1

⇒ 2x³ – 3x² + 5x + 1 can be written as 2(x³ – x² + x + )

⇒ x³ – x² + x + = (x² – 1)(x – a) + b

= x³ – ax² – x + a + b

∴ a = and a + b =

⇒ b = – 1