Algebra

i. 2x⁵ – x³ + x – 6

There is only one variable ‘x’ with whole number power. So, this is polynomial in one variable.

ii. 3x² – 2x + 1

There is only one variable ‘x’ with whole number power. So, this is polynomial in one variable.

iii. y³ + 2√3

There is only one variable ‘y’ with whole number power. So, this is polynomial in one variable.

iv.

⇒

There is only one variable ‘x’ with whole number power. So, this is polynomial in one variable.

v. 3√ t + 2t

There is only one variable ‘t’ but in 3√t, power of t is which is not a whole number. So, this is not a polynomial in one variable.

vi. x³ + y³ + z³

There are three variables x, y and z, but power is whole number. So, this is not a polynomial in one variable.

i 2 + 3x – 4x² + x³

Co-efficient of x² = -4

Co-efficient of x = 3

ii √3x + 1

Co-efficient of x² = 0

Co-efficient of x = √3

iii x³ + √2x² + 4x – 1

Co-efficient of x² = √2

Co-efficient of x = 4

iv

= x² + 3x + 6 = 0

Co-efficient of x² = 1

Co-efficient of x = 3

i. 4 – 3x²

Degree of the polynomial = 2

ii. 5y + √2

Degree of the polynomial = 1

iii. 12 – x + 4x³

Degree of the polynomial = 3

iv. 5

Degree of the polynomial = 0

i. 3x² + 2x +1

Since, the highest degree of polynomial is 2

It is a quadratic polynomial.

ii. 4x³ – 1

Since, the highest degree of polynomial is 3.

It is a cubic polynomial.

iii. y + 3

Since, the highest degree of polynomial is 1

It is a linear polynomial

iv. y² – 4

Since, the highest degree of polynomial is 2

It is a quadratic polynomial.

v. 4x³

Since, the highest degree of polynomial is 3.

It is a cubic polynomial.

vi. 2x

Since, the highest degree of polynomial is 1

It is a linear polynomial

Binomial means having two terms. So binomial of degree 27 is x²⁷ + y.

Monomial means having one term. So, monomial of degree is x⁴⁹.

Trinomial means having three term. So, trinomial of degree is x³⁶ + y + 2.

i. p(x) = 4x – 1

⇒

⇒

Hence, is the zero of p(x).

ii. p(x) = 3x + 5

⇒

⇒

Hence, is the zero of p(x).

iii. p(x) = 2x

p(0) = 2(0) = 0

Hence, 0 is the zero of p(x).

iv. p(x) = x + 9

p(-9) = -9 + 9 = 0

Hence, -9 is the zero of p(x).

i. x – 5 = 0

⇒ x = 5

∴ x = 5 is a root of x – 5 = 0

ii. 5x – 6 = 0

⇒ 5x = 6

⇒

∴ is a root of 5x – 6 = 0

iii. 11x + 1 = 0

⇒ 11x = -1

⇒

∴ is a root of 11x + 1 = 0.

iv. -9x = 0

⇒

⇒ x = 0

∴ x = 0 is a root of -9x = 0.

x² – 5x + 6 = 0

Let p(x) = x² – 5x + 6

p(2) = (2)² – 5(2) + 6

= 4 – 10 + 6

= 10 – 10 = 0

∴ x = 2 is a root of x² – 5x + 6 = 0

p(x) = x² – 5x + 6

p(3) = (3)² – 5(3) + 6

= 9 – 15 + 6

= 15 – 15 = 0

∴ x = 3 is a root of x² – 5x + 6 = 0

x² + 4x + 3 = 0

let p(x) = x² + 4x + 3

p(-1) = (-1)² + 4(-1) + 3

= 1 – 4 + 3

= 4 – 4 = 0

∴ x = -1 is a root of x² + 4x + 3 = 0

p(x) = x² + 4x + 3 = 0

p(2) = (2)² + 4(2) + 3

= 4 + 8 + 3

= 11 + 4 = 15 ≠ 0

∴ x = 2 is not a root of x² + 4x + 3 = 0.

x³ – 2x² – 5x + 6 = 0

let p(x) = x³ – 2x² – 5x + 6

p(1) = (1)³ – 2(1)² – 5(1) + 6

= 1 – 2 × 1 – 5 + 6

= 1 – 2 – 5 + 6

= 7 – 7 = 0

∴ x = 1 is a root of x³ – 2x² – 5x + 6 = 0.

p(x) = x³ – 2x² – 5x + 6

p(-2) = (-2)³ – 2(-2)² – 5(-2) + 6

= -8 – 2 × 4 – 5 × 2 + 6

= -8 – 8 + 10 + 6

= -16 + 16 = 0

∴ x = -2 is a root of x³ – 2x² – 5x + 6 = 0.

p(x) = x³ – 2x² – 5x + 6 = 0

p(3) = (3)³ – 2(3)² – 5(3) + 6

= 27 – 2 × 9 – 5 × 3 + 6

= 27 – 18 – 15 + 6

= 33 – 33 = 0

∴ x = 3 is a root of x³ – 2x² – 5x + 6 = 0.

x³ – 2x² – x + 2 = 0

p(x) = x³ – 2x² – x + 2 = 0

p(-1) = (-1)³ – 2(-1)² - (-1) + 2

= -1 – 2 × 1 + 1 + 2

= -1 – 2 + 1 + 2

= -3 + 3 = 0

∴ x = -1 is a root of x³ – 2x² – x + 2 = 0

p(x) = x³ – 2x² – x + 2 = 0

p(2) = (2)³ – 2(2)² – (2) + 2

= 8 – 2 × 4 – 2 + 2

= 8 – 8 – 2 + 2

= 10 – 10 = 0

∴ x = 2 is a root of x³ – 2x² – x + 2 = 0.

p(x) = x³ – 2x² – x + 2 = 0

p(3) = (3)³ – 2(3)² – (3) + 2

= 27 – 2 × 9 – 3 + 2

= 27 – 18 – 3 + 2

= 29 – 21 = 8 ≠ 0

∴ x = 3 is not a root of x³ – 2x² – x + 2 = 0.

(5x³ – 8x² + 5x – 7) ÷ (x – 1)

We see that the equation is already arranged in descending order.

Now we need to divide (5x³ – 8x² + 5x – 7) by (x – 1).

Now we need to find out by how much should we multiple “x” to get a value as much as 5x³.

To get x³, we need to multiply x×x².

Therefore, we need to multiply with 5x² × (x – 1) and we get (5x³ – 5x²) now subtract (5x³ – 5x²) from 5x³ – 8x² + 5x – 7 so we get – 3x².

Now we carry 5x – 7 along with – 3x², as shown below

So, in same way we have keep dividing till we get rid of x as shown below.

here (x – 1) × (– 3x)

= – 3x² + 3x

here (x – 1) × 2

= 2x – 2

Therefore, we got the quotient = 5x² – 3x + 2 and

Remainder = –5

(2x² – 3x – 14) ÷ (x + 2)

We see that the equation is already arranged in descending order.

Now we need to divide (2x² – 3x – 14) by (x + 2).

Now we need to find out by how much should we multiple “x” to get a value as much as 2x².

To get x², we need to multiply x×x.

Therefore, we need to multiply with 2x × (x + 2) and we get (2x² + 4x) now subtract (2x² + 4x) from 2x² – 3x – 14 so we get –7x.

Now we carry 14 along with –7x, as shown below

So, in same way we have keep dividing till we get rid of x as shown below.

here (x + 2) × (– 7)

= – 7x – 14

Therefore, we got the quotient = 2x – 7 and

Remainder = 0

(9 + 4x + 5x² + 3x³) ÷ (x + 1)

We see that the equation is not arranged in descending order, so we need first arrange it in descending order of the power of x.

Therefore it becomes,

(3x³ + 5x² + 4x + 9) ÷ (x + 1)

Now we need to divide (3x³ + 5x² + 4x + 9) by (x + 1).

Now we need to find out by how much should we multiple “x” to get a value as much as 3x³.

To get x³, we need to multiply x×x².

Therefore, we need to multiply with 3x² × (x + 1) and we get (3x³ + 3x²) now subtract (3x³ + 3x²) from 3x³ + 5x² + 4x + 9 so we get 2x².

Now we carry 4x + 9 along with 2x², as shown below

So, in same way we have keep dividing till we get rid of x as shown below.

here (x + 1) × ( 2x)

= 2x² + 2x

here (x + 1) × 2

= 2x + 2

Therefore, we got the quotient = 3x² + 2x + 2 and

Remainder = 7

(4x³ – 2x² + 6x + 7) ÷ (3 + 2x)

We see that the equation is not arranged in descending order, so we need first arrange it in descending order of the power of x.

Therefore it becomes,

(4x³ – 2x² + 6x + 7) ÷ (2x + 3)

Now we need to divide (4x³ – 2x² + 6x + 7) by (2x + 3)

Now we need to find out by how much should we multiple “x” to get a value as much as 4x³.

To get x³, we need to multiply x×x².

Therefore we need to multiply with 2x² × (2x + 3) and we get (4x³ + 6x²) now subtract (4x³ + 6x²) from 4x³ – 2x² + 6x + 7so we get – 8x².

Now we carry 6x + 7along with 4x², as shown below

So, in same way we have keep dividing till we get rid of x as shown below.

here (2x + 3) × (–4x)

= – 8x² – 12x

here (2x + 3) × 9

= 18x + 27

Therefore, we got the quotient = 2x² – 4x + 9 and

Remainder = –20

(−18 – 9x + 7x²) ÷ (x – 2)

We see that the equation is not arranged in descending order, so we need first arrange it in descending order of the power of x.

Therefore it becomes,

(7x² – 9x – 18) ÷ (x – 2)

Now we need to divide (7x² – 9x – 18) by (x – 2).

Now we need to find out by how much should we multiple “x” to get a value as much as 7x².

To get x², we need to multiply x×x.

Therefore, we need to multiply with 7x × (x – 2) and we get (7x² – 14x) now subtract (7x² – 14x) from 7x² – 9x – 18 so we get 5x.

Now we carry 18 along with 5x, as shown below

So, in same way we have keep dividing till we get rid of x as shown below.

here (x – 2) × 5

= 5x – 10

Therefore, we got the quotient = 7x + 5 and

Remainder = – 8

3x³ + 4x² – 5x + 8 is divided by x – 1

Remainder theorem states that if p(x) is any polynomial and a is any real number and If p(x) is divided by the linear polynomial (x – a) , then the remainder is p(a).

Let p(x) = 3x³ + 4x² – 5x + 8 and we have (x – 1)

The zero of (x – 1) is 1

Now using Remainder theorem,

p(x) = 3x³ + 4x² – 5x + 8 is divided by x – 1 then, p(1) is the remainder

p(1) = 3(1)³ + 4(1)² – 5(1) + 8

= 3 + 4 – 5 + 8

= 10

Remainder = 10

5x³ + 2x² – 6x + 12 is divided by x + 2

Remainder theorem states that if p(x) is any polynomial and a is any real number and If p(x) is divided by the linear polynomial (x – a) , then the remainder is p(a).

Let p(x) = 5x³ + 2x² – 6x + 12 and we have (x + 2)

The zero of (x + 2) is – 2

Now using Remainder theorem,

p(x) = 5x³ + 2x² – 6x + 12 is divided by x + 2 then, p(–2) is the remainder

p(–2) = 5(–2)³ + 2(–2)² – 6(–2) + 12

= 5×(–8) + 2×4 – (– 12) + 12

= – 40 + 8 + 12 + 12

= – 40 + 32

= – 8

Remainder = –8

2x³ – 4x² + 7x + 6 is divided by x – 2

Remainder theorem states that if p(x) is any polynomial and a is any real number and If p(x) is divided by the linear polynomial (x – a) , then the remainder is p(a).

Let p(x) = 2x³ – 4x² + 7x + 6 and we have (x – 2)

The zero of (x – 2) is 2

Now using Remainder theorem,

p(x) = 2x³ – 4x² + 7x + 6 is divided by x – 2 then, p(2) is the remainder

p(2) = 2(2)³ – 4(2)² + 7(2) + 6

= 16 – 16 + 14 +6

= 20

Remainder = 20

4x³ – 3x² + 2x – 4 is divided by x + 3

Remainder theorem states that if p(x) is any polynomial and a is any real number and If p(x) is divided by the linear polynomial (x – a) , then the remainder is p(a).

Let p(x) = 4x³ – 3x² + 2x – 4 and we have (x + 3)

The zero of (x + 3) is – 3

Now using Remainder theorem,

p(x) = 4x³ – 3x² + 2x – 4 is divided by x + 3 then, p(– 3) is the remainder

p(– 3) = 4(–3)³ – 3(–3)² + 2(–3) – 4

= – 108 – 27 – 6 – 4

= – 145

Remainder = –145

4x³ – 12x² + 11x – 5 is divided by 2x – 1

Remainder theorem states that if p(x) is any polynomial and a is any real number and If p(x) is divided by the linear polynomial (x – a) , then the remainder is p(a).

Let p(x) = 4x³ – 12x² + 11x – 5 and we have (2x – 1)

The zero of (2x – 1) is

Now using Remainder theorem,

p(x) = 4x³ – 12x² + 11x – 5 is divided by 2x – 1 then, is the remainder

Remainder = –2

8x⁴ + 12x³ – 2x² – 18x + 14 is divided by x + 1

Remainder theorem states that if p(x) is any polynomial and a is any real number and If p(x) is divided by the linear polynomial (x – a) , then the remainder is p(a).

Let p(x) = 8x⁴ + 12x³ – 2x² – 18x + 14 and we have (x + 1)

The zero of (x + 1) is – 1

Now using Remainder theorem,

p(x) = 8x⁴ + 12x³ – 2x² – 18x + 14 is divided by x + 1 then, p(– 1) is the remainder

p(– 1) = 8(–1)⁴ + 12(–1)³ – 2(–1)² – 18(–1) + 14

= 8 – 12 – 2 +18 + 14

= 26

Remainder = 26

x³ – ax² – 5x + 2a is divided by x – a

Remainder theorem states that if p(x) is any polynomial and a is any real number and If p(x) is divided by the linear polynomial (x – a) , then the remainder is p(a).

Let p(x) = x³ – ax² – 5x + 2a and we have (x – a)

The zero of (x – a) is a

Now using Remainder theorem,

p(x) = x³ – ax² – 5x + 2a is divided by x – a then, p(a) is the remainder

p(a) = (a)³ – a(a)² – 5(a) + 2a

= a³ – a³ – 5a+ 2a

= – 3a

Remainder = –3a

2x³ – ax² + 9x – 8 is divided by x – 3 and remainder = 28

Remainder theorem states that if p(x) is any polynomial and a is any real number and If p(x) is divided by the linear polynomial (x – a) , then the remainder is p(a).

Let p(x) = 2x³ – ax² + 9x – 8 and we have (x – 3)

The zero of (x – 3) is 3

Now using Remainder theorem,

p(x) = 2x³ – ax² + 9x – 8 is divided by x – a then, p(3) is the remainder which is 28

p(3) = 2x³ – ax² + 9x – 8 =28

= 2(3)³ – a(3)² + 9(3) – 8 =28

= 54 – 9a + 27 – 8 = 28

= 73 – 9a = 28

= 9a = 73 –28

= 9a = 45

a = 5

x³ – 6x² + mx + 60 divided by (x + 2) and remainder = 2

Remainder theorem states that if p(x) is any polynomial and a is any real number and If p(x) is divided by the linear polynomial (x – a) , then the remainder is p(a).

Let p(x) = x³ – 6x² + mx + 60 and we have (x + 2)

The zero of (x + 2) is –2

Now using Remainder theorem,

p(x) = x³ – 6x² + mx + 60 is divided by x + 2 then, p(–2) is the remainder which is 2

p(–2) = x³ – 6x² + mx + 60 = 2

= (–2)³ – 6(–2)² + m(–2) + 60 =2

= – 8 – 24 – 2m + 60 = 2

= – 32 – 2m + 60 = 2

= 28 – 2m = 2

= 2m = 28 – 2

= 2m = 26

m = 13

mx³ – 2x² + 25x – 26 is divided by (x – 1) without remainder that means remainder = 0

Remainder theorem states that if p(x) is any polynomial and a is any real number and If p(x) is divided by the linear polynomial (x – a) , then the remainder is p(a).

Let p(x) mx³ – 2x² + 25x – 26 and we have (x – 1)

The zero of (x – 1) is 1

Now using Remainder theorem,

p(x) = mx³ – 2x² + 25x – 26 is divided by x – 1 then, p(1) is the remainder which is 0

p(1) = mx³ – 2x² + 25x – 26 = 0

= m(1)³ – 2(1)² + 25(1) – 26 = 0

= m – 2 + 25 – 26 = 0

= m – 3 = 0

m = 3

x³ + 3x² – m and 2x³ – mx + 9 is divided by (x – 2) and the remainder is same.

Now let p(x) = x³ + 3x² – m is divided by x – 2 then, p(2) is the remainder

p(2) = (2)³ + 3(2)² – m

= 8 + 12 – m

= 20 – m

Now let q(x) = 2x³ – mx + 9 is divided by x – 2 then, q(2) is the remainder

q(2) = 2(2)³ – m(2) + 9

= 16 – 2m + 9

= 25 – 2m

Now, as the question says that the remainder for p(x) and q(x) is same

Therefore, p(2) = q(2)

20 – m = 25 – 2m

2m – m = 25 – 20

m = 5

Remainder = p(2) = 20 – m

= 15

Let f(x) = 6x⁴ + 7x³ – 5x – 4

By factor theorem,

x + 1 = 0 ;x = – 1

If f(– 1) = 0 then (x + 1) is a factor of f(x)

∴f(– 1) = 6(– 1)⁴ + 7(– 1)³ – 5(– 1) – 4

= 6 – 7 + 5 – 4 = 11 – 11 = 0

∴(x + 1) is a factor of f(x) = 6x⁴ + 7x³ – 5x – 4

Let f(x) = 2x⁴ + 9x³ + 2x² + 10x + 15

By factor theorem,

x + 1 = 0 ;x = – 1

If f(– 1) = 0 then (x + 1) is a factor of f(x)

∴f(– 1) = 2(– 1)⁴ + 9(– 1)³ + 2(– 1)² + 10(– 1) + 15

= 2 – 9 + 2 – 10 + 15 = 19 – 19 = 0

∴(x + 1) is a factor of f(x) = 2x⁴ + 9x³ + 2x² + 10x + 15

Let f(x) = 3x³ + 8x² + 6x – 5

By factor theorem,

x + 1 = 0 ;x = – 1

If f(– 1) = 0 then (x + 1) is a factor of f(x)

∴f(– 1) = 3(– 1)³ + 8(– 1)² – 6(– 1) – 5

= – 3 + 8 + 6 – 5 = 6(not equal to 0)

∴(x + 1) is not a factor of f(x) = 3x³ + 8x² + 6x – 5

Let f(x) = x³ – 14x² + 3x + 12

By factor theorem,

x + 1 = 0 ;x = – 1

If f(– 1) = 0 then (x + 1) is a factor of f(x)

∴f(– 1) = (– 1)³ – 14(– 1)² + 3(– 1) + 12

= – 1 – 14 – 3 + 12 = – 6(not equal to 0)

∴(x + 1) is not a factor of f(x) = x³ – 14x² + 3x + 12

Let f(x) = x³ + 3x² – 5x + 36.

By factor theorem,

x + 4 = 0: x = – 4

If f(– 4) = 0, then (x + 4) is a factor

∴f(– 4) = (– 4)³ + 3(– 4)² – 5(– 4) + 36

= – 64 + 48 + 20 + 36

= – 64 + 104 = 40

∴f(– 4) is not equal to 0

So, (x + 4) is not a factor of f(x).

f(x) = 4x³ – 6x² + 9x – 7

By factor theorem,

(x – 1) = 0 ; x = 1

Since, (x – 1) is a factor of f(x)

Therefore, f(1) = 0

f(1) = 4(1)³ – 6(1)² + 9(1) – 7 = 4 – 6 + 9 – 7 = 13 – 13 = 0

∴(x – 1) is a factor of f(x)

Let f(x) = 4x³ + 4x² – x – 1

By factor Theorem,

2x + 1 = 0 ; x = – 1/2

∴f(– 1/2) = 4(– 1/2)³ + 4(– 1/2)² – (– 1/2) – 1

= 4(– 1/8) + 4(1/4) + (1/2) – 1

= (– 1/2) + 1 + (1/2) – 1 = 0

∴f(– 1/2) = 0

So, (2x + 1) is a factor of f(x).

Let f(x) = x³ + 3x² – px + 24.

By factor theorem,

x + 3 = 0;x = – 3

∴(x + 3) is a factor of f(x)

So, f(– 3) = 0.

f(– 3) = (– 3)³ – 3(– 3)² – p(– 3) + 24 = 0

= >27 – 27 + 3p + 24 = 0

= > – 59 + 24 + 3p = 0

∴3p – 30 = 0

⇒ p = 30/3

⇒p = 10

2x³ – 3x² – 2x + 3: Coefficient of x² = – 3

Coefficient of x = – 2

Degree of polynomial = Highest power of x in the polynomial = 3

Given polynomial has degree = 1

Given polynomial has degree = 2

Given : 2x – 5 = 0

∴ x = 5/2

So, zero of polynomial = 5/2

Given polynomial equation: 3x – 1 = 0

∴ x = 1/3

So, root = 1/3

Given polynomial equation :

x² + 2x = 0

∴x(x + 2) = 0

x = 0, x + 2 = 0

x = 0, x = – 2

So, the roots are 0 and – 2

f(x) = ax + b

Therefore, by Remainder Theorem, f(x) = 0

ax + b = 0

x = – b/a ∴Remainder = p(x) = p(– b/a)

Given f(x) = x³ – ax² + ax – 2

By Remainder Theorem,

x – a = 0

x = a

∴Remainder = f(a) = a³ – a³ + 2a – a = a

As (ax – b) is a factor of p(x)

ax – b = 0

∴x = b/a So, p(x) = 0 ;

p(b/a) = 0

x² – 3x – 10 = 0

x² – 5x + 2x – 10 = 0

x(x – 5) + 2(x – 5) = 0

(x – 5)(x + 2) = 0 Hence, (x – 5) is a factor.

Given: f(x) = x³ – 2x² + 2x – 1 = 0

By hit and trial method,

put x = 1

f(1) = 1 – 2 + 2 – 1 = 0

∴(x – 1) is a factor.