Algebra Class 9th Mathematics Term 1 Tamilnadu Board Solution

Class 9th Mathematics Term 1 Tamilnadu Board Solution
Exercise 3.1
  1. State whether the following expressions are polynomials in one variable or not.…
  2. Write the coefficient of x^2 and x in each of the following i. 2 + 3x - 4x^2 +…
  3. Write the degree of each of the following polynomials. i. 4 - 3x^2 ii. 5y + √2…
  4. Classify the following polynomials based on their degree. i. 3x^2 + 2x + 1 ii.…
  5. Give one example of a binomial of degree 27 and monomial of degree 49 and…
Exercise 3.2
  1. Find the zeros of the following polynomials. i. p(x)= 4x - 1 ii. p(x) = 3x + 5…
  2. Find the roots of the following polynomial equations. i. x - 3 = 0 ii. 5x - 6 =…
  3. x^2 - 5x + 6 = 0; x = 2, 3 Verify Whether the following are roots of the…
  4. x^2 + 4x + 3 = 0; x = −1, 2 Verify Whether the following are roots of the…
  5. x^3 - 2x^2 - 5x + 6 = 0; x = 1, −2, 3 Verify Whether the following are roots of…
  6. x^3 - 2x^2 - x + 2 = 0; x = −1, 2, 3 Verify Whether the following are roots of…
Exercise 3.3
  1. (5x^3 - 8x^2 + 5x - 7) ÷ (x - 1) Find the quotient the and remainder of the…
  2. (2x^2 - 3x - 14) ÷ (x + 2) Find the quotient the and remainder of the following…
  3. (9 + 4x + 5x^2 + 3x^3) ÷ (x+ 1) Find the quotient the and remainder of the…
  4. (4x^3 - 2x^2 + 6x + 7) ÷ (3 + 2x) Find the quotient the and remainder of the…
  5. (−18 - 9x + 7x^2) ÷ (x - 2) Find the quotient the and remainder of the…
Exercise 3.4
  1. 3x^3 + 4x^2 - 5x + 8 is divided by x - 1 Find the remainder using remainder…
  2. 5x^3 + 2x^2 - 6x + 12 is divided by x + 2 Find the remainder using remainder…
  3. 2x^3 - 4x^2 + 7x + 6 is divided by x - 2 Find the remainder using remainder…
  4. 4x^3 - 3x^2 + 2x - 4 is divided by x + 3 Find the remainder using remainder…
  5. 4x^3 - 12x^2 + 11x - 5 is divided by 2x - 1 Find the remainder using remainder…
  6. 8x^4 + 12x^3 - 2x^2 - 18x + 14 is divided by x + 1 Find the remainder using…
  7. x^3 - ax^2 - 5x + 2a is divided by x - a Find the remainder using remainder…
  8. When the polynomial 2x^3 - 2x^2 + 9x - 8 is divided by x - 3 the remainder is…
  9. Find the value of m if x^3 - 6x^2 + mx + 60 leaves the remainder 2 when divided…
  10. If (x - 1) divides mx^3 - 2x^2 + 25x - 26 without remainder find the value of m…
  11. If the polynomials x^3 + 3x^2 - m and 2x^3 - mx + 9 leaves the same remainder…
Exercise 3.5
  1. 6x^4 + 7x^3 - 5x - 4 Determine whether (x + 1) is a factor of the following…
  2. 2x^4 + 9x^3 + 2x^2 + 10x + 15 Determine whether (x + 1) is a factor of the…
  3. 3x^3 + 8x^2 + 6x - 5 Determine whether (x + 1) is a factor of the following…
  4. x^3 - 14x^2 + 3x + 12 Determine whether (x + 1) is a factor of the following…
  5. Determine whether (x + 4) is a factor of x^3 + 3x^2 - 5x + 36.
  6. Using factor theorem show that (x - 1) is a factor of 4x^3 - 6x^2 + 9x - 7.…
  7. Determine whether (2x + 1) is a factor of 4x^3 + 4x^2 - x - 1.
  8. Determine the value of p if (x + 3) is a factor of x^3 - 3x^2 - px + 24.…
Exercise 3.6
  1. The coefficient of x^2 x in 2x^3 - 3x^2 - 2x + 3 are respectively:A. 2, 3 B. -…
  2. The degree of polynomial 4x^2 - 7x^3 + 6x + 1 is:A. 2 B. 1 C. 3 D. 0…
  3. The polynomial 3x - 2 is a :A. Linear polynomial B. Quadratic polynomial C.…
  4. The polynomial 4x^2 + 2x - 2 is a :A. Linear polynomial B. Quadratic polynomial…
  5. The zero of the polynomial 2x - 5:A. 5/2 B. - 5/2 C. 2/5 D. - 2/5…
  6. The root of polynomial equation 3x - 1 is:A. - 1/3 B. 1/3 C. 1 D. 3…
  7. The root of polynomial equation x^2 + 2x = 0:A. 0, 2 B. 1, 2 C. 1, - 2 D. 0, - 2…
  8. If a polynomial p(x) is divided by (ax + b), then the remainder is:A. p(b/a) B.…
  9. If a polynomial x^3 - ax^2 + ax - a is divided by (x - a), then the remainder…
  10. If (ax - b) is a factor of p(x) then,A. p(b) = 0 B. p(- b/a) = 0 C. p(a) = 0 D.…
  11. One of the factor of x^2 - 3x - 10 is :A. x - 2 B. x + 5 C. x - 5 D. x - 3…
  12. One of the factor of x^3 - 2x^2 + 2x - 1 is :A. x - 1 B. x + 1 C. x - 2 D. x +…

Exercise 3.1
Question 1.

State whether the following expressions are polynomials in one variable or not. Give reasons for your answer.

i. 2x5 – x3 + x – 6

ii. 3x2 – 2x + 1

iii. y3 + 2√3

iv. 

v. 

vi. x3 + y3 + z6


Answer:

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.



Question 2.

Write the coefficient of x2 and x in each of the following

i. 2 + 3x – 4x2 + x3

ii. √3 x + 1

iii. x3 + √2 x2 + 4x – 1

iv. 


Answer:

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



Question 3.

Write the degree of each of the following polynomials.

i. 4 – 3x2

ii. 5y + √2

iii. 12 – x + 4x3

iv. 5


Answer:

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



Question 4.

Classify the following polynomials based on their degree.

i. 3x2 + 2x + 1

ii. 4x3 – 1

iii. y + 3

iv. y2 – 4

v. 4x3

vi. 2x


Answer:

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



Question 5.

Give one example of a binomial of degree 27 and monomial of degree 49 and trinomial of degree 36.


Answer:

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.




Exercise 3.2
Question 1.

Find the zeros of the following polynomials.

i. p(x)= 4x – 1

ii. p(x) = 3x + 5

iii. p(x) = 2x

v. p(x) = x + 9


Answer:

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).



Question 2.

Find the roots of the following polynomial equations.

i. x – 3 = 0

ii. 5x – 6 = 0

iii. 11x + 1 = 0

iv. −9x = 0


Answer:

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.



Question 3.

Verify Whether the following are roots of the polynomial equations indicated against them.

x2 – 5x + 6 = 0; x = 2, 3


Answer:

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



Question 4.

Verify Whether the following are roots of the polynomial equations indicated against them.

x2 + 4x + 3 = 0; x = −1, 2


Answer:

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.



Question 5.

Verify Whether the following are roots of the polynomial equations indicated against them.

x3 – 2x2 – 5x + 6 = 0; x = 1, −2, 3


Answer:

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.



Question 6.

Verify Whether the following are roots of the polynomial equations indicated against them.

x3 – 2x2 – x + 2 = 0; x = −1, 2, 3


Answer:

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.




Exercise 3.3
Question 1.

Find the quotient the and remainder of the following division.

(5x3 – 8x2 + 5x – 7) ÷ (x – 1)


Answer:

(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



Question 2.

Find the quotient the and remainder of the following division.

(2x2 – 3x – 14) ÷ (x + 2)


Answer:

(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



Question 3.

Find the quotient the and remainder of the following division.

(9 + 4x + 5x2 + 3x3) ÷ (x+ 1)


Answer:

(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



Question 4.

Find the quotient the and remainder of the following division.

(4x3 – 2x2 + 6x + 7) ÷ (3 + 2x)


Answer:

(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



Question 5.

Find the quotient the and remainder of the following division.

(−18 – 9x + 7x2) ÷ (x – 2)


Answer:

(−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




Exercise 3.4
Question 1.

Find the remainder using remainder theorem, when

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


Answer:

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



Question 2.

Find the remainder using remainder theorem, when

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


Answer:

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



Question 3.

Find the remainder using remainder theorem, when

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


Answer:

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



Question 4.

Find the remainder using remainder theorem, when

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


Answer:

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



Question 5.

Find the remainder using remainder theorem, when

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


Answer:

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



Question 6.

Find the remainder using remainder theorem, when

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


Answer:

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



Question 7.

Find the remainder using remainder theorem, when

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


Answer:

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



Question 8.

When the polynomial 2x3 – 2x2 + 9x – 8 is divided by x – 3 the remainder is 28. Find the value of a.


Answer:

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



Question 9.

Find the value of m if x3 – 6x2 + mx + 60 leaves the remainder 2 when divided by (x + 2).


Answer:

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



Question 10.

If (x – 1) divides mx3 – 2x2 + 25x – 26 without remainder find the value of m


Answer:

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



Question 11.

If the polynomials x3 + 3x2 – m and 2x3 – mx + 9 leaves the same remainder when they are divided by (x – 2), find the value of m. Also find the remainder


Answer:

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




Exercise 3.5
Question 1.

Determine whether (x + 1) is a factor of the following polynomials:

6x4 + 7x3 – 5x – 4


Answer:

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



Question 2.

Determine whether (x + 1) is a factor of the following polynomials:

2x4 + 9x3 + 2x2 + 10x + 15


Answer:

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



Question 3.

Determine whether (x + 1) is a factor of the following polynomials:

3x3 + 8x2 + 6x – 5


Answer:

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



Question 4.

Determine whether (x + 1) is a factor of the following polynomials:

x3 – 14x2 + 3x + 12


Answer:

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



Question 5.

Determine whether (x + 4) is a factor of x3 + 3x2 – 5x + 36.


Answer:

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).



Question 6.

Using factor theorem show that (x – 1) is a factor of 4x3 – 6x2 + 9x – 7.


Answer:

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)



Question 7.

Determine whether (2x + 1) is a factor of 4x3 + 4x2 – x – 1.


Answer:

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).



Question 8.

Determine the value of p if (x + 3) is a factor of x3 – 3x2 – px + 24.


Answer:

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




Exercise 3.6
Question 1.

The coefficient of x2 & x in 2x3 – 3x2 – 2x + 3 are respectively:
A. 2, 3

B. – 3, – 2

C. – 2, – 3

D. 2, – 3


Answer:


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

Coefficient of x = – 2


Question 2.

The degree of polynomial 4x2 – 7x3 + 6x + 1 is:
A. 2

B. 1

C. 3

D. 0


Answer:

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


Question 3.

The polynomial 3x – 2 is a :
A. Linear polynomial

B. Quadratic polynomial

C. Cubic polynomial

D. constant polynomial


Answer:

Given polynomial has degree = 1


Question 4.

The polynomial 4x2 + 2x – 2 is a :
A. Linear polynomial

B. Quadratic polynomial

C. Cubic polynomial

D. constant polynomial


Answer:

Given polynomial has degree = 2


Question 5.

The zero of the polynomial 2x – 5:
A. 5/2

B. – 5/2

C. 2/5

D. – 2/5


Answer:

Given : 2x – 5 = 0


∴ x = 5/2


So, zero of polynomial = 5/2


Question 6.

The root of polynomial equation 3x – 1 is:
A. – 1/3

B. 1/3

C. 1

D. 3


Answer:

Given polynomial equation: 3x – 1 = 0


∴ x = 1/3


So, root = 1/3


Question 7.

The root of polynomial equation x2 + 2x = 0:
A. 0, 2

B. 1, 2

C. 1, – 2

D. 0, – 2


Answer:

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


Question 8.

If a polynomial p(x) is divided by (ax + b), then the remainder is:
A. p(b/a)

B. p(– b/a)

C. p(a/b)

D. p(– a/b)


Answer:

f(x) = ax + b


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


ax + b = 0


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


Question 9.

If a polynomial x3 – ax2 + ax – a is divided by (x – a), then the remainder is:
A. a3

B. a2

C. a

D. – a


Answer:

Given f(x) = x3 – ax2 + ax – 2


By Remainder Theorem,


x – a = 0


x = a


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


Question 10.

If (ax – b) is a factor of p(x) then,
A. p(b) = 0

B. p(– b/a) = 0

C. p(a) = 0

D. p(b/a) = 0


Answer:

As (ax – b) is a factor of p(x)


ax – b = 0


∴x = b/a So, p(x) = 0 ;


p(b/a) = 0


Question 11.

One of the factor of x2 – 3x – 10 is :
A. x – 2

B. x + 5

C. x – 5

D. x – 3


Answer:

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.


Question 12.

One of the factor of x3 – 2x2 + 2x – 1 is :
A. x – 1

B. x + 1

C. x – 2

D. x + 2


Answer:

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.