Polynomials Solution of TS & AP Board Class 10 Mathematics

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Exercise 3.1

Question 1.

If p(x) = 5x⁷ – 6x⁵ + 7x - 6, find

(i) coefficient of x⁵

(ii) degree of p(x)

(iii) constant term.

Answer:

(i) A coefficient is a multiplicative factor in some term of a polynomial. It is the constant (-6 here) written before the variable (x⁵ here).

∴ The coefficient of x⁵ is -6.

(ii) Degree of p(x) is the highest power of x in p(x).

∴ The degree of p(x) is 7.

(iii) The term that is not attached to a variable (i.e., x) is -6.

∴ The constant term is -6.

Question 2.

State which of the following statements are true and which are false? Give reasons for your choice.

(i) The degree of the polynomial  is 

(ii) The coefficient of x2 in the polynomial p(x) = 3x³ – 4x² + 5x + 7 is 2.

(iii) The degree of a constant term is zero.

(iv)  is a quadratic polynomial.

(v) The degree of a polynomial is one more than the number of terms in it.

Answer:

(i) False

Since, Degree of a polynomial is the highest power of x in the polynomial which is 2 in.

(ii) False

A coefficient is a multiplicative factor in some term of a polynomial. It is the constant written before the variable.

∴ The coefficient of x² is -4.

(iii) True

Since, the power of x of a constant term is 0.

∴ the degree of a constant term is 0.

(iv) False

For an expression to be a polynomial term, any variables in the expression must have whole-number powers (i.e., x⁰, x¹, x²,……)

Since, the power in the expression  has powers -2 and -1 which are not whole numbers, therefore, is not a polynomial.

(v) False

Degree of a polynomial is the highest power of x in the polynomial. The degree of polynomial is not related to the number of terms in the polynomial.

Therefore, the statement is false.

Question 3.

If p(t) t³ – 1, find the values of p(1),p(-1),P(0),p(2),p(-2).

Answer:

P(t) = t³ – 1

Therefore,

P(1) = 1³ – 1

⇒ P(1) = 1 – 1

⇒ P(1) = 0

P(-1) = (-1)³ – 1

⇒ P(-1) = -1 – 1

⇒ P(-1) = -2

P(0) = 0³ – 1

⇒ P(0) = 0 – 1

⇒ P(0) = -1

P(2) = 2³ – 1

⇒ P(2) = 8 – 1

⇒ P(2) = 7

P(-2) = (-2)³ – 1

⇒ P(2) = -8 – 1

⇒ P(2) = -9

Question 4.

Check whether -2 and 2 are the zeroes of the polynomial x⁴ - 16

Answer:

A zero or root of a polynomial function is a number that, when plugged in for the variable, makes the function equal to zero.

Now,

P(x) = x⁴ – 16

Therefore,

P(-2) = (-2)⁴ – 16

⇒ P(-2) = 16 – 16

⇒ P(-2) = 0

P(2) = 2⁴ – 16

⇒ P(2) = 16 – 16

⇒ P(2) = 0

Hence, Yes, -2 and -2 are zeroes of the polynomial x⁴ – 16.

Question 5.

Check whether 3 and -2 are the zeroes of the polynomial when p(x) = x² – x - 6

Answer:

A zero or root of a polynomial function is a number that, when plugged in for the variable, makes the function equal to zero.

Now,

P(x) = x² – x – 6

Therefore,

P(-2) = (-2)² – (-2) – 6

⇒ P(-2) = 4 + 2 – 6

⇒ P(-2) = 0

P(3) = 3² – 3 – 6

⇒ P(3) = 9 – 3 – 6

-⇒ P(3) = 0

Hence, Yes, 3 and -2 are zeroes of the polynomial x² – x – 6.

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Exercise 3.2

Question 1.

The graphs of y = p(x) are given in the figure below, for some polynomials p(x) In each case, find the number of zeroes of p(x)

Answer:

(i) Since, the graph does not intersect with x-axis at any point therefore, it has no zeroes.

(ii) Since, the graph intersects with x-axis at only one point therefore, it has 1 number of zeroes.

(iii) Since, the graph intersects with x-axis three points therefore, it has 3 number of zeroes.

(iv) Since, the graph intersects with x-axis two points therefore, it has 2 number of zeroes.

(v) Since, the graph intersects with x-axis at four points therefore, it has 4 number of zeroes.

(vi) Since, the graph intersects with x-axis at three points therefore, it has 3 number of zeroes.

Question 2.

Find the zeroes of the given polynomials.

p(x) = 3x

Answer:

A zero or root of a polynomial function is a number that, when plugged in for the variable, makes the function equal to zero.

Therefore, to find zeroes put p(x)=0.

p(x) = 0

⇒ 3x = 0

⇒ x = 0

Hence, x=0 is the zero of the polynomial.

Question 3.

Find the zeroes of the given polynomials.

p(x) = x² + 5x + 6

Answer:

p(x) = 0

⇒ x² + 5x + 6 = 0

⇒ x² + 3x + 2x + 6 = 0

⇒ x(x + 3) + 2(x + 3) = 0

⇒ (x + 3)(x + 2) = 0

⇒ x = -3 and x = -2

Hence, x=-3 and x=-2 are the zeroes of the polynomial.

Question 4.

Find the zeroes of the given polynomials.

p(x)- (x+2)(x+3)

Answer:

p(x) = 0

⇒ (x + 3)(x + 2)

⇒ x = -3 and x = -2

Hence, x=-3 and x=-2 are the zeroes of the polynomial.

Question 5.

Find the zeroes of the given polynomials.

p(x)=x⁴ - 16

Answer:

p(x) = 0

⇒ x⁴ – 16 = 0

⇒ (x² – 4)(x² + 4)= 0

⇒ (x + 2)(x – 2)(x² + 4)= 0

⇒ x = -2, x = 2 and x² = 4

⇒ x = -2, x = 2 and x = ±2

Hence, x=2 and x=-2 are the zeroes of the polynomial.

Question 6.

Draw the graphs of the given polynomial and find the zeroes. Justify the answer.

p(x) = x² – x - 12

Answer:

p(x) = x² – x – 12

Clearly, the graph intersects with x – axis at x = -3 and x = 4

Hence, the zeroes of the polynomial x² – x – 12 are x = -3 and x=4.

Question 7.

Draw the graphs of the given polynomial and find the zeroes. Justify the answer.

p(x) = x² – 6x + 9

Answer:

p(x) = x² – 6x + 9

Clearly, the graph intersects with x – axis at x = 4

Hence, the zeroes of the polynomial x² – 6x + 9 are x=3.

Question 8.

Draw the graphs of the given polynomial and find the zeroes. Justify the answer.

p(x) = x² – 4x + 5

Answer:

p(x) = x² – 4x + 5

Clearly, the graph does not intersect with x – axis at any point.

Hence, there are no zeroes of the polynomial x² – 4x + 5.

Question 9.

Draw the graphs of the given polynomial and find the zeroes. Justify the answer.

p(x) = x² + 3x - 4

Answer:

p(x) = x² + 3x – 4

Clearly, the graph intersects with x – axis at x = -4 and x = 1.

Hence, the zeroes of the polynomial x² + 3x – 4 are x = -4 and x= 1

Question 10.

Draw the graphs of the given polynomial and find the zeroes. Justify the answer.

p(x) = x² - 1

Answer:

p(x) = x² – 1

Clearly, the graph intersects with x – axis at x = -1 and x = 1.

Hence, the zeroes of the polynomial x² – 1 are x = -1 and x= 1

Question 11.

Why are 1/4 and -1 zeroes of the polynomials p(x) – 4x² + 3x – 1?

Answer:

A zero or root of a polynomial function is a number that, when plugged in for the variable, makes the function equal to zero.

Therefore, if p(x)=0, for a given x then the value of x is zero of polynomial.

p(x) = 4x² + 3x – 1

For 1/4,

p(1/4) = 4(1/4)² + 3(1/4) – 1

⇒ p(1/4) = 1/4 + 3/4 – 1

⇒ p(1/4) = 1 – 1

⇒ p(1/4) = 0

Therefore, x = 1/4 is a zero of the polynomial 4x² + 3x – 1.

For -1,

p(-1) = 4(-1)² + 3(-1) – 1

⇒ p(-1) = 4 – 3 – 1

⇒ p(-1) = 4 – 4

⇒ p(-1) = 0

Therefore, x = -1 is a zero of the polynomial 4x² + 3x – 1.

Hence, x = 1/4 and x = -1 are zeroes of the polynomial 4x² + 3x – 1.

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Exercise 3.3

Question 1.

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

x² – 2x – 8

Answer:

p(x) = 0

⇒ x² – 2x – 8 = 0

⇒ x² + 2x – 4x – 8 = 0

⇒ x(x + 2) – 4(x + 2) = 0

⇒ (x + 2)(x – 4) = 0

⇒ x = -2 and x = 4

Hence, -2 and 4 are zeroes of the polynomial x² – 2x – 8.

Now,

Sum of zeroes = -2 + 4

⇒ Sum of zeroes = 2

Product of zeroes = -2 × 4

⇒ Product of zeroes = -8

Question 2.

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

4s² – 4s + 1

Answer:

p(s) = 0

⇒ 4s² – 4s + 1 = 0

⇒ 4s² – 2s – 2s + 1 = 0

⇒ 2s(2s – 1) – 1(2s – 1) = 0

⇒ (2s – 1) (2s – 1) = 0

⇒ x = 1/2 and x = 1/2

Hence, 1/2 and 1/2 are zeroes of the polynomial 4s² – 4s + 1.

Now,

Sum of zeroes = 1/2 + 1/2

⇒ Sum of zeroes = 1

Product of zeroes = 1/2 × 1/2

⇒ Product of zeroes = 1/4

Question 3.

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

6x² – 3 – 7x

Answer:

p(x) = 0

⇒ 6x² – 3 – 7x = 0

⇒ 6x² – 7x – 3 = 0

⇒ 6x² – 9x + 2x – 3 = 0

⇒ 3x(2x – 3) + 1(2x – 3) = 0

⇒ (2x – 3) (3x + 1) = 0

⇒  and 

Hence,  and  are zeroes of the polynomial 6x² – 3 – 7x.

Now,

⇒ Product of zeroes = -1/2

Question 4.

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

4u² + 8u

Answer:

p(u) = 0

⇒ 4u² + 8u = 0

⇒ 4u(u + 2) = 0

⇒ u = 0 and u = -2

Hence, -2 and 0 are zeroes of the polynomial 4u² + 8u.

Now,

Sum of zeroes = -2 + 0

⇒ Sum of zeroes = -2

Product of zeroes = -2 × 0

⇒ Product of zeroes = 0

Question 5.

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

t² - 15

Answer:

p(t) = 0

⇒ t² – 15 = 0

⇒ (t – √15) (t + √15) = 0

⇒ t = -√15 and t = √15

Hence, -√15 and √15 are zeroes of the polynomial t² – 15.

Now,

Sum of zeroes = -√15 + √15

⇒ Sum of zeroes = -√15 + √15

⇒ Sum of zeroes = 0

Product of zeroes = -√15 × √15

⇒ Product of zeroes = -15

Question 6.

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

3x² – x – 4

Answer:

p(x) = 0

⇒ 3x² – x – 4 = 0

⇒ 3x² + 3x – 4x – 4 = 0

⇒ 3x(x + 1) –4(x + 1) = 0

⇒ (x + 1)(3x – 4)= 0

Hence, -1 and are zeroes of the polynomial 3x² – x – 4.

Now,

Question 7.

Find the quadratic polynomial in each case, with the given numbers as the sum and product of its zeroes respectively.

1/4, -1

Answer:

Given: α + β = 1/4

αβ = -1

Let the quadratic polynomial be ax² + bx + c …(1)

Where, a≠0

And zeroes of the polynomial are α and β .

Now we know that,

 …(2)

And,

If a = 4

⇒ c = -4 …(3)

From (2) and (3),

a = 4, b = -1 and c = -4

Hence, the polynomial is 4x² – x – 4

Question 8.

Find the quadratic polynomial in each case, with the given numbers as the sum and product of its zeroes respectively.

Answer:

Given: α + β = √2

Let the quadratic polynomial be ax² + bx + c …(1)

Where, a≠0

And zeroes of the polynomial are α and β .

Now we know that,

And,

If a = 3

⇒ b = -3√2 …(2)

⇒c = 1 …(3)

From (2) and (3),

a = 3, b = -3 √2 and c = 1

Hence, the polynomial is 3x² – 3√2x + 1

Question 9.

Find the quadratic polynomial in each case, with the given numbers as the sum and product of its zeroes respectively.

Answer:

Given: α + β = 0

αβ = √5

Let the quadratic polynomial be ax² + bx + c …(1)

Where, a≠0

And zeroes of the polynomial are α and β .

Now we know that,

 …(2)

And,

If a = 1

⇒c = √5 …(3)

From (2) and (3),

a = 1, b = 0 and c = √5

Hence, the polynomial is x² +√5

Question 10.

Find the quadratic polynomial in each case, with the given numbers as the sum and product of its zeroes respectively.

1, 1

Answer:

Given: α + β = 1

αβ = 1

Let the quadratic polynomial be ax² + bx + c …(1)

Where, a≠0

And zeroes of the polynomial are α and β .

Now we know that,

And,

If a = 1

⇒ b = -1 …(2)

⇒c = 1 …(3)

From (2) and (3),

a = 1, b = -1 and c = 1

Hence, the polynomial is x² – x + 1

Question 11.

Find the quadratic polynomial in each case, with the given numbers as the sum and product of its zeroes respectively.

1/4, 1/4

Answer:

Given: α + β = -1/4

αβ = 1/4

Let the quadratic polynomial be ax² + bx + c …(1)

Where, a≠0

And zeroes of the polynomial are α and β .

Now we know that,

And,

If a = 4

⇒ b = 1 …(2)

⇒c = 1 …(3)

From (2) and (3),

a = 4, b = 1 and c = 1

Hence, the polynomial is 4x² + x + 1

Question 12.

Find the quadratic polynomial in each case, with the given numbers as the sum and product of its zeroes respectively.

4, 1

Answer:

Given: α + β = 4

αβ = 1

Let the quadratic polynomial be ax² + bx + c …(1)

Where, a≠0

And zeroes of the polynomial are α and β.

Now we know that,

And,

If a = 1

⇒ b = -4 …(2)

⇒c = 1 …(3)

From (2) and (3),

a = 1, b = -4 and c = 1

Hence, the polynomial is x² - 4x + 1

Question 13.

Find the quadratic polynomial, for the zeroes given in each case.

2,-1

Answer:

Let the quadratic polynomial be ax² + bx + c

And, its zeroes be α and β

α = 2

β = -1

If a = 1,

⇒ b = -1

⇒ c = -2

Hence, the polynomial is x² – x – 2

Question 14.

Find the quadratic polynomial, for the zeroes given in each case.

Answer:

Let the quadratic polynomial be ax² + bx + c

And, its zeroes be α and β

α = √3

β = -√3

If a = 1,

⇒ b = 0

⇒ c = -3

Hence, the polynomial is x² – 3

Question 15.

Find the quadratic polynomial, for the zeroes given in each case.

1/4, -1

Answer:

Let the quadratic polynomial be ax² + bx + c

And, its zeroes be α and β

α = 1/4

β = -1

If a = 4,

⇒ b = 3

⇒ c = -1

Hence, the polynomial is 4x² + 3x – 1.

Question 16.

Find the quadratic polynomial, for the zeroes given in each case.

1/2, 3/2

Answer:

Let the quadratic polynomial be ax² + bx + c

And, its zeroes be α and β

If a = 4,

⇒ b = -8

⇒ c = 3

Hence, the polynomial is 4x² – 8x + 3.

Question 17.

Verify that 1, -1 and -3 are the zeroes of the cubic polynomial x³ + 3x² – x - 3 and check the relationship between zeroes and the coefficients.

Answer:

P(x) = x³ + 3x² – x – 3

For x = 1,

⇒ P(1) = 1³ + 3(1)² – 1 – 3

⇒ P(1) = 1+ 3 – 1 – 3

⇒ P(1) = 0

For x = -1,

⇒ P(-1) = (-1)³ + 3(-1)² – (-1) – 3

⇒ P(-1) = -1+ 3 + 1 – 3

⇒ P(-1) = 0

For x = -3,

⇒ P(-3) = (-3)³ + 3(-3)² – (-3) – 3

⇒ P(-3) = -27+ 27 + 3 – 3

⇒ P(-3) = 0

Now,

Sum of zeroes = 1 + (-1) + (-3) = -3

Hence,

And,

Product of zeroes = 1 × (-1) × (-3) = -3

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Exercise 3.4

Question 1.

Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

p(x) = x³ – 3x² + 5x – 3, g(x) = x² - 2

Answer:

p(x) = x³ – 3x² + 5x – 3

_{g(x) = x}² – 2

On dividing them,

The quotient is x – 3.

And,

The remainder is 7x – 9.

Question 2.

Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:
p(x) = x⁴ – 3x² + 4x + 5, g(x) = x² + 1 - x

Answer:

p(x) = x⁴ – 3x² + 4x + 5

g(x) = x² + 1 – x

On dividing them,

The quotient is x² + x - 3

And,

The remainder is 8

Question 3.

Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

p(x) = x⁴ – 5x + 6, g(x) = 2 - x²

Answer:

p(x) = x⁴ – 5x + 6

_{g(x) = 2 – x}²

On dividing them,

The quotient is –x² – 2

And,

The remainder is -5x + 10

Question 4.

Check in which case the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial.

t² – 3, 2t⁴ + 3t³ – 2t² – 9t - 12

Answer:

p(x) = 2t⁴ + 3t³ – 2t² - 9t – 12

_{g(x) = t}² – 3

On dividing them,

The remainder is 0

Hence, yes t² – 3 is a factor of 2t⁴ + 3t³ – 2t² - 9t – 12

Question 5.

Check in which case the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial.

x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2

Answer:

p(x) = 3x⁴ + 5x³ – 7x² + 2x + 2

_{g(x) = x}² + 3x + 1

On dividing them,

The remainder is 0

Hence, yes x² + 3x + 1 is a factor of 3x⁴ + 5x³ – 7x² + 2x + 2

Question 6.

Check in which case the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial.

x² - 3x + 1, x⁵ - 4x³ + x² + 3x + 1

Answer:

p(x) = x⁵ – 4x³ + x² + 3x + 1

_{g(x) = x}³ – 3x + 1

On dividing them,

The remainder is 2 ≠ 0

_{Hence, no x}³ – 3x + 1 is not a factor of x⁵ – 4x³ + x² + 3x + 1

Question 7.

Obtain all other zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5,

if two of its zeroes are  and 

Answer:

Two zeroes are  and 

is a factor

is a factor

is a factor

Now,

Therefore, 3x² + 6x +3 is also a factor

Dividing 3x² + 6x +3 by 3,

We get,

x² + 2x +1

Factorising x² + 2x +1,

x² + 2x +1 = 0

⇒ x² + x + x +1 = 0

⇒ x(x+ 1) + 1(x +1) = 0

⇒ (x+ 1)(x +1) = 0

⇒ x= -1, -1

Therefore, the other zeroes are -1 and -1.

Question 8.

On dividing x³ – 3x² + x + 2 by a polynomial g(x), the quotient and remainder were x - 2 and – 2x + 4 respectively. Find g(x)

Answer:

Dividend = x³ – 3x² + x + 2

Quotient = x – 2

Remainder = -2x + 4

Dividend = Divisor × Quotient + Remainder

⇒ x³ – 3x² + x + 2 = Divisor ×(x – 2) + (-2x + 4)

⇒ x³ – 3x² + x + 2 – (-2x + 4) = Divisor ×(x – 2)

⇒ x³ – 3x² + x + 2 + 2x – 4 = Divisor ×(x – 2)

⇒ x³ – 3x² + 3x – 2 = Divisor ×(x – 2)

Therefore, g(x) = x² – x + 1

Question 9.

Give examples of polynomials p(x), g(x), q(x) and r(x) which satisfy the division algorithm and

deg p (x) = deg q (x)

Answer:

Let g(x) = 2

And p(x) = 2x² – 2x + 14

Then, dividing p(x) by g(x) gives.

q(x) = x² – x + 7

and,

r(x) = 0

Deg(p(x)) = Deg(q(x)) = 2

Hence,

g(x) = 2

p(x) = 2x² – 2x + 14

q(x) = x² – x + 7

r(x) = 0

Question 10.

Give examples of polynomials p(x), g(x), q(x) and r(x) which satisfy the division algorithm and

deg q (x) = deg r (x)

Answer:

Let g(x) = x + 1

And p(x) = x² + 3

Then, dividing p(x) by g(x) gives

q(x) = x

and,

r(x) = -x + 3

Deg(r(x)) = Deg(q(x)) = 1

Hence,

g(x) = x + 1

p(x) = x² + 3

q(x) = x

r(x) = -x + 3

Question 11.

Give examples of polynomials p(x), g(x), q(x) and r(x) which satisfy the division algorithm and

deg r (x) = 0

Answer:

Let g(x) = 3

And p(x) = 3x + 3

Then, dividing p(x) by g(x) gives

q(x) = x + 1

and,

r(x) = 0

Deg(r(x)) = 0

Hence,

g(x) = 3

p(x) = 3x + 3

q(x) = x + 1

r(x) = 0