RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions is part of RBSE Solutions for Class 9 Maths. Here we have given RBSE Rajasthan Board Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions.

Board |
RBSE |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 3 |

Chapter Name |
Polynomial |

Exercise |
Additional Questions |

Number of Questions Solved |
36 |

Category |
RBSE Solutions |

## RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions

**Multiple Choice Questions**

Question 1.

A polynomial (RBSESolutions.com) of degree n in x has atmost:

(A) n terms

(B) (n – 1) terms

(C) (n + 1) terms

(D) \(\frac { n }{ 2 }\) terms

Solution

C

Question 2.

The zeroes of the (RBSESolutions.com) polynomial p(x) = x(x² – 1) are:

(A) 0, 1

(B) 0, – 1

(C) 0, – 1, 1

(D) ± 1

Solution

C

Question 3.

If

is equal to

Solution

B

Question 4.

√7 is a polynomial (RBSESolutions.com) of degree:

(A) 0

(B) 7

(C) \(\frac { 1 }{ 2 }\)

(D) 2

Solution

A

Question 5.

The coefficient of x in (x – 3)(x – 4) is:

(A) 7

(B) 1

(C) – 7

(D) 12

Solution

C

Question 6.

If p(x) = x² -3√2x + 1, then p(3√2) is equal to:

(A) 3√2

(B) 3√2 – 1

(C) 6√2 – 1

(D) 1

Solution

D

Question 7.

Which of the following is (RBSESolutions.com) a polynomial in one variable:

Solution

A

Question 8.

The degree of the zero polynomial is:

(A) 0

(B) 1

(C) any real number

(D) not exist

Solution

D

Question 9.

If x^{91} + 91 is divided by x + 1, then (RBSESolutions.com) the remainder is:

(A) 0

(B) 90

(C) 92

(D) None of these

Solution

B

Question 10.

Zero of the polynomial p(x) = √3x + 3 is

Solution

A

**Very Short Answer Type Questions**

Question 1.

Show that \(\frac { 1 }{ 2 }\) is zero (RBSESolutions.com) of the polynomial 2x^{2} + 7x – 4.

Solution.

Let p(x) = 2x^{2} + 7x – 4

Question 2.

Find the (RBSESolutions.com) remainder when

x^{4} + x^{3} – 2x^{2} + x + 1 is divided by x – 1.

Solution.

Let p(x) = x^{4} + x^{3} – 2x^{2} + x + 1

zero of x – 1 is 1

So, P(1) = (1)^{4} + (1)^{3} – 2(1)^{2} + (1) + 1 = 1 + 1 – 2 + 2 = 2

Hence, 2 is the remainder, when x^{4} + x^{3} – 2x^{2} + x + 1 is divided by x – 1.

Question 3.

If x – 1 is a factor of p(x) = 2x^{2} + kx + √2 then find (RBSESolutions.com) the value of k.

Solution.

x – 1 is a factor of p(x) = 2x^{2} + kx + √2

p(1) = 0 (by factor theorem)

⇒ 2(1)^{2} + k(1) + √2 = 0

⇒ 2 + k + √2 = 0

k = -(2 + √2)

Question 4.

Find the (RBSESolutions.com) value of m, if (x + 3) is a factor of 3x^{2} + mx + 6.

Solution.

Let p(x) = 3x^{2} + mx + 6

If x + 3 is a factor of p(x), then

p(-3) = 0

p(-3) = 3(- 3)^{2} + m(-3) + 6 = 0

⇒ 3 x 9 – 3m + 6 = 0

⇒ 27 – 3m + 6 = 0

⇒ 3m = 33

⇒ m = 11

Question 5.

Verify that 1 is not a zero (RBSESolutions.com) of the polynomial 4x^{4} – 3x^{3} + 2x^{2} – 5x + 1.

Solution.

Let p(x) = 4x^{4} – 3x^{3} + 2x^{2} – 5x + 1

Now

P(1) = 4(1)^{4} – 3(1)^{3} + 2(1)^{2} – 5(1) + 1 = 4 – 3 + 2 – 5 + 1 = -1

p(1) ≠ 0

1 is not a zero of p(x).

Remainder Theorem Calculator. Various types of online remainder theorem calculators are available online.

Question 6.

Using remainder theorem, find the (RBSESolutions.com) value of k so that (4x^{2} + kx – 1) leaves the remainder 2 when divided by (x – 3).

Solution.

Let p(x) = 4x^{2} + kx – 1

Here we are given that p(3) = 2

i.e. 4(3)^{2} + k(3) -1 = 2

⇒ 4 x 9 + 3k – 1 = 2

⇒ 36 + 3k = 3

⇒ 3k = -33

⇒ k = -11

Question 7.

If (x – 2) is a factor of the (RBSESolutions.com) polynomial x^{4} – 2x^{3} + ax – 1, then find the value of a.

Solution.

(x – 2) is a factor of the polynomial p(x) = x^{4} – 2x^{3} + ax – 1, then according to factor theorem p(2) = 0

⇒ (2)^{4} – 2(2)^{3} + a(2) – 1 = 0

⇒ 2a – 1 = 0

⇒ a = \(\frac { 1 }{ 2 }\)

Question 8.

Find the value of

(x – y)^{3} + (y – z)^{3} + (z – x)^{3}.

Solution.

Let a = x – y, b = y – z, c = z – x

Here a + b + c = x – y + y – z + z – x = 0

a^{3} + b^{3} + c^{3} = 3abc

⇒ (x – y)^{3} + (y – z)^{3} + (z – x)^{3} = 3(x – y)(y – z)(z – x).

Question 9.

If a, b, c are all non-zero and a + b + c = 0 then find the (RBSESolutions.com) value of \(\frac { { a }^{ 2 } }{ bc } +\frac { { b }^{ 2 } }{ ca } +\frac { { c }^{ 2 } }{ ab }\)

Solution.

We know that a^{3} + b^{3} + c^{3} – 3abc = (a + b + c)(a^{2} + b^{2} + c^{2} – ab – bc – ca)

a + b + c then a^{3} + b^{3} + c^{3} = 3abc …(1)

Dividing equation (1) by abc, we get

Question 10.

Solution.

**Short Answer Type Questions**

Question 1.

Find the remainder (RBSESolutions.com) obtained on dividing p(x) = x^{3} + 1 by x + 1.

Solution.

By long division method

Here we find that remainder is zero.

Also p(-1) = (-1)^{3} + 1 = -1 + 1 = 0,

which is equal to the (RBSESolutions.com) remainder obtained by actual division.

Question 2.

Find the value of

4a^{2} + b^{2} + 25c^{2} + 4ab – 10bc – 20ca when a = 1, b = 2 and c = 3.

Solution.

We have,

4a^{2} + b^{2} + 25c^{2} + 4ab – 10bc – 20ca = (2a)^{2} + (b)^{2} + (-5c)^{2} – 2(2a)(b) + 2(b)(-5c) + 2(-5c)(2a) = (2a + b – 5c)^{2}

It is given that a = 1, b = 2 and c = 3,

so (2a + b – 5c)^{2} = (2 x 1 + 2 – 5 x 3)^{2} = (2 + 2 – 15)^{2} = (4 – 15)^{2} = (-11)^{2} = 121

Question 3.

If 3a – 2b = 11 and ab = 12, then find (RBSESolutions.com) the value of 27a^{3} – 8b^{3}.

Solution.

We know that

(a – b)^{3} = a^{3} – b^{3} – 3ab(a – b)

Using this identity,

(3a – 2b)^{3} = (3a)^{3} – (2b)^{3} -3 x 3a x 2b (3a – 2b)

⇒ (3a – 2b)^{3} = 27a^{3} – 8b^{3} – 18ab(3a – 2b)

Now substituting 3a – 2b = 11 and ab = 12,

we get

(11)^{2} = 27a^{3} – 8b^{3} – 18 x 12 x 11

⇒ 1331 = 27a^{3} – 8b^{3} – 2376

⇒ 27a^{3} – 8b^{2} = 1331 + 2376

⇒ 27a^{3} – 8b^{3} = 3707

Question 4.

Use factor (RBSESolutions.com) theorem, show that (x + √2) is a factor of (2√2x^{2} + 5x + √2).

Solution.

Let p(x) = (2√2x^{2} + 5x + √2)

If x + √2 is a factor of p(x), then according to factor theorem p(√2) = 0.

p(-√2) = 2√2(-√2)^{2} + 5(-√2) + √2 = 2√2 x 2 – 5√2 + √2 = 5√2 – 5√2 = 0

p(-√2) = 0

(x + √2) is a factor p(x) i.e. (2√2 x^{2} + 5x + √2).

Question 5.

If a + b + c = 9 and ab + bc + ca = 23 then find (RBSESolutions.com) the value of a^{2} + b^{2} + c^{2}.

Solution.

We have, a + b + c = 9 Squaring both sides, we get

(a + b + c)^{2} = (9)^{2}

⇒ a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca = 81

⇒ a^{2} + b^{2} + c^{2} + 2(ab + bc + ca) = 81

⇒ a^{2} + b^{2} + c^{2} + 2 x 23 = 81

⇒ a^{2} + b^{2} + c^{2} = 81 – 46

⇒ a^{2} + b^{2} + c^{2} = 35 Thus, a^{2} + b^{2} + c^{2} = 35.

Question 6.

Simplify by using (RBSESolutions.com) suitable identity

Solution.

Question 7.

Factorise

(i) 216a^{3} – 125

(ii) a^{3} – b^{3} – a + b

Solution.

(i) we have,

216a^{3} – 125 = (6a)^{3} – (5)^{3}

[a^{3} – b^{3} = (a – b)(a^{2} + ab + b^{2})]

= (6a – 5) {(6a)^{2} + 6a x 5 + (5)^{2}}

= (6a – 5)(36a^{2} + 30a + 25)

(ii) We have,

a^{3} – b^{3} – a + b = a^{3} – b^{3} – (a – b)

= (a – b)(a^{2} + ab + b^{2}) – (a – b)

[a^{3} – b^{3} = (a – b)(a^{2} + ab + b^{2})]

= (a – b)(a^{2} + ab + b^{2} – 1)

Question 8.

If p = 4 – q show that p^{3} + q^{3} + 12pq = 64.

Solution.

We have,

p = 4 – q ⇒ p + q = 4 …(i)

Cubing both sides, (RBSESolutions.com) we get

(p + q)^{3} = (4)^{3}

⇒ p^{3} + 3p2q + 3pq^{2} + q^{3} = 64

⇒ p^{3} + q^{3} + 3pq(p + q) = 64

⇒ p^{3} + q^{3} + 3pq x 4 = 64 [using (i)]

⇒ p^{3} + q^{3} + 12pq = 64

Question 9.

Solution.

**Long Answer Type Questions**

Question 1.

If (x – 3) and (x – \(\frac { 1 }{ 2 }\)) are both (RBSESolutions.com) factors of ax^{2} + 5x + b, show that a = b.

Solution.

Let p(x) = ax^{2} + 5x + b

x – 3 is a factor of p(x).

p(3) = 0

Question 2.

For what values of a and b so that (RBSESolutions.com) the polynomial x^{3} + 10x^{2} + ax – 6 is exactly divisible by (x – 1) and (x + 2).

Solution.

Let p(x) = x^{3} + 10x^{2} + ax + b

p(x) i.e. x^{3} + 10x^{2} + ax + 6 is exactly divisible by (x – 1) and (x + 2)

Therefore, p(1) and p(-2) must equal to zero.

p(1) = (1)^{3} + 10(1)^{2} + a x 1 + b = 0

⇒ 1 + 10 + a + b = 0

⇒ a + b = – 11 …(i)

Also, p(-2) = 0

(-2)^{3} + 10(-2)^{2} + a x (- 2) + b = 0

-8 + 40 – 2a + b = 0

⇒ – 2a + b = – 32 …(ii)

Solving (i) and (ii), we get

a = 7 and b = -18.

Question 3.

If ax^{3} + bx^{2} + x – 6 has x + 2 as a factor and leaves (RBSESolutions.com) a remainder 4 when divided by (x – 2), find the values of a and b.

Solution.

Let p(x) = ax^{3} + bx^{2} + x – 6

(x + 2) is a factor of p(x)

⇒ p(-2) = 0 [∴ x + 2 = 0 ⇒ x = -2]

⇒ a(-2)^{3} + b(-2)^{2} + (-2) – 6 = 0

⇒ -8a + 4b – 2 – 6 = 0

⇒ -8a + 4b = 8

⇒ -2a + b = 2 …(i)

It is given that p(x) leaves the (RBSESolutions.com) remainder 4 when it divided by (x – 2) i.e. p(2) = 4.

⇒ a(2)^{3} + b(2)^{2} + (2) – 6 = 4

⇒ 8a + 4b – 4 = 4

⇒ 8a + 46 = 8

⇒ 2a + b = 2 …(ii)

Solving (i) and (ii), we get a = 0 and b = 2.

Question 4.

Evaluate

Solution.

In the given expression, we see that both numerator and denominator (RBSESolutions.com) is in the form a^{3} + b^{3} + c^{3} = 3abc because a + b + c = 0.

From numerator, we see that a^{2} – b^{2} + b^{2} – c^{2} + c^{2} – a^{2} = 0

⇒ (a^{2} – b^{2})^{3} + (b^{2} – c^{2})^{3} + (c^{2} – a^{2})^{3} = 3 (a^{2} – b^{2})(b^{2} – c^{2})(c^{2} – a^{2})

Similarly, from denominator,

a – b + b – c + c – a = 0

(a – b)^{3} + (b – c)^{3} + (c – a)^{3} = 3(a – b)(b – c)(c – a)

Value of the expression

Question 5.

If the polynomials (3x^{3} + ax^{2} + 3x + 5) and (4x^{3} + x^{2} – 2x + a) leaves the same (RBSESolutions.com) remainder when divided by (x – 2), find the value of a. Also the find the remainder in each-case.

Solution.

Let the given polynomials are p(x) = 3x^{3} + ax^{2} + 3x + 5 and f(x) = 4x^{3} + x^{2} – 2x + a

According to question p(2) = f(2)

3(2)^{3} + a(2)^{2} + 3(2) + 5 = 4(2)^{3} + (2)^{2} – 2(2) + a

⇒ 3 x 8 + 4a + 6 + 5 = 32 + 4 – 4 + a

⇒ 24 + 4a + 11 = 32 + a

⇒ 4a – a = 32 – 35

⇒ 3a = -3

⇒ a = -1

As the remainder is same, so p(2) or f(2) would (RBSESolutions.com) be same when a = – 1

p(2) = 3(2)^{3} + (-1)(2)^{2} + 3 x 2 + 5 [∴ a = -1]

= 24 – 4 + 6 + 5

= 35 – 4 = 31

Thus, a = – 1 and p(2) or f(2) = 31

Question 6.

Prove that (a + b + c)^{3} – a^{3} – b^{3} – c^{3} = 3(a + b)(b + c)(c + a).

Solution.

We have,

L.H.S.

= (a + b + c)^{3} – a^{3} -b^{3} – c^{3}

= {(a + b + c)^{3} – (a)^{3}} – (b^{3} + c^{3})

= (a + b + c – a){(a + b + c)^{2} + a(a + b + c) + a^{2}} – (b + c){b^{2} – bc + c^{2})

[∵ x^{3} – y^{3} = (x – y)(x^{2} + xy + y^{2}) and x^{3} + y^{3} = (x + y)(x^{2} – xy + y^{2})]

= (b + c){a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca + a^{2} + ab + ac + a^{2}} – (b + c)(b^{2} – bc + c^{2})

= (b + c)[3a^{2} + b^{2} + c^{2} + 3ab + 2bc + 3ca – b^{2} + bc – c^{2}]

= (b + c)[3a^{2} + 3ab + 3bc + 3ca]

= 3(b + c)[a^{2} + ab + bc + ca]

= 3(b + c)[a(a + b) + c(a + b)]

= 3(b + c)(a + b)(a + c)

= 3(a + b)(b + c)(c + a)

= R.H.S.

Question 7.

(i) For what (RBSESolutions.com) value of m is x^{3} – 2mn^{2} + 16 divisible by (x + 2)?

(ii) Show that (2x – 3) is a factor of x + 2x^{3} – 9x + 12.

Solution.

(i) Let p(x) = x^{3} – 2mn^{2} + 16

p(x) will be divisible by (x + 2) if

p(-2)= 0

p(-2) = (-2)^{3} – 2m(-2)^{2} + 16 = -8 – 8m + 16 = – 8m + 8

p(-2) = 0

⇒ -8m + 8 = 0

⇒ 8m = 8

⇒ m = 1

(ii) We know that (2x – 3) will be a factor of x + 2x^{3} – 9x + 12 if p(x) on dividing by 2x – 3, leaves (RBSESolutions.com) a remainder zero

So, the remainder obtained (RBSESolutions.com) on dividing p(x) by 2x – 3 is zero.

Hence, (2x – 3) is a factor of x + 2x^{3} – 9x + 12

We hope the given RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions will help you. If you have any query regarding RBSE Rajasthan Board Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions, drop a comment below and we will get back to you at the earliest.

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