RD sharma class 10 Chapter 2 Polynomials

Chapter 2 Polynomials Download Pdf of RD sharma class 10 Chapter 2 Polynomials

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Questions on polynomials Class 9 SET-3

Q1. Find the zeros of the polynomial

1. x² + 7x +  10
2. x² - 25

Q2. Factorize: 

1. 999² - 1
2. (10.2)³
3. 1002 × 998

Q3. Factorize:

1. x³ - 3x² - 9x - 5
2. x³ + 7x² - 21x - 27

Q4. Factorize:

1. \(3{x^2} + 27{y^2} + {z^2} - 18xy + 6\sqrt 3 yz - 2\sqrt 3 zx\)
2. 27x³ + 125y³
3. \(\dfrac{1}{{64}}{a^2} + {b^3} + 125{c^3} - \dfrac{{15}}{4}abc\)
4. \({x^3} - \dfrac{1}{{{x^3}}}\)
5. 8x³ - (2x - y)³
6. a⁶ - b⁶

Q5. Using factor theorem, show that (a - b) is the factor of a(b² - c²) + b(c² - a²) + c(a² - b²).

Q6. Factorize:

1. \(4\sqrt 3 {x^2} + 5x - 2\sqrt 3 \)
2. \(21{x^2} - 2x + \dfrac{1}{{21}}\)
3. 9(2a - b)² - 4(2a - b) - 13

Q7. Simplify and factorize (a + b + c)² - (a - b - c)² + 4b² - 4c² 

Q8. Factorize: (a² - b²)³ + (b² - c²)³ + (c² - a²)³

Q9. For what value of a is 2x³ + ax² + 11x + a + 3 exactly divisible by (2x - 1).

Q10. If x  - 2 is a factor of a polynomial f(x) = x⁵ - 3x⁴ - ax³ + 2ax + 4, then find the value of a.

Q11. Find the value of a and b so that x² - 4 is a factor of ax⁴ + 2x³ - 3x² + bx - 4

Q12. If x = 2 and x = 0 are zeroes of the polynomial 2x³ - 3x² + px + q, then find the value of p and q.

Q13. Find the value of a and b, so that x³ - ax² - 13x + b is exactly divisible by (x - 1) as well as (x + 3).

Q14. The polynomial x³ - mx² + 4x + 6 when divided by (x + 2) leaves remainder 14, find m.

Q15. If the polynomial ax³ + 3x² - 13 and 2x³  - 15x + a, when divided by (x - 2)  leave the same remainder. Find the value of a.

Q16. If both (x -  2) and \(\left( {x - \dfrac{1}{2}} \right)\)  are the factors of px² + 5x + r, show that p = r.

Q17. If f(x) = x⁴ - 2x³ + 3x² - ax + b is divided by x - 1 and x  + 1 the remainders are 5 and 19 respectively, then find a and b.

Q18. If A and B be the remainders when the polynomials x³ + 2x² - 5ax - 7 and x³ + ax² - 12x + 6 are divided by (x + 1) and (x - 2) respectively and 2A + B = 6, find the value of a.

Q19. Show that x + 1 and 2x - 3 are factors of 2x³ - 9x² + x + 12.

Q20. If sum of remainders obtained by dividing ax³ - 3ax² + 7x + 5 by (x + 1) is -36. find a.

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Questions on Polynomials class 9 SET -1

Extra Questions on Polynomials Class 9

Q1. Find the remainder when \({y^3} + {y^2} - 2y + 5\) is divided by y - 5.

Q2. Determine the remainder when p(x) = \({x^3} + 3{x^2} - 6x + 15\) is divided by x - 2.

Q3.  When \(f(x) = {x^4} - 2{x^3} + 3b{x^2} - ax\) is divided by x+1 and x - 1, we get  remainder as 19 and 5 respectively. Find the remainder if f(x) is divided by x - 3.

Q4.  What must be subtracted from \(4{x^4} - 2{x^3} - 6{x^2} + x - 5\) so that the result is exactly divisible by \(2{x^2} + 3x - 2\) ?

Q5.  If (x + 1) and (x - 1) both are factors of \(a{x^3} + {x^2} - 2x + b\), find a and b.

Q6.  Factorize each of the following expressions:

1. \(48{x^3} - 36{x^2}\)
2. \(5{x^2} - 15xy\)
3. \(15{x^3}{y^2}z - 25x{y^2}{z^2}\)

Q7. Factorize:

1. \(2{x^2}(x + y) - 3(x + y)\)
2. \(5xy(5x + y) - 5y(5x + y)\)
3. \(x({x^2} + {y^2} - {z^2}) + y({x^2} + {y^2} - {z^2}) + z({x^2} + {y^2} - {z^2})\)
4. \(ab({a^2} + {b^2} - {c^2}) + bc({a^2} + {b^2} - {c^2}) + ca({a^2} + {b^2} - {c^2})\)

Q8. Factorize each of the following expressions:

1. \(25{x^2}{y^2} - 20x{y^2}z + 4{y^2}{z^2}\)
2. \(4{x^2} - 4\sqrt 7 x + 7\)
3. \(\dfrac{{{a^2}}}{{{b^2}}} + 2 + \dfrac{{{b^2}}}{{{a^2}}}\)
4. \(4{a^2} + 12ab + 9{b^2} - 8a - 12b\)

Q9. Factorize each of the following:

1. \(25{x^2} - 36{y^2}\)
2. \(2ab - {a^2} - {b^2} + 1\)
3. \(36{a^2} - 12a + 1 - 25{b^2}\)
4. \({a^4} - 81{b^4}\)
5. \({a^{12}}{b^4} - {a^4}{b^{12}}\)
6. \(4{x^2} - 9{y^2} - 2x - 3y\)

Q10. Factorize by completing the square.

1. \({a^4} + {a^2} + 1\)
2. \({y^4} + 5{y^2} + 9\)
3. \({x^4} + 4\)
4. \({x^4} + 4{x^2} + 3\)

Q11. Factorize by completing the square.

1. \({a^3} - 27\)
2. \(1 - 27{x^3}\)
3. \(8{x^3} - {(2x - 3y)^3}\)
4. \({a^8} - {a^2}{b^6}\)
5. \({a^3} - 5\sqrt 5 {b^3}\)

Q12. Factorize the following:

1. \(16{p^3}{q^2} + 54{r^3}\)
2. \(\dfrac{{{a^3}}}{8} + 8{b^3}\)
3. \(2\sqrt 2 {a^3} + 3\sqrt 3 {b^3}\)
4. \(8{a^4}b + \dfrac{1}{{125}}a{b^4}\)
5. \({a^7} - 64a\)

Q13.Q13. Factorize:

1. \({x^3} + 9{x^2} + 27x + 27\)
2. \({x^3} - 9{x^2}y + 27x{y^2} - 27{y^3}\)

Q14. Using identities, find the value of

1. 101²
2. 98²
3. (0.98)²
4. 101 × 99
5. 190 × 190 - 10 × 10

Q15. Expand using suitable identity

1. (x + 5y + 6z)²
2. (2a - 3b + 4c)²
3. ( - a + 6b + 5c)²
4. (- p + 4q - 3r)²

Q16. Expand using suitable identity

1. (2x + 5y)³
2. (5p - 3q)³
3. (- a + 2b)³

Q17. Evaluate using identities

1. 102³
2. 99³

Q18. Simplify : 

1. (2a + b)³ + (2a - b)³
2. (4x + 5y)³ - (4x - 5y)³

Q19. Factorize:

1. 30x³y + 24x²y² - 6xy
2. 5x(a - b) + 6y(a - b)

Q20. Factorize:

1. 9x² - y²
2. (3 - x)² - 36x²
3. (2x - 3y)² - (3x + 4y)²
4. 16x⁴ - y⁴

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Class 9 Chapter 2 (Polynomials) Class Notes

 

Notes of Polynomials

Topics in the Chapter

• Polynomials
• Classification of polynomials on the basis of number of terms
• Degree of a Polynomial
• Values of polynomials at different points
• Zeroes of a polynomial
• Identity: (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx
• Remainder Theorem
• Factor Theorem
• Identity: x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx)

Polynomials

→ An algebraic expression in which the exponents of the variables are non-negative integers
are called polynomials.
For example: 3x⁴ + 2x³ + x + 9, 3x² etc.

Constant polynomial
→ A constant polynomial is of the form p(x) = k, where k is a real number.
For example, –9, 10, 0 are constant polynomials.

Zero polynomial
A constant polynomial ‘0’ is called zero polynomial.

General form of a polynomial
A polynomial of the form where are p(x) = a_nxⁿ + a_n-1x^n-1 + .... + a₁x + a₀x where a₀, a₁... a_r, are constant and a_n≠ 0.
Here, a₀, a₁... a_n are the respective coefficients of x⁰,x¹, x² ... xⁿ and n is the power of
the variable x.
a_nxⁿ + a_n-1x^n-1 - a₀ and are called the terms of p(x).

Classification of polynomials on the basis of number of terms

• A polynomial having one term is called a monomial. e.g. 3x, 25t³ etc.

• A polynomial having one term is called a monomial. e.g. 2t-6, 3x⁴ +2x etc.

• A polynomial having one term is called a monomial. e.g. 3x⁴ + 8x + 7 etc.

Degree of a Polynomial

• The degree of a polynomial is the highest exponent of the variable of the polynomial.
For example, the degree of polynomial 3x⁴ + 2x³ + x + 9 is 4.

• The degree of a term of a polynomial is the value of the exponent of the term.

Classification of polynomial according to their degrees 

• A polynomial of degree one is called a linear polynomial e.g. 3x+2, 4x, x+9.

• A polynomial of degree is called a quadratic polynomial. e.g. x²+ 9, 3x² + 4x + 6

• A polynomial of degree three is called a cubic polynomial e.g. 10x³+ 3, 9x³

Note: The degree of a non-zero constant polynomial is zero and the degree of a zero polynomial is not defined.

Values of polynomials at different points

• A polynomial is made up of constants and variables. Hence, the value of the polynomial changes
as the value of the variable in the polynomial changes.
• Thus, for the different values of the variable x, we get different values of the polynomial.


Zeroes of a polynomial

• A real number a is said to be the zero of polynomial p(x) if p(a) = 0, In this case, a is also called the root of the equation p(x) = 0.
Note:
• The maximum number of roots of a polynomial is less than or equal to the degree of the polynomial.
• A non-zero constant polynomial has no zeroes.
• A polynomial can have more than one zero.

Division of a polynomial by a monomial using long division method

Example: Divide x⁴ -2x³- 2x²+7x-15 by x-2

Steps of the Division of a Polynomial with a Non –Zero Polynomial

Divide x² - 3x -10 by 2 + x

Step 1:  Write the dividend and divisor in the descending order i.e. in the standard form. x² - 3x -10 and x + 2

Divide the first term of the dividend with the first term of the divisor.

x²/x = x this will be the first term of the quotient.

Step 2: Now multiply the divisor with this term of the quotient and subtract it from the dividend.

Step 3: Now the remainder is our new dividend so we will repeat the process again by dividing the dividend with the divisor.

Step 4: – (5x/x) = – 5 

Step 5: 

The remainder is zero.

Hence x² - 3x – 10 = (x + 2)(x - 5) + 0

Dividend = (Divisor × Quotient) + Remainder

Remainder Theorem says that if p(x) is any polynomial of degree greater than or equal to one and let ‘t’ be any real number and p (x) is divided by the linear polynomial x – t, then the remainder is p(t).

As we know that

P(x) = g(x) q(x) + r(x)

If p(x) is divided by (x-t) then

If x = t

P (t) = (t - t).q (t) + r = 0

To find the remainder or to check the multiple of the polynomial we can use the remainder theorem.

Example:

What is the remainder if a⁴ + a³ – 2a² + a + 1 is divided by a – 1.

Solution:

P(x) = a⁴ + a³ – 2a² + a + 1

To find the zero of the (a – 1) we need to equate it to zero.

a -1 = 0

a = 1

p (1) = (1)⁴ + (1)³ – 2(1)² + (1) + 1

= 1 + 1 – 2 + 1 + 1

= 2

So by using the remainder theorem, we can easily find the remainder after the division of polynomial.

Factor Theorem

Factor theorem says that if p(y) is a polynomial with degree n≥1 and t is a real number, then

1. (y - t) is a factor of p(y), if p(t) = 0, and

2. P (t) = 0 if (y – t) is a factor of p(y).

Example: 1

Check whether g(x) = x – 3 is the factor of p(x) = x³ - 4x² + x + 6 using factor theorem.

Solution:

According to the factor theorem if x - 3 is the factor of p(x) then p(3) = 0, as the root of x – 3 is 3.

P (3) = (3)³ - 4(3)² + (3) + 6

= 27 – 36 + 3 + 6 = 0

Hence, g(x) is the factor of p(x).

Example: 2

Find the value of k, if x – 1 is a factor of p(x) = kx² – √2x + 1

Solution:

As x -1 is the factor so p(1) = 0

Factorization of Polynomials

Factorization can be done by three methods

1. By taking out the common factor

If we have to factorize x² –x then we can do it by taking x common.

x(x – 1) so that x and x-1 are the factors of x² – x.

2. By grouping

ab + bc + ax + cx = (ab + bc) + (ax + cx)

= b(a + c) + x(a + c)

= (a + c)(b + x)

3. By splitting the middle term

x² + bx + c = x² + (p + q) + pq

= (x + p)(x + q)

This shows that we have to split the middle term in such a way that the sum of the two terms is equal to ‘b’ and the product is equal to ‘c’.

Example: 1

Factorize 6x² + 17x + 5 by splitting the middle term.

Solution:

If we can find two numbers p and q such that p + q = 17 and pq = 6 × 5 = 30, then we can get the factors.

Some of the factors of 30 are 1 and 30, 2 and 15, 3 and 10, 5 and 6, out of which 2 and 15 is the pair which gives p + q = 17.

6x² + 17x + 5 =6 x² + (2 + 15) x + 5

= 6 x² + 2x + 15x + 5

= 2 x (3x + 1) + 5(3x + 1)

= (3x + 1) (2x + 5)

Algebraic Identities

1. (x + y)2 = x2 + 2xy + y2

2. (x - y)2 = x2 - 2xy + y2

3. (x + y) (x - y) = x2 - y2

4. (x + a) (x + b) = x2 + (a + b)x + ab

5. (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

6. (x + y)3 = x3 + y3 + 3xy(x + y) = x3+ y3 + 3x2y + 3xy2

7. (x - y)3 = x3- y3 - 3xy(x - y) = x3 - y3 - 3x2y + 3xy2

8. x3 + y3 = (x + y)(x2 – xy + y2)

9. x3 - y3 = (x - y)(x2 + xy + y2)

10. x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz - zx)       x3 + y3 + z3 = 3xyz if x + y + z = 0

Example: 2

Factorize 8x³ + 27y³ + 36x²y + 54xy²

Solution:

The given expression can be written as

= (2x)³ + (3y)³ + 3(4x²) (3y) + 3(2x) (9y²)

= (2x)³ + (3y)³ + 3(2x)²(3y) + 3(2x)(3y)²

= (2x + 3y)³ (Using Identity VI)

= (2x + 3y) (2x + 3y) (2x + 3y) are the factors.

Example: 3

Factorize 4x² + y² + z² – 4xy – 2yz + 4xz.

Solution:

4x² + y² + z² – 4xy – 2yz + 4xz = (2x)² + (–y)² + (z)² + 2(2x) (-y)+ 2(–y)(z) + 2(2x)(z)

= [2x + (- y) + z]² (Using Identity V)

= (2x – y + z)² = (2x – y + z) (2x – y + z)

Identity: (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

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