CBSE Class 9 Mathematics Polynomials Worksheet Set 05

Chapter - 2 (Polynomials): Key Concepts

Constants: A quantity or symbol retaining an unchanging numerical value is called a constant.
Example: \( 7, 3, -2, \frac{3}{7} \) are all constants.

Variables: A symbol that can take on varying numeric quantities is known as a variable.
Example: In the formula for the circumference of a circle, \( C = 2\pi r \), \( 2 \) and \( \pi \) represent constant values, whereas \( C \) (circumference) and \( r \) (radius) act as variables.

Algebraic Expressions: A mathematical collection of constants and variables linked via basic arithmetic operations like addition (\( + \)), subtraction (\( - \)), multiplication (\( \times \)), and division (\( \div \)) is termed an algebraic expression.
Example: \( 4 + 9x - 5x^2y + \frac{3}{8}xy \).

Terms: The individual sections of an algebraic expression demarcated by plus (\( + \)) or minus (\( - \)) symbols are referred to as the terms of that expression.
Example: The expression \( x^3 + 2x^2y + 4xy^2 + y^3 + 7 \) contains 5 separate terms: \( x^3 \), \( 2x^2y \), \( 4xy^2 \), \( y^3 \), and \( 7 \).

Polynomials: An algebraic expression in which all variable exponents consist exclusively of non-negative integers (whole numbers) is termed a polynomial.
(i) \( 5x^3 - 4x^2 - 6x - 3 \) is a polynomial in one variable \( x \).
(ii) \( 5 + 8x^{3/2} + 4x^{-2} \) is an algebraic expression, but not a polynomial because its powers include a fraction and a negative integer.
Polynomials are commonly represented as \( p(x) \), \( q(x) \), and \( r(x) \).

Coefficients: In the polynomial \( x^3 + 3x^2 + 3x + 1 \), the numerical factors multiplying \( x^3 \), \( x^2 \), and \( x \) are \( 1 \), \( 3 \), and \( 3 \) respectively, while \( +1 \) serves as the constant term.

Degree of a Polynomial in One Variable: In a single-variable polynomial, the highest exponent associated with the variable determines the degree of that polynomial.

Classification of Polynomials on the Basis of Degree:

Classification of Polynomials on the Basis of Number of Terms:

Constant Polynomial: A polynomial featuring only one single term that is a constant is designated a constant polynomial. The degree of any non-zero constant polynomial is zero.

Zero Polynomial: A polynomial made up of solely the single constant zero is identified as the zero polynomial. The degree of a zero polynomial is not defined.

Zeroes of a Polynomial: Let \( p(x) \) be a polynomial. If \( p(\alpha) = 0 \), then the real number \( \alpha \) is defined as a zero of \( p(x) \).
Remark: Discovering the zeroes of \( p(x) \) corresponds to finding solutions to the equation \( p(x) = 0 \).

Remainder Theorem: Let \( f(x) \) represent a polynomial possessing degree \( n \ge 1 \), and suppose \( a \) is any real number. Whenever \( f(x) \) is divided by the linear polynomial \( (x - a) \), the resulting remainder is \( f(a) \).

Factor Theorem: Let \( f(x) \) be a polynomial of degree \( n \ge 1 \) and \( a \) be any real number.
(i) If \( f(a) = 0 \), then \( (x - a) \) is a factor of \( f(x) \).
(ii) If \( (x - a) \) is a factor of \( f(x) \), then \( f(a) = 0 \).
Factor: A polynomial \( p(x) \) is described as a factor of \( q(x) \) whenever \( p(x) \) divides \( q(x) \) leaving zero remainder.

Factorization: Decomposing a given polynomial into a product of polynomials, each possessing a strictly lower degree such that none of these factors can be broken down further, is defined as factorization.
Example: \( x^2 - 16 = (x + 4)(x - 4) \).

Methods of Factorization:
1. Extracting common factors: \( 36q^3b - 60a^2bc = 12a^2b(3a - 5c) \)
2. Grouping terms:
\( ab + bc + ax + cx = (ab + bc) + (ax + cx) \)
\( = b(a + c) + x(a + c) \)
\( = (a + c)(b + x) \)
3. Splitting the middle term of quadratic trinomials:
\( x^2 + (p + q)x + pq = (x + p)(x + q) \)

Identity: An identity is an equality (algebraic or trigonometric) that holds true regardless of the values substituted for its variables.

Standard Algebraic Identities:
(i) \( (x + y)^2 = x^2 + 2xy + y^2 \)
(ii) \( (x - y)^2 = x^2 - 2xy + y^2 \)
(iii) \( x^2 - y^2 = (x - y)(x + y) \)
(iv) \( (x + a)(x + b) = x^2 + (a + b)x + ab \)
(v) \( (x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx \)
(vi) \( (x + y)^3 = x^3 + y^3 + 3xy(x + y) \)
(vii) \( (x - y)^3 = x^3 - y^3 - 3xy(x - y) \)
(viii) \( x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) \)
Special Case: If \( x + y + z = 0 \), then \( x^3 + y^3 + z^3 = 3xyz \).

Section - A

Question 1. Which of the following expressions is polynomial?
(i) \( x^5 - 2x^3 + x + 7 \)
(ii) \( y^3 - \sqrt{3}y \)
(iii) \( 5\sqrt{z} - 6 \)
(iv) \( x - \frac{1}{x} \)
(v) \( x^{108} - 1 \)
(vi) \( \sqrt[3]{x} - 27 \)
(vii) \( x^{-2} + 2x^{-1} + 3 \)
Answer: An expression qualifies as a polynomial when the exponents of every variable term are strictly non-negative integers (whole numbers).
(i) \( x^5 - 2x^3 + x + 7 \): The exponents \( 5, 3, 1 \) are all non-negative integers. Hence, it is a polynomial.
(ii) \( y^3 - \sqrt{3}y \): The powers of \( y \) are \( 3 \) and \( 1 \), which are non-negative integers (the coefficient \( \sqrt{3} \) being irrational does not affect the variable's power). Hence, it is a polynomial.
(iii) \( 5\sqrt{z} - 6 = 5z^{1/2} - 6 \): The power \( \frac{1}{2} \) is a fraction, not an integer. Hence, it is not a polynomial.
(iv) \( x - \frac{1}{x} = x - x^{-1} \): The exponent \( -1 \) is negative. Hence, it is not a polynomial.
(v) \( x^{108} - 1 \): The exponent \( 108 \) is a whole number. Hence, it is a polynomial.
(vi) \( \sqrt[3]{x} - 27 = x^{1/3} - 27 \): The power \( \frac{1}{3} \) is fractional. Hence, it is not a polynomial.
(vii) \( x^{-2} + 2x^{-1} + 3 \): The variable possesses negative powers (\( -2, -1 \)). Hence, it is not a polynomial.
Therefore, expressions (i), (ii), and (v) are polynomials.
In simple words: Expressions count as polynomials only when the variable powers are positive whole numbers or zero.

Exam Tip: Be mindful of square roots on variables versus constants: roots over variables disqualify a polynomial, but irrational constant coefficients are completely allowed.

Question 2. Write the degree of each of the following polynomial.
(i) \( 2x - \sqrt{5} \)
(ii) \( 3 - x + x^2 - 6x^3 \)
(iii) \( 9 \)
(iv) \( 8x^4 - 36x + 5x^7 \)
(v) \( x^9 - x^5 + 3x^{10} + 8 \)
(vi) \( 2 - 3x^2 \)
Answer: The degree represents the maximum power of the variable present in the given polynomial:
(i) In \( 2x^1 - \sqrt{5} \), the highest power of \( x \) is \( 1 \). Degree = \( 1 \).
(ii) In \( 3 - x + x^2 - 6x^3 \), the greatest exponent is \( 3 \). Degree = \( 3 \).
(iii) The constant \( 9 = 9x^0 \) has variable power \( 0 \). Degree = \( 0 \).
(iv) In \( 8x^4 - 36x + 5x^7 \), the largest exponent is \( 7 \). Degree = \( 7 \).
(v) In \( x^9 - x^5 + 3x^{10} + 8 \), the highest power is \( 10 \). Degree = \( 10 \).
(vi) In \( 2 - 3x^2 \), the maximum power is \( 2 \). Degree = \( 2 \).
In simple words: Find the biggest exponent written on the variable to determine the polynomial's degree.

Exam Tip: The terms may not be arranged in descending order; scan the entire expression so you do not mistake the first exponent for the highest.

Question 3. (i) Give an example of a binomial of degree 27.
(ii) Give an example of a monomial of degree 16.
(iii) Give an example of trinomial of degree 3.
Answer:
(i) A binomial possesses two non-zero terms. A suitable binomial of degree 27 is:
\( x^{27} - 5 \) (or \( 3x^{27} + 2x \)).
(ii) A monomial contains only one term. A suitable monomial of degree 16 is:
\( 5x^{16} \) (or \( -y^{16} \)).
(iii) A trinomial has three non-zero terms. A suitable trinomial of degree 3 is:
\( 2x^3 + x^2 - 4 \) (or \( x^3 - 3x + 1 \)).
In simple words: Write expressions matching the required number of terms while ensuring the highest variable power matches the requested degree.

Exam Tip: Ensure that the terms in your example cannot be combined through simplification, otherwise a supposed trinomial may collapse into a binomial.

Section - B

Question 4. If \( p(x) = 5 - 4x + 2x^2 \) find (i) \( p(0) \) (ii) \( p(3) \) (iii) \( p(-2) \)
Answer: Substitute the given input values into \( p(x) = 5 - 4x + 2x^2 \):
(i) For \( x = 0 \):
\( p(0) = 5 - 4(0) + 2(0)^2 = 5 - 0 + 0 = 5 \).
(ii) For \( x = 3 \):
\( p(3) = 5 - 4(3) + 2(3)^2 = 5 - 12 + 2(9) = 5 - 12 + 18 = 11 \).
(iii) For \( x = -2 \):
\( p(-2) = 5 - 4(-2) + 2(-2)^2 = 5 + 8 + 2(4) = 5 + 8 + 8 = 21 \).
In simple words: Put each given number in place of x and carry out the arithmetic step by step.

Exam Tip: Pay special attention to signs when evaluating negative inputs: \( (-2)^2 = +4 \) and \( -4(-2) = +8 \).

Question 5. Find the zeros of the polynomials given below :
(i) \( p(x) = x - 5 \)
(ii) \( q(x) = x + 4 \)
(iii) \( h(x) = 6x - 1 \)
(iv) \( p(x) = ax + b \)
(v) \( r(x) = x^2 + 3x \)
(vi) \( l(x) = x^2 + 2x + 1 \)
Answer: The zeroes of a polynomial are obtained by setting the polynomial equal to zero and solving for \( x \):
(i) \( x - 5 = 0 \implies x = 5 \).
(ii) \( x + 4 = 0 \implies x = -4 \).
(iii) \( 6x - 1 = 0 \implies 6x = 1 \implies x = \frac{1}{6} \).
(iv) \( ax + b = 0 \implies ax = -b \implies x = -\frac{b}{a} \) (where \( a \neq 0 \)).
(v) Factorize \( r(x) \):
\( x^2 + 3x = 0 \implies x(x + 3) = 0 \).
So, \( x = 0 \) or \( x + 3 = 0 \implies x = -3 \). Zeroes are \( 0, -3 \).
(vi) Factorize \( l(x) \):
\( x^2 + 2x + 1 = 0 \implies (x + 1)^2 = 0 \).
So, \( x + 1 = 0 \implies x = -1, -1 \). The repeated zero is \( -1 \).
In simple words: Equate each expression to zero and solve the resulting equation for the variable.

Exam Tip: A quadratic polynomial has two zeroes; do not forget to state both roots for items (v) and (vi).

Question 6. Find the remainder when \( f(x) = 12x^3 - 13x^2 - 5x + 7 \) is divided by \( (3x + 2) \)?
Answer: Applying the Remainder Theorem, the remainder when dividing \( f(x) \) by \( (3x + 2) \) equals \( f\left(-\frac{2}{3}\right) \).
Find the zero of the divisor:
\( 3x + 2 = 0 \)
\( \implies x = -\frac{2}{3} \)
Now calculate \( f\left(-\frac{2}{3}\right) \):
\( f\left(-\frac{2}{3}\right) = 12\left(-\frac{2}{3}\right)^3 - 13\left(-\frac{2}{3}\right)^2 - 5\left(-\frac{2}{3}\right) + 7 \)
\( = 12\left(-\frac{8}{27}\right) - 13\left(\frac{4}{9}\right) + \frac{10}{3} + 7 \)
Simplify the fractions:
\( = -\frac{32}{9} - \frac{52}{9} + \frac{10}{3} + 7 \)
Combine the terms with common denominator \( 9 \):
\( = \frac{-32 - 52 + 30 + 63}{9} \)
\( = \frac{-84 + 93}{9} = \frac{9}{9} = 1 \).
Therefore, the remainder is \( 1 \).
In simple words: Substitute \( x = -2/3 \) into the polynomial to compute the remainder without doing long division.

Exam Tip: Reducing \( 12\left(-\frac{8}{27}\right) \) to \( -\frac{32}{9} \) simplifies finding the common denominator significantly.

Question 7. Show that \( (x + 5) \) is a factor of the polynomial \( f(x) = x^3 + x^2 + 3x + 115 \)
Answer: In accordance with the Factor Theorem, \( (x + 5) \) will be a factor of \( f(x) \) if evaluating at \( x = -5 \) yields \( f(-5) = 0 \).
Zero of divisor \( x + 5 = 0 \implies x = -5 \).
Substitute \( x = -5 \) into \( f(x) \):
\( f(-5) = (-5)^3 + (-5)^2 + 3(-5) + 115 \)
\( = -125 + 25 - 15 + 115 \)
Combine positive and negative values:
\( = (-125 - 15) + (25 + 115) \)
\( = -140 + 140 = 0 \).
Since \( f(-5) = 0 \), it is proved that \( (x + 5) \) is a factor of the given polynomial.
In simple words: Because inserting -5 gives an outcome of zero, the Factor Theorem confirms that \( x + 5 \) divides it evenly.

Exam Tip: Always state "By Factor Theorem, since \( f(-5) = 0 \), \( (x + 5) \) is a factor" to conclude proof questions cleanly.

Question 8. Find the value of a for which \( (x - a) \) is a factor of the polynomial. \( f(x) = x^5 - a^2x^3 + 2x + a - 3 \)
Answer: Under the Factor Theorem, if \( (x - a) \) divides \( f(x) \) completely, then \( f(a) \) must equal \( 0 \).
Set \( f(a) = 0 \):
\( (a)^5 - a^2(a)^3 + 2(a) + a - 3 = 0 \)
\( a^5 - a^5 + 2a + a - 3 = 0 \)
\( 3a - 3 = 0 \)
\( 3a = 3 \)
\( \implies a = 1 \).
Thus, the value of \( a \) is \( 1 \).
In simple words: Replace x with a, set the entire expression equal to zero, and solve the remaining simple equation for a.

Exam Tip: Notice how the higher powers \( a^5 - a^5 \) eliminate each other, reducing the problem to a straightforward linear equation.

Section - C

Question 9. Factorize the following expressions.
(i) \( 5x^2 - 20xy \)
(ii) \( 5a(b + c) - 7b(b + c) \)
(iii) \( x(x - y)^2 + 3x^2y(x - y) \)
(iv) \( 6ab - b^2 + 12ac - 2bc \)
(v) \( x^2 + \frac{1}{x^2} + 2 - 2x - \frac{2}{x} \)
Answer:
(i) Factor out the common term \( 5x \):
\( 5x^2 - 20xy = 5x(x - 4y) \).

(ii) Factor out the shared binomial \( (b + c) \):
\( 5a(b + c) - 7b(b + c) = (b + c)(5a - 7b) \).

(iii) Take out the highest common factor \( x(x - y) \):
\( x(x - y)^2 + 3x^2y(x - y) = x(x - y)[(x - y) + 3xy] \)
\( = x(x - y)(x - y + 3xy) \).

(iv) Group the terms in pairs:
\( (6ab + 12ac) - (b^2 + 2bc) \)
\( = 6a(b + 2c) - b(b + 2c) \)
Factor out \( (b + 2c) \):
\( = (b + 2c)(6a - b) \).

(v) Group the terms strategically:
\( \left(x^2 + \frac{1}{x^2} + 2\right) - \left(2x + \frac{2}{x}\right) \)
The first bracket is the expansion of \( \left(x + \frac{1}{x}\right)^2 \):
\( = \left(x + \frac{1}{x}\right)^2 - 2\left(x + \frac{1}{x}\right) \)
Take out the common factor \( \left(x + \frac{1}{x}\right) \):
\( = \left(x + \frac{1}{x}\right)\left(x + \frac{1}{x} - 2\right) \).
In simple words: Pull out common multipliers or group matching pairs together to factorize each expression.

Exam Tip: Recognize \( x^2 + \frac{1}{x^2} + 2 \) as \( \left(x + \frac{1}{x}\right)^2 \) immediately; this is a frequently tested grouping pattern.

Question 10. Factorize :
(i) \( 9x^2 - 16y^2 \)
(ii) \( x^3 - x \)
Answer:
(i) Apply the difference of two squares identity, \( a^2 - b^2 = (a - b)(a + b) \):
\( 9x^2 - 16y^2 = (3x)^2 - (4y)^2 \)
\( = (3x - 4y)(3x + 4y) \).

(ii) First extract the common variable \( x \):
\( x^3 - x = x(x^2 - 1) \)
Now apply the identity \( a^2 - b^2 = (a - b)(a + b) \) to \( (x^2 - 1^2) \):
\( = x(x - 1)(x + 1) \).
In simple words: Rewrite expressions as differences of perfect squares and apply \( a^2 - b^2 = (a - b)(a + b) \).

Exam Tip: In item (ii), remember to factor out the common factor \( x \) first before trying to apply any algebraic identity.

Question 11. Factorize: \( a(a - 1) - b(b - 1) \)
Answer: Expand the brackets and regroup terms:
\( a(a - 1) - b(b - 1) = a^2 - a - b^2 + b \)
Rearrange into pairs:
\( = (a^2 - b^2) - (a - b) \)
Factor the difference of squares in the first part:
\( = (a - b)(a + b) - 1(a - b) \)
Extract the common binomial factor \( (a - b) \):
\( = (a - b)(a + b - 1) \).
In simple words: Open the brackets, group the squared terms together, and factor out \( (a - b) \).

Exam Tip: Be careful with the sign change when factoring out \( -1 \): \( -a + b = -(a - b) \).

Question 12. Factorize following expressions.
(i) \( x^2 + 9x + 18 \)
(ii) \( x^2 - 4x - 21 \)
(iii) \( x^2 - 9x + 18 \)
(iv) \( x^2 - 19x + 78 \)
Answer: Split the middle term by finding two integers whose sum equals the middle coefficient and whose product equals the constant term:
(i) For \( x^2 + 9x + 18 \): Numbers are \( 6 \) and \( 3 \) (sum = \( 9 \), product = \( 18 \)).
\( x^2 + 6x + 3x + 18 = x(x + 6) + 3(x + 6) = (x + 6)(x + 3) \).

(ii) For \( x^2 - 4x - 21 \): Numbers are \( -7 \) and \( 3 \) (sum = \( -4 \), product = \( -21 \)).
\( x^2 - 7x + 3x - 21 = x(x - 7) + 3(x - 7) = (x - 7)(x + 3) \).

(iii) For \( x^2 - 9x + 18 \): Numbers are \( -6 \) and \( -3 \) (sum = \( -9 \), product = \( 18 \)).
\( x^2 - 6x - 3x + 18 = x(x - 6) - 3(x - 6) = (x - 6)(x - 3) \).

(iv) For \( x^2 - 19x + 78 \): Numbers are \( -13 \) and \( -6 \) (sum = \( -19 \), product = \( 78 \)).
\( x^2 - 13x - 6x + 78 = x(x - 13) - 6(x - 13) = (x - 13)(x - 6) \).
In simple words: Find two numbers that multiply to give the last number and add up to give the middle number.

Exam Tip: If the constant term is positive while the linear term is negative, both factors must be negative.

Question 13. Calculate \( (997)^2 \) using algebraic identities.
Answer: Express \( 997 \) as \( (1000 - 3) \) and employ the identity \( (a - b)^2 = a^2 - 2ab + b^2 \):
Here \( a = 1000 \) and \( b = 3 \).
\( (997)^2 = (1000 - 3)^2 \)
\( = (1000)^2 - 2(1000)(3) + (3)^2 \)
\( = 1000000 - 6000 + 9 \)
\( = 994000 + 9 = 994009 \).
In simple words: Rewrite 997 as 1000 minus 3 and expand using the formula for \( (a - b)^2 \).

Exam Tip: Always show the identity used; direct multiplication without showing algebraic steps will forfeit marks.

Question 14. Calculate 103 X 107 using algebraic identities.
Answer: Rewrite the factors as \( (100 + 3)(100 + 7) \) and apply the identity:
\( (x + a)(x + b) = x^2 + (a + b)x + ab \)
Here, \( x = 100 \), \( a = 3 \), and \( b = 7 \).
\( 103 \times 107 = (100 + 3)(100 + 7) \)
\( = (100)^2 + (3 + 7)(100) + (3 \times 7) \)
\( = 10000 + 10(100) + 21 \)
\( = 10000 + 1000 + 21 = 11021 \).
In simple words: Represent the numbers as 100 plus 3 and 100 plus 7, then evaluate with the standard product formula.

Exam Tip: Verify the product by checking the last digit (\( 3 \times 7 = 21 \)), so the final answer must end in 21.

Question 15. Expand \( (2a + 3b + 4c)^2 \).
Answer: Use the trinomial square identity:
\( (x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx \)
Here, take \( x = 2a \), \( y = 3b \), and \( z = 4c \):
\( (2a + 3b + 4c)^2 = (2a)^2 + (3b)^2 + (4c)^2 + 2(2a)(3b) + 2(3b)(4c) + 2(4c)(2a) \)
\( = 4a^2 + 9b^2 + 16c^2 + 12ab + 24bc + 16ca \).
In simple words: Square each term, then add twice the product of every combination of two terms.

Exam Tip: Don't forget to square the numerical coefficients along with the variables, e.g., \( (3b)^2 = 9b^2 \), not \( 3b^2 \).

Question 16. Factorize \( 4x^2 + y^2 + z^2 - 4xy - 2yz + 4xz \).
Answer: Compare the given polynomial with the expansion \( (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \):
Notice the negative signs appear on the terms containing \( y \) (namely \( -4xy \) and \( -2yz \)), while the term \( +4xz \) is positive. This reveals that the \( y \)-term carries a negative sign.
Rewrite the terms as:
\( = (2x)^2 + (-y)^2 + (z)^2 + 2(2x)(-y) + 2(-y)(z) + 2(2x)(z) \)
This corresponds to the square of \( (2x - y + z) \):
\( = (2x - y + z)^2 \) (or \( (2x - y + z)(2x - y + z) \)).
In simple words: Identify which variable is negative by looking at the negative product terms, then write the trinomial square.

Exam Tip: Writing \( (-2x + y - z)^2 \) is also completely correct since \( (-A)^2 = A^2 \).

Question 17. Expand (i) \( (4a + 5b)^3 \) (ii) \( (5x - 3y)^3 \)
Answer:
(i) Use the identity \( (x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3 \):
Here, \( x = 4a \) and \( y = 5b \):
\( (4a + 5b)^3 = (4a)^3 + 3(4a)^2(5b) + 3(4a)(5b)^2 + (5b)^3 \)
\( = 64a^3 + 3(16a^2)(5b) + 3(4a)(25b^2) + 125b^3 \)
\( = 64a^3 + 240a^2b + 300ab^2 + 125b^3 \).

(ii) Use the identity \( (x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3 \):
Here, \( x = 5x \) and \( y = 3y \):
\( (5x - 3y)^3 = (5x)^3 - 3(5x)^2(3y) + 3(5x)(3y)^2 - (3y)^3 \)
\( = 125x^3 - 3(25x^2)(3y) + 3(5x)(9y^2) - 27y^3 \)
\( = 125x^3 - 225x^2y + 135xy^2 - 27y^3 \).
In simple words: Apply the cubic expansion formulas for \( (x + y)^3 \) and \( (x - y)^3 \).

Exam Tip: Remember that in \( (x - y)^3 \), the signs alternate: positive, negative, positive, negative.

Question 18. Evaluate (i) \( (95)^3 \) (ii) \( (106)^3 \)
Answer:
(i) Write \( 95 \) as \( (100 - 5) \) and apply \( (x - y)^3 = x^3 - y^3 - 3xy(x - y) \):
\( (95)^3 = (100 - 5)^3 \)
\( = (100)^3 - (5)^3 - 3(100)(5)(100 - 5) \)
\( = 1000000 - 125 - 1500(95) \)
\( = 1000000 - 125 - 142500 \)
\( = 1000000 - 142625 = 857375 \).

(ii) Write \( 106 \) as \( (100 + 6) \) and apply \( (x + y)^3 = x^3 + y^3 + 3xy(x + y) \):
\( (106)^3 = (100 + 6)^3 \)
\( = (100)^3 + (6)^3 + 3(100)(6)(100 + 6) \)
\( = 1000000 + 216 + 1800(106) \)
\( = 1000000 + 216 + 190800 \)
\( = 1191016 \).
In simple words: Decompose the base numbers into \( 100 \pm k \) and calculate using the binomial cube identity.

Exam Tip: Using the compact form \( 3xy(x \pm y) \) simplifies multi-digit mental multiplication compared to using expanded powers.

Question 19. Factorize (i) \( x^3 + 64 \) (ii) \( 27x^3 + 125y^3 \) (iii) \( 8a^3 - 27b^3 \) (iv) \( 1 - 64a^3 \)
Answer: Utilize the sum and difference of cubes identities:
\( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \)
\( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \)

(i) \( x^3 + 64 = (x)^3 + (4)^3 \)
\( = (x + 4)(x^2 - x(4) + 4^2) \)
\( = (x + 4)(x^2 - 4x + 16) \).

(ii) \( 27x^3 + 125y^3 = (3x)^3 + (5y)^3 \)
\( = (3x + 5y)((3x)^2 - (3x)(5y) + (5y)^2) \)
\( = (3x + 5y)(9x^2 - 15xy + 25y^2) \).

(iii) \( 8a^3 - 27b^3 = (2a)^3 - (3b)^3 \)
\( = (2a - 3b)((2a)^2 + (2a)(3b) + (3b)^2) \)
\( = (2a - 3b)(4a^2 + 6ab + 9b^2) \).

(iv) \( 1 - 64a^3 = (1)^3 - (4a)^3 \)
\( = (1 - 4a)(1^2 + 1(4a) + (4a)^2) \)
\( = (1 - 4a)(1 + 4a + 16a^2) \).
In simple words: Represent each term as a cube and factorize using \( a^3 \pm b^3 \).

Exam Tip: Do not confuse \( (a + b)^3 \) with \( a^3 + b^3 \); the latter factors into a linear binomial and an irreducible quadratic trinomial.

Section - D

Question 20. Factorize \( a^3 + 27b^3 + 8c^3 - 18abc \)
Answer: Express the terms in the form \( x^3 + y^3 + z^3 - 3xyz \):
\( a^3 + (3b)^3 + (2c)^3 - 3(a)(3b)(2c) \)
Using the identity:
\( x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) \)
Substitute \( x = a \), \( y = 3b \), and \( z = 2c \):
\( = (a + 3b + 2c)\left(a^2 + (3b)^2 + (2c)^2 - a(3b) - (3b)(2c) - (2c)a\right) \)
\( = (a + 3b + 2c)(a^2 + 9b^2 + 4c^2 - 3ab - 6bc - 2ca) \).
In simple words: Write each cubed term as a base and apply the three-variable cubic identity.

Exam Tip: Check that \( 3(a)(3b)(2c) = 18abc \) to ensure the identity matches the given problem perfectly.

Question 21. Factorize \( (p - q)^3 + (q - r)^3 + (r - p)^3 \)
Answer: Let \( x = p - q \), \( y = q - r \), and \( z = r - p \).
Calculate the sum of these quantities:
\( x + y + z = (p - q) + (q - r) + (r - p) = 0 \).
Recall the algebraic property: if \( x + y + z = 0 \), then:
\( x^3 + y^3 + z^3 = 3xyz \).
Substituting the original terms back gives:
\( (p - q)^3 + (q - r)^3 + (r - p)^3 = 3(p - q)(q - r)(r - p) \).
In simple words: Since the three expressions sum up to zero, the sum of their cubes equals three times their product.

Exam Tip: Verifying \( x + y + z = 0 \) must be written out explicitly as a step before stating the final factored form.

Question 22. Find the product \( (3x - 5y - 4)(9x^2 + 25y^2 + 15xy + 12x - 20y + 16) \)
Answer: Express the given product in the form of the identity:
\( (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) = a^3 + b^3 + c^3 - 3abc \)
Let \( a = 3x \), \( b = -5y \), and \( c = -4 \).
Examine the second factor:
\( a^2 + b^2 + c^2 - ab - bc - ca \)
\( = (3x)^2 + (-5y)^2 + (-4)^2 - (3x)(-5y) - (-5y)(-4) - (-4)(3x) \)
\( = 9x^2 + 25y^2 + 16 + 15xy - 20y + 12x \)
This matches the second factor exactly. Therefore, the product simplifies to:
\( = a^3 + b^3 + c^3 - 3abc \)
\( = (3x)^3 + (-5y)^3 + (-4)^3 - 3(3x)(-5y)(-4) \)
\( = 27x^3 - 125y^3 - 64 - 180xy \).
In simple words: This product matches the special three-variable identity, giving the sum of cubes minus three times the product.

Exam Tip: Mind the signs: since three negative values are being multiplied in \( -3(3x)(-5y)(-4) \), the resulting sign remains negative (\( -180xy \)).

Question 23. If \( x + y + z = 9 \) and \( xy + yz + zx = 23 \) then find the value of \( (x^3 + y^3 + z^3 - 3xyz) \)?
Answer: We use the identity:
\( x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - (xy + yz + zx)) \)
First, find \( x^2 + y^2 + z^2 \) using the expansion of \( (x + y + z)^2 \):
\( (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx) \)
Substitute the given values:
\( (9)^2 = x^2 + y^2 + z^2 + 2(23) \)
\( 81 = x^2 + y^2 + z^2 + 46 \)
\( x^2 + y^2 + z^2 = 81 - 46 = 35 \).
Now substitute this into the cubic identity:
\( x^3 + y^3 + z^3 - 3xyz = (9)(35 - 23) \)
\( = 9 \times 12 = 108 \).
Thus, the value is \( 108 \).
In simple words: Square the sum to find \( x^2 + y^2 + z^2 \), then substitute everything into the identity formula.

Exam Tip: Calculate \( x^2 + y^2 + z^2 \) as an essential first step before plugging into the factorized cubic identity.

Self Evaluation

Question 24. Which of the following expression is a polynomial?
(a) \( \sqrt{x} - 1 \)
(b) \( \frac{x - 1}{x + 1} \)
(c) \( x^2 - \frac{2}{x^2} + 5 \)
(d) \( x^2 + \frac{2x^{3/2}}{\sqrt{x}} + 6 \)
Answer: (d) \( x^2 + \frac{2x^{3/2}}{\sqrt{x}} + 6 \)
In simple words: The term \( \frac{2x^{3/2}}{\sqrt{x}} \) simplifies to \( 2x \), making all exponents whole numbers and hence a polynomial.

Exam Tip: Always simplify variable fractions before checking powers: \( \frac{x^{3/2}}{x^{1/2}} = x^{3/2 - 1/2} = x^1 \), which is a non-negative integer.

Question 25. Degree of zero polynomial is
(a) 1
(b) 0
(c) not defined
(d) none of these
Answer: (c) not defined
In simple words: The zero polynomial can be written with any power of x multiplied by 0, so its degree cannot be fixed or defined.

Exam Tip: Be careful not to confuse a constant non-zero polynomial (degree 0) with the zero polynomial (degree not defined).

Question 26. For what value of k is the polynomial \( p(x) = 2x^3 - kx^2 + 3x + 10 \) exactly divisible by \( (x + 2) \)?
(a) \( \frac{-1}{3} \)
(b) \( \frac{1}{3} \)
(c) 3
(d) -3
Answer: (d) -3
In simple words: Substitute \( x = -2 \) into the polynomial, set the remainder equal to 0, and solve for \( k \).

Exam Tip: Watch the signs: \( (-2)^3 = -8 \) giving \( 2(-8) = -16 \), while \( -k(-2)^2 = -4k \).

Question 27. The zeroes of the polynomial \( p(x) = 3x^2 - 1 \) are
(a) \( \frac{1}{3} \)
(b) \( \frac{1}{\sqrt{3}} \)
(c) \( \frac{-1}{\sqrt{3}} \)
(d) \( \frac{1}{\sqrt{3}} \text{ and } \frac{-1}{\sqrt{3}} \)
Answer: (d) \( \frac{1}{\sqrt{3}} \text{ and } \frac{-1}{\sqrt{3}} \)
In simple words: Setting \( 3x^2 - 1 = 0 \) gives \( x^2 = 1/3 \), which yields both positive and negative roots.

Exam Tip: Quadratic equations always yield two solutions; taking the square root requires both \( + \) and \( - \) signs.

Question 28. If \( \frac{x}{y} + \frac{y}{x} = -1 \) where \( x \neq 0, y \neq 0 \) then find the value of \( x^3 - y^3 \).
Answer: Simplify the given equation by taking an LCM of the denominators:
\( \frac{x^2 + y^2}{xy} = -1 \)
\( \implies x^2 + y^2 = -xy \)
\( \implies x^2 + xy + y^2 = 0 \).
Recall the factoring identity for the difference of cubes:
\( x^3 - y^3 = (x - y)(x^2 + xy + y^2) \)
Substitute \( x^2 + xy + y^2 = 0 \) into the identity:
\( x^3 - y^3 = (x - y)(0) = 0 \).
Thus, the value of \( x^3 - y^3 \) is \( 0 \).
In simple words: Clearing the fraction yields \( x^2 + xy + y^2 = 0 \), which directly forces \( x^3 - y^3 \) to be zero.

Exam Tip: Recognize the quadratic factor of \( x^3 - y^3 \) hiding inside the fraction equation to solve this within two lines.

Question 29. If \( (x + 2) \) and \( (x - 1) \) are factors of \( (x^3 + 10x^2 + mx + n) \) then find value of m & n?
Answer: Let \( p(x) = x^3 + 10x^2 + mx + n \).
Since \( x - 1 \) is a factor, \( p(1) = 0 \):
\( (1)^3 + 10(1)^2 + m(1) + n = 0 \)
\( 1 + 10 + m + n = 0 \)
\( m + n = -11 \) ...(1)
Since \( x + 2 \) is a factor, \( p(-2) = 0 \):
\( (-2)^3 + 10(-2)^2 + m(-2) + n = 0 \)
\( -8 + 10(4) - 2m + n = 0 \)
\( -8 + 40 - 2m + n = 0 \)
\( 32 - 2m + n = 0 \implies -2m + n = -32 \) ...(2)
Subtract equation (2) from equation (1):
\( (m + n) - (-2m + n) = -11 - (-32) \)
\( 3m = 21 \)
\( \implies m = 7 \)
Substitute \( m = 7 \) into equation (1):
\( 7 + n = -11 \)
\( \implies n = -11 - 7 = -18 \).
Therefore, \( m = 7 \) and \( n = -18 \).
In simple words: Set up two linear equations by evaluating the polynomial at 1 and -2, then solve simultaneously for m and n.

Exam Tip: Eliminating \( n \) by subtracting the two linear equations is the fastest method to solve for \( m \).

Question 30. Find the value of \( (369)^2 - (368)^2 \)
Answer: Apply the difference of two squares identity, \( a^2 - b^2 = (a + b)(a - b) \):
Here \( a = 369 \) and \( b = 368 \).
\( (369)^2 - (368)^2 = (369 + 368)(369 - 368) \)
\( = (737)(1) = 737 \).
Hence, the value is \( 737 \).
In simple words: Because the numbers differ by 1, the difference of their squares is simply their sum.

Exam Tip: For consecutive integers, \( (n+1)^2 - n^2 = 2n + 1 = (n+1) + n \).

Question 31. Find value of 104 X 96
Answer: Express the factors symmetrically about 100 as \( (100 + 4)(100 - 4) \) and use the identity \( (a + b)(a - b) = a^2 - b^2 \):
\( 104 \times 96 = (100 + 4)(100 - 4) \)
\( = (100)^2 - (4)^2 \)
\( = 10000 - 16 = 9984 \).
Thus, the product is \( 9984 \).
In simple words: Represent the product as \( (100 + 4)(100 - 4) \) and subtract 16 from 10000.

Exam Tip: Avoid direct multiplication when symmetric values close to 100 are given; applying \( a^2 - b^2 \) saves time and prevents arithmetic mistakes.

Question 32. If \( a + b + c = 0 \) find value of \( \left(\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab}\right) \)?
Answer: Combine the three fractions by taking the common denominator \( abc \):
\( \frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab} = \frac{a^3 + b^3 + c^3}{abc} \)
We know the identity: when \( a + b + c = 0 \), then \( a^3 + b^3 + c^3 = 3abc \).
Substitute this into the fraction:
\( = \frac{3abc}{abc} = 3 \).
Hence, the value is \( 3 \).
In simple words: Adding the fractions yields \( (a^3 + b^3 + c^3) / abc \). Since the cubes sum to \( 3abc \), dividing by \( abc \) leaves 3.

Exam Tip: State the conditional identity \( a^3 + b^3 + c^3 = 3abc \) whenever \( a + b + c = 0 \) to justify the numerator substitution.