About Polynomials – Class 9 Maths
The Polynomials chapter introduces students to the fascinating world of algebraic expressions. With our comprehensive Class 9, class-9 maths notes, you can understand terms, coefficients, degrees, and types of polynomials in a simple, structured manner. Our NCERT solutions for class 9 Maths help you solve all textbook exercises, from zeroes of a polynomial to factorisation methods, with clear explanations. Our Class 9 Maths tuition ensures every student can handle polynomial identities and applications confidently. Regular problem-solving sessions and quizzes make the learning process engaging and effective. By mastering this chapter, students build the foundation required for advanced algebra in higher classes.
Definition
A polynomial is an algebraic expression having one or more terms involving powers of the variable.
An expression p(x) of the form: p(x) = a₀xⁿ + a₁xⁿ⁻¹ + a₂xⁿ⁻² + ... + aₙ where all a₀, a₁, a₂...aₙ are real numbers and n is non-negative integer (i.e. whole numbers) is called a polynomial.
Examples:
- x² - b²
- px² + gx + r
- x⁴ - 3x² + 2x + 1
- y⁵ - 8y + 6
If the power of x or y be in either increasing or decreasing order, the polynomial in x or y is said to be in standard form.
For example: 5x⁴ + 2x³ + 7x² + 8x + 6 or -6 + 8x + 7x² + 2x³ + 5x⁴
Classification by Number of Terms
- Monomials: Polynomials having only one term, e.g. 2x, 5x²
- Binomials: Polynomials having only two terms, e.g. x + 1, x² - x
- Trinomials: Polynomials having only three terms, e.g. x² + x, 2 - x + x²
Example: Which of the following expressions are polynomials?
(i) x³ - 5x + 2
(ii) y² + √(2y) - √5
(iii) 2√x + 7
(iv) 4t² + (1/6)t³ + 2√3
(v) ³√(y) + 4
Solution:
(i) x³ - 5x + 2 is an expression having only non-negative integral powers of x. So, it is a polynomial.
(ii) y² + √(2y) - √5 is an expression having only non-negative integral power of y. So, it is a polynomial.
(iii) 2√x + 7 is an expression in which one term, namely 2√x, has rational power of x. So, it is not a polynomial.
(iv) 4t² + (1/6)t³ + 2√3 is an expression having only non-negative integral powers of t. So, it is a polynomial in t.
(v) The given expression may be written as y^(1/3) + 4. It contains a term containing rational power of y. So, it is not a polynomial.
Terms And Their Coefficients
If f(x) = a₀xⁿ + a₁xⁿ⁻¹ + a₂xⁿ⁻² + ... + aₙ is a polynomial in variable x, then a₀xⁿ, a₁xⁿ⁻¹, a₂xⁿ⁻² and aₙ are known as the terms of polynomial f(x) and a₀, a₁, ..., aₙ respectively are known as their coefficients, aₙ is called the constant term.
Example: In the polynomial g(x) = 2x⁴ - 5x² + 2x - 8, the coefficient of x³ is 0 whereas the constant term is -8.
Example: Write the coefficients of x³ in each of the following
(i) 3 + x³ + x
(ii) 5 - 2x³ + x²
(iii) ax³ - bx² - cx + dx³
Solution:
(i) Coefficient of x³ is 1.
(ii) Coefficient of x³ is -2.
(iii) Polynomial can be written as (a + d)x³ - bx² - cx, coefficient of x³ is a + d.
Degree Of The Polynomial
Highest Index of x in algebraic expression is called the degree of the polynomial, here a₀, a₁x, a₂x²,..., aₙxⁿ are called the terms of the polynomial and a₀, a₁, a₂,..., aₙ are called various coefficients of the polynomial f(x).
Note: A polynomial in x is said to be in standard form when the terms are written either in increasing order or decreasing order of the indices of x in various terms.
Different Types Of Polynomials
Generally, we divide the polynomials in the following categories.
(i) Based on degrees
There are four types of polynomials based on degrees:
(A) Linear Polynomials: A polynomial of degree one is called a linear polynomial. The general formula of linear polynomial is ax + b, where a and b are any real constant and a ≠ 0.
(B) Quadratic Polynomials: A polynomial of degree two is called a quadratic polynomial. The general form of a quadratic polynomial is ax² + bx + c, where a ≠ 0.
(C) Cubic Polynomials: A polynomial of degree three is called a cubic polynomial. The general form of a cubic polynomial is ax³ + bx² + cx + d, where a ≠ 0.
(D) Biquadratic (or quadric) Polynomials: A polynomial of degree four is called a biquadratic (quadratic) polynomial. The general form of a biquadratic polynomial is ax⁴ + bx³ + cx² + dx + e, where a ≠ 0.
Note: A polynomial of degree five or more than five does not have any particular name. Such a polynomial usually called a polynomial of degree five or six or ....etc.
(ii) Based on number of terms
There are three types of polynomials based on number of terms:
(A) Monomial: A polynomial is said to be monomial if it has only one term. e.g. x, 9x², 5x³ all are monomials.
(B) Binomial: A polynomial is said to be binomial if it contains two terms e.g. 2x² + 3x, x³ + 5x³, -8x³ + 3, all are binomials.
(C) Trinomials: A polynomial is said to be a trinomial if it contains three terms. e.g. 3x³ - 8 + 5/2, √(7x¹⁰)·8x⁴ - 3x², 5 - 7x + 8x⁹, are all trinomials.
Note: A polynomial having four or more than four terms does not have particular Name. These are simply called polynomials.
Example: Give one example each of a binomial of degree 53 and a monomial of degree 90.
Solution:
(i) A binomial of degree 53 is ax⁵³ + b for all 'a' is non-zero real number and a is called coefficient of x⁵³ also b ≠ 0. One such example is 2x⁵³ + 100.
(ii) A monomial of degree 90 is ax⁹⁰, for all 'a' in a non-zero real number 'a' is called coefficient of x⁹⁰. One such example -7x⁹⁰.
Example: Classify the following as linear, quadratic and cubic polynomials:
(i) 2x³ + 3
(ii) 4t
(iii) 5 - x - x²
Solution:
(i) 2x³ + 3 is a polynomial of degree 3. So, it is a cubic polynomial.
(ii) 4t is a polynomial of degree 1. So, it is a linear polynomial.
(iii) Arranging in standard form -x² - x + 5. It is a polynomial of degree 2. So, it is a quadratic polynomial.
(iii) Zero degree polynomial
Any non-zero number (constant) is regarded as polynomial of degree zero or zero degree polynomial. i.e. f(x) = a, where a ≠ 0 is a zero degree polynomial, since we can write f(x) = a as f(x) = ax⁰.
(iv) Zero polynomial
A polynomial whose all coefficients are zeros is called as zero polynomial i.e. f(x) = 0, we cannot determine the degree of zero polynomial.
Algebraic Identity
An identity is an equality which is true for all values of the variables.
Some important identities are:
(i) (a + b)² = a² + 2ab + b²
(ii) (a - b)² = a² - 2ab + b²
(iii) a² - b² = (a + b)(a - b)
(iv) a³ + b³ = (a + b)(a² - ab + b²)
(v) a³ - b³ = (a - b)(a² + ab + b²)
(vi) (a + b)³ = a³ + b³ + 3ab(a + b)
(vii) (a - b)³ = a³ - b³ - 3ab(a - b)
(viii) a⁴ + a²b² + b⁴ = (a² + ab + b²)(a² - ab + b²)
(ix) a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ac)
Special case: if a + b + c = 0 then a³ + b³ + c³ = 3abc.
(a) Value Form:
(i) a² + b² = (a + b)² - 2ab, if a + b and ab are given
(ii) a² + b² = (a - b)² + 2ab if a - b and ab are given
(iii) a + b = √[(a - b)² + 4ab] if a - b and ab are given
(iv) a - b = √[(a + b)² - 4ab] if a + b and ab are given
(v) a² + 1/a² = (a + 1/a)² - 2 if a + 1/a is given
(vi) a² + 1/a² = (a - 1/a)² + 2 if a - 1/a is given
(vii) a³ + b³ = (a + b)³ - 3ab(a + b) if (a + b) and ab are given
(viii) a³ - b³ = (a - b)³ + 3ab(a - b) if (a - b) and ab are given
(ix) x³ + 1/a³ = (a + 1/a)³ - 3(a + 1/a) if a + 1/a is given
(x) a³ - 1/a³ = (a - 1/a)³ + 3(a - 1/a), if (a - 1/a) is given
(xi) a⁴ + b⁴ = (a² + b²)² - 2a²b² = [(a + b)² - 2ab]² - 2a²b², if (a + b) and ab are given
(xii) a² - b² = (a² + b²)(a² - b²) = [(a + b)² - 2ab](a + b)(a - b)
(xiii) a⁵ + b⁵ = (a³ + b³)(a² + b²) - a²b²(a + b)
Example: What is the zero of the polynomial p(x) = 2x - 3r.
Solution: For zero of polynomial put p(x) = 0
i.e. 2x - 3r = 0
or 2x = 3r
∴ x = 3r/2
Example: Find zeros of the polynomial q(t) = 3t - 4.
Solution: Zeros of polynomial q(t) = 3t - 4 is q(t) = 0
or 3t - 4 = 0
∴ 3t = 4
∴ t = 4/3
Example: Find the value of each of the following polynomials at the indicated value of variables:
(i) p(x) = 5x² - 3x + 7 at x = 1.
(ii) q(y) = 3y³ - 4y + √19 at y = -2.
(iii) p(t) = 4t⁴ + 5t³ - t² + 8 at t = a.
Solution:
(i) p(x) = 5x² - 3x + 7
The value of the polynomial p(x) at x = 1 is given by
∴ p(1) = 5(1)² - 3(1) + 7
= 5 - 3 + 7 = 9
(ii) q(y) = 3y³ - 4y + √19
The value of the polynomial q(y) at y = -2 is given by
q(-2) = 3(-2)³ - 4(-2) + √19 = -24 + 8 + √19 = -16 + √19
(iii) p(t) = 4t⁴ + 5t³ - t² + 8
The value of the polynomial p(t) at t = a is given by
p(a) = 4a⁴ + 5a³ - a² + 8
Example: Find the value of:
(i) 36x² + 49y² + 84xy, when x = 3, y = 6
(ii) 25x² + 16y² - 40xy, when x = 6, y = 7
Solution:
(i) 36x² + 49y² + 84xy = (6x)² + (7y)² + 2 × (6x) × (7y)
= (6x + 7y)²
= (6 × 3 + 7 × 6)² [When x = 3, y = 6]
= (18 + 42)²
= (60)²
= 3600. Ans.
(ii) 25x² + 16y² - 40xy = (5x)² + (4y)² - 2 × (5x) × (4y)
= (5x - 4y)²
= (5 × 6 - 4 × 7)² [When x = 6, y = 7]
= (30 - 28)²
= 2²
= 4 Ans.
Factors Of A Polynomial
If a polynomial f(x) can be written as a product of two or more other polynomial f₁(x), f₂(x), f₃(x),...., then each of the polynomials f₁(x), f₂(x),..., is called a factor of polynomial f(x). The method of finding the factors of a polynomials is called factorisations.
(a) Factorisation by Making a Trinomial a Perfect Square:
Example: 81a²b²c² + 64a⁶b² - 144a⁴b²c
Solution: 81a²b²c² + 64a⁶b² - 144a⁴b²c
= [9abc]² - 2[9abc][8a³b] + [8a³b]²
= [9abc - 8a³b]² = a²b²[9c - 8a²]² Ans.
Example: (3a - 1/b)² - 6(3a - 1/b) + 9 + c + (3a - 1/b - 3)²
Solution: (3a - 1/b)² - 6(3a - 1/b) + 9 + c + (3a - 1/b - 3)²
= (3a - 1/b)² - 2·3·(3a - 1/b) + (3)² + c + (3a - 1/b - 3)²
= (3a - 1/b - 3)² + (c + 1/b - 2a)·(3a - 1/b - 3)
= (3a - 1/b - 3)[3a - 1/b + 3 + 1/b - 2a]
= (3a - 1/b - 3)[a + c - 3] Ans.
(b) Factorisation by Using the Formula for the Difference of Two Squares:
a² - b² = (a + b)(a - b)
Example: Factorise: 4(2a + 3b - 4c)² - (a - 4b + 5c)²
Solution: = 4(2a + 3b - 4c)² - (a - 4b + 5c)²
= [2(2a + 3b - 4c)]² - (a - 4b + 5c)²
= [4a + 6b - 8c + a - 4b + 5c][4a + 6b - 8c - a + 4b - 5c]
= [5a + 2b - 3c][3a + 10b - 13c] Ans.
Example: Factorise: 4x² + 1/4x² - 2 - 9y².
Solution: 4x² + 1/4x² - 2 - 9y²
= (2x)² + (1/2x)² - 2·(2x)·(1/2x) - (3y)²
= (2x + 1/2x)² - (3y)²
= (2x + 1/2x + 3y)(2x + 1/2x - 3y) Ans.
(c) Factorisation by Using Formula of a³ + b³ and a³ - b³:
Example: Factorize: 64a¹³b + 343ab¹³.
Solution: 64a¹³b + 343ab¹³ = ab[64a¹² + 343b¹²]
= ab[(4a⁴)³ + (7b⁴)³]
= ab[4a⁴ + 7b⁴][(4a⁴)² - (4a⁴)(7b⁴) + (7b⁴)²]
= ab[4a⁴ + 7b⁴][16a⁸ - 28a⁴b⁸ + 49b⁸] Ans.
Example: Factorize: p³q²x⁴ + 3p²qx³ + 3px² + x/q - q²r³x
Solution: In above question, If we take common then it may become in the form of a³ + b³.
∴ p³q²x⁴ + 3p²qx³ + 3px² + x/q - q²r³x
= (x/q)[p³q³x³ + 3p²q²x² + 3pqx + 1 - q³r³]
= (x/q)[(pqx)³ + 3(pqx)²·1 + 3pqx·(1)² + (1)³ - q³r³]
Let pqx = A & 1 = B
= (x/q)[A³ + 3A²B + 3AB² + B³ - q³r³]
= (x/q)[(pqx + 1)³ - (qr)³] = (x/q)[pqx + 1 - qr][(pqx + 1)² + (pqx + 1)qr + (qr)³]
= (x/q)[pqx + 1 - qr][p²q²x² + 1 + 2pqx + pq²xr + qr + q²r²] Ans.
Example: Factorize: x³ - 6x² + 32
Solution: x³ + 32 - 6x²
= x³ + 8 + 24 - 6x²
= [(x)³ + (2)³] + 6[4 - x²]
= (x + 2)[x² - 2x + 4] + 6[2 + x][2 - x]
= (x + 2)[x² - 2x + 4 + 6(2 - x)]
= (x + 2)[x² - 2x + 4 + 12 - 6x]
= (x + 2)[x² - 8x + 16]
= (x + 2)(x - 4)² Ans.
Remainder Theorem
Statement: Let p(x) be any polynomial of degree greater than or equal to 1 and let a be any real number. If p(x) is divided by the linear polynomial x - a, then the remainder is p(a).
Proof: Let p(x) be any polynomial with degree greater than or equal to one. Suppose that when p(x) is divided by x - a, the quotient is q(x) and the remainder is r(x), i.e.,
p(x) = (x - a)q(x) + r(x) ...(i)
Since the degree of x - a is one and the degree of r(x) is less than the degree of x - a, it implies that the degree of r(x) = 0. This means that r(x) is a constant, say r.
∴ So, for every value of x we have r(x) = r.
Therefore equation (i) becomes
p(x) = (x - a)q(x) + r
In particular, if x = a, above equation gives us
p(a) = (a - a)q(a) + r
= 0·q(a) + r
p(a) = r
which proves the theorem.
Example: Find the remainder when f(x) = x³ - 6x² + 2x - 4 is divided by g(x) = 1 - 2x.
Solution: 1 - 2x = 0 ⇒ 2x = 1 ⇒ x = 1/2
f(1/2) = (1/2)³ - 6(1/2)² + 2(1/2) - 4
= 1/8 - 6/4 + 1 - 4
= (1 - 12 + 8 - 32)/8
= -35/8 Ans.
Example: The polynomials ax³ + 3x² - 13 and 2x³ - 5x + a are divided by x + 2 if the remainder in each case is the same, find the value of a.
Solution: p(x) = ax³ + 3x² - 13 and q(x) = 2x³ - 5x + a
when p(x) & q(x) are divided by x + 2 = 0 ⇒ x = -2
p(-2) = q(-2)
⇒ a(-2)³ + 3(-2)² - 13 = 2(-2)³ - 5(-2) + a
⇒ -8a + 12 - 13 = -16 + 10 + a
⇒ -9a = -5
⇒ a = 5/9 Ans.
Factor Theorem
Statement: If p(x) is a polynomial of degree n ≥ 1 and a is any real number, then (i) x - a is a factor of p(x), if p(a) = 0, and (ii) p(a) = 0, if x - a is a factor of p(x).
Proof: By the Remainder Theorem we have, p(x) = (x - a)q(x) + p(a).
(i) If p(a) = 0, then p(x) = (x - a)q(x), which indicates that x - a is a factor of p(x).
(ii) Since x - a is a factor of p(x), p(x) = (x - a)g(x) for some polynomial g(x). In this case, p(a) = (a - a)g(a) = 0.
Example: Show that x + 1 and 2x - 3 are factors of 2x³ - 9x² + x + 12.
Solution: To prove that (x + 1) and (2x - 3) are factors of 2x³ - 9x² + x + 12 it is sufficient to show that p(-1) and p(3/2) both are equal to zero.
p(-1) = 2(-1)³ - 9(-1)² + (-1) + 12 = -2 - 9 - 1 + 12 = -12 + 12 = 0
And, p(3/2) = 2(3/2)³ - 9(3/2)² + (3/2) + 12
= 27/4 - 81/4 + 3/2 + 12
= (27 - 81 + 6 + 48)/4 = 0
Hence, (x + 1) and (2x - 3) are the factors 2x³ - 9x² + x + 12. Ans.
Example: Find α and β if x + 1 and x + 2 are factors of p(x) = x³ + 3x² - 2αx + β.
Solution: When we put x + 1 = 0 or x = -1 and x + 2 = 0 or x = -2 in p(x)
Then, p(-1) = 0 & p(-2) = 0
Therefore, p(-1) = (-1)³ + 3(-1)² - 2α(-1) + β = 0
⇒ -1 + 3 + 2α + β = 0 ⇒ β = -2α - 2 ...(i)
And, p(-2) = (-2)³ + 3(-2)² - 2α(-2) + β = 0
⇒ -8 + 12 + 4α + β = 0 ⇒ β = -4α - 4 .....(ii)
From equation (i) and (ii)
-2α - 2 = -4α - 4 ⇒ 2α = -2 ⇒ α = -1
Put α = -1 in equation (i) ⇒ β = -2(-1) - 2 = 2 - 2 = 0.
Hence, α = -1 β = 0. Ans.
Factorisation Quadratic Polynomial
For factorisation of a quadratic expression ax² + bx + c where a ≠ 0, there are two methods.
(a) By Method of Completion of Square:
In the form ax² + bx + c where a ≠ 0, firstly we take 'a' common in the whole expression then factorise by converting the expression a[x² + (b/a)x + c/a] as the difference of two squares.
Example: Factorize x² - 31x + 220.
Solution: x² - 31x + 220
= x² - 2·(31/2)·x + (31/2)² - (31/2)² + 220
= (x - 31/2)² - 961/4 + 220
= (x - 31/2)² - 81/4
= (x - 31/2 - 9/2)(x - 31/2 + 9/2)
= (x - 11)(x - 20)
Example: Factorize: -10x² + 31x - 24
Solution: -10x² + 31x - 24
= -[10x² - 31x + 24] = -10[x² - 31/10·x + 24/10]
= -10[x² - 2·(31/20)·x + (31/20)² - (31/20)² + 24/10]
= -10[(x - 31/20)² - 961/400 + 24/10]
= -10[(x - 31/20)² - 1/400]
= -10(x - 31/20 - 1/20)(x - 31/20 + 1/20)
= -((2x - 3)/2)((5x - 8)/5) = -(2x - 3)(5x - 8)
(b) By Splitting the Middle Term:
In the quadratic expression ax² + bx + c, where a is the coefficient of x², b is the coefficient of x and c is the constant term. In the quadratic expression of the form x² + bx + c, a = 1 is the multiple of x² and another terms are the same as above.
There are four types of quadratic expression:
(i) ax² + bx + c
(ii) ax² - bx + c
(iii) ax² - bx - c
(iv) ax² + bx - c
Example: Factorize: 2x² + 12√2x + 35.
Solution: 2x² + 12√2x + 35
Product ac = 70 & b = 12√2
Split the middle term as 7√2 & 5√2
∴ 2x² + 12√2x + 35 = 2x² + 7√2x + 5√2x + 35
= √2x(√2x + 7) + 5(√2x + 7)
= (√2x + 5)(√2x + 7)
Example: Factorize: x² - 14x + 24.
Solution: Product ac = 24 & b = -14
Split the middle term as -12 & -2
∴ x² - 14x + 24 = x² - 12x - 2x + 24
⇒ x(x - 12) - 2(x - 12) = (x - 12)(x - 2)
Example: Factorize: x² - 13x/24 - 1/12.
Solution: (1/24)[24x² - 13x - 2]
Product ac = -48 & b = -13 ∴ We split the middle term as -16x + 3x.
= (1/24)[24x² - 16x + 3x - 2]
= (1/24)[8x(3x - 2) + 1(3x - 2)]
= (1/24)(3x - 2)(8x + 1)
Example: Factorize: (3/2)x² - 8x - 35/2.
Solution: (1/2)(3x² - 21x + 5x - 35) = (1/2)[3x(x - 7) + 5(x - 7)]
= (1/2)(x - 7)(3x + 5) Ans.
(c) Integral Root Theorem:
If f(x) is a polynomial with integral coefficient and the leading coefficient is 1, then any integer root of f(x) is a factor of the constant term. Thus if f(x) = x³ - 6x² + 11x - 6 has an Integral root, then it is one of the factors of 6 which are ±1, ±2, ±3, ±6.
Now Infect f(1) = (1)³ - 6(1)² + 11(1) - 6 = 1 - 6 + 11 - 6 = 0
f(2) = (2)³ - 6(2)² + 11(2) - 6 = 8 - 24 + 22 - 6 = 0
f(3) = (3)³ - 6(3)² + 11(3) - 6 = 27 - 54 + 33 - 6 = 0
Therefore Integral roots of f(x) are 1,2,3.
(d) Rational Root Theorem:
Let b/c be a rational fraction in lowest terms. If b/c is a root of the polynomial f(x) = a₀xⁿ + a₁xⁿ⁻¹ + ... + aₙ₋₁x + aₙ, aₙ ≠ 0 with integral coefficients. Then b is a factor of constant term a₀ and c is a factor of the leading coefficient aₙ.
For example: If b/c is a rational root of the polynomial f(x) = 6x³ + 5x² - 3x - 2, then the values of b are limited to the factors of -2 which are ±1, ±2 and the value of c are limited to the factors of 6, which are ±1, ±2, ±3, ±6. Hence, the possible rational roots of f(x) are ±1, ±2, ±1/2, ±1/3, ±1/6, ±2/3, ±3. Infect -1 is a Integral root and -1/2, -2/3 are the rational roots of f(x) = 6x³ + 5x² - 3x - 2.
NOTE:
(i) An nth degree polynomial can have at most n real roots.
(ii) Finding a zero or root of polynomial f(x) means solving the polynomial equation f(x) = 0. It follows from the above discussion that if f(x) = ax + b, a ≠ 0 is a linear polynomial, then it has only one root given by f(x) = 0 i.e. f(x) = ax + b = 0
⇒ ax = -b
⇒ x = -b/a
Thus a = -b/a is the only root of f(x) = ax + b.
Example: If f(x) = 2x³ - 13x² + 17x + 12 then find out the value of f(-2) & f(3).
Solution: f(x) = 2x³ - 13x² + 17x + 12
f(-2) = 2(-2)³ - 13(-2)² + 17(-2) + 12
= -16 - 52 - 34 + 12 = -90 Ans.
f(3) = 2(3)³ - 13(3)² + 17(3) + 12
= 54 - 117 + 51 + 12 = 0 Ans.
(e) Factorisation of an Expression Reducible to A Quadratic Expression:
Example: Factorize:- -8 + 9(a - b)⁶ - (a - b)¹²
Solution: -8 + 9(a - b)⁶ - (a - b)¹²
Let (a - b)⁶ = x
Then -8 + 9x - x² = -(x² - 9x + 8) = -(x² - 8x - x + 8)
= -(x - 8)(x - 1)
= -[(a - b)⁶ - 8][(a - b)⁶ - 1]
= [1 - (a - b)⁶][(a - b)⁶ - 8]
= [(1)³ - {(a - b)²}³][{(a - b)²}³ - (2)³]
= [1 - (a - b)²][1 + (a - b)⁴ + (a - b)²][(a - b)² - 2][(a - b)⁴ + 4 + 2(a - b)²] Ans.
Example: Factorize: 6x² - 5xy - 4y² + x + 17y - 15
Solution: 6x² + x[1 - 5y] - [4y² - 17y + 15]
= 6x² + x[1 - 5y] - [4y² - 17y + 15]
= 6x² + x[1 - 5y] - [4y(y - 3) - 5(y - 3)]
= 6x² + x[1 - 5y] - (4y - 5)(y - 3)
= 6x² + 3(y - 3)x - 2(4y - 5)x - (4y - 5)(y - 3)
= 3x[2x + y - 3] - (4y - 5)(2x + y - 3)
= (2x + y - 3)(3x - 4y + 5) Ans.
Polynomial Summary
Definitions:
- An expression p(x) = a₀xⁿ + a₁xⁿ⁻¹ + a₂xⁿ⁻² + aₙ is a polynomial
- Where a₀, a₁, ...aₙ are real numbers and n is non-negative integer.
- Degree of a polynomial is the greatest exponent of the variable in the polynomial.
- Constant polynomial is a polynomial of degree zero. The constant polynomial f(x) = 0 is called zero polynomial.
- Degree of zero polynomial is not defined.
Classification:
- Polynomials having only one term are known as monomials.
- Polynomials having two terms are known as binomials.
- Polynomials having three terms are known as trinomials.
- A polynomial of degree one is called a linear polynomial e.g. ax + b where a ≠ 0.
- A polynomial of degree two is called a quadratic polynomial e.g. ax² + bx + c where a ≠ 0.
- A polynomial of degree 3 is called a cubic polynomial e.g. px³ + qx² + rx + s, p ≠ 0.
- A polynomial of degree 4 is called a biquadratic polynomial
Algebraic Identities:
- (x + y)² = x² + 2xy + y²
- (x - y)² = x² - 2xy + y²
- x² - y² = (x + y)(x - y)
- (x + a)(x + b) = x² + (a + b)x + ab
- (x + y)³ = x³ + y³ + 3xy(x + y)
- (x - y)³ = x³ - y³ - 3xy(x - y)
- x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)
- If x + y + z = 0, then x³ + y³ + z³ = 3xyz
Value of a polynomial:
Value of a polynomial p(x) at x = a is p(a).
Zero of a polynomial:
Zero of a polynomial p(x) is a number 'a' such that p(a) = 0.
Remainder Theorem:
Let p(x) is a polynomial of degree greater than or equal to 1 and a is any real number, if p(x) is divided by the linear polynomial x - a then the remainder is p(a).
Solved Questions
1. Write the following in standard form.
(i) 8x⁷ + 6x² + 88 - 3x⁶ - 9x
(ii) 6x + 9x⁵ - 4x³ + 1
Sol. The standard forms of the given polynomial are
(i) 8x⁷ - 3x⁶ + 6x² - 9x + 88
(ii) 9x⁵ - 4x³ + 6x + 1
2. Give one example each of a binomial of degree 53 and a monomial of degree 90.
Sol. (i) A binomial of degree 53 is ax⁵³ + b for all 'a' is non-zero real number and a is called coefficient of x⁵³ also b ≠ 0. One such example is 2x⁵³ + 100.
(ii) A monomial of degree 90 is ax⁹⁰, for all 'a' in a non-zero real number 'a' is called coefficient of x⁹⁰. One such example -7x⁹⁰.
3. Classify the following as linear, quadratic and cubic polynomials:
(i) 2x³ + 3
(ii) 4t
(iii) 5 - x - x²
Sol. (i) 2x³ + 3 is a polynomial of degree 3. So, it is a cubic polynomial.
(ii) 4t is a polynomial of degree 1. So, it is a linear polynomial.
(iii) Arranging in standard form -x² - x + 5.
It is a polynomial of degree 2. So, it is a quadratic polynomial
4. Find the degree of each of the polynomials given below:
(i) x⁵ - x⁴ + 3
(ii) 2 - y² - y³ + 2y⁸
(iii) 2
Sol. (i) The highest power of the variable is 5. So, the degree of the polynomial is 5.
(ii) Arranging in standard form 2y⁸ - y³ - y² + 2. The highest power of the variable is 8. So, the degree is 8.
(iii) The only term here is 2 which can be written as 2x⁰ So the exponent of x is 0. Therefore, the degree of the polynomial is 0.
5. What is the degree of polynomial p(x) = a³ + 3x² + 4x + 7.
Sol. Degree of a polynomial is the exponent of the variable with the highest power.
For given polynomial highest power of x is 2, hence its degree is 2.
6. Find the zeros of the polynomial in each of the following cases:
(i) f(x) = x - 5
(ii) p(x) = ax + d, a ≠ 0
Sol. We know that the zeros of a polynomial f(x) are given by solving the polynomial equation f(x) = 0.
Therefore,
(i) Zeros of f(x) = x - 5 is given by
f(x) = 0
⇒ x - 5 = 0
or x = 5
Thus, x = 5 is a zero of f(x) = x - 5
(ii) Zeros of p(x) = ax + d are given by
p(x) = 0 ⇒ ax + d = 0 ⇒ ax = -d or x = -d/a
Thus, x = -d/a is the zero of p(x).
Questions (continued):
- Check whether the given value is a zero of the polynomial P(x) = (9/4)x² - 9, x = -2.
- Determine the remainder when the polynomial p(x) = x⁴ - 3x² + 2x + 1 is divided by x - 1.
- Find the remainder when p(x) = 4x³ - 12x² + 14x - 3 is divided by g(x) = x - 1/2.
- Factorise y² - 5y + 6 by using the Factor Theorem.
- Factorize: x² + 5x - 24.
- Factorize: x² + 3√3x + 6 by splitting the middle term.
- Factorize: 4√3x² + 5x - 2√3.
Answers:
- x = -2 is a zero of (9/4)x² - 9.
- p(1) = 1
- p(1/2) = 3/2
- (y - 2)(y - 3)
- (x + 8)(x - 3).
- (x + 2√3)(x + √3)
- (√3x + 2)(4x - √3)
7. Check whether the given value is a zero of the polynomial P(x) = (9/4)x² - 9, x = -2.
Sol. Given P(x) = (9/4)x² - 9.
Put x = -2.
i.e. P(-2) = (9/4)(-2)² - 9
= 9 - 9 = 0
Thus, x = -2 is a zero of (9/4)x² - 9.
8. Determine the remainder when the polynomial p(x) = x⁴ - 3x² + 2x + 1 is divided by x - 1.
Sol. By remainder theorem, required remainder is equal to p(1).
Now, p(x) = x⁴ - 3x² + 2x + 1
∴ p(1) = (1)⁴ - 3(1)² + 2(1) + 1 = 1 - 3 + 2 + 1 = 1
Hence, required remainder = p(1) = 1
9. Find the remainder when p(x) = 4x³ - 12x² + 14x - 3 is divided by g(x) = x - 1/2.
Sol. By remainder theorem, p(x) when divided by g(x) = x - 1/2 gives a remainder which is equal to p(1/2).
Now, p(x) = 4x³ - 12x² + 14x - 3
∴ p(1/2) = 4(1/2)³ - 12(1/2)² + 14(1/2) - 3 = 4/8 - 12/4 + 7 - 3 = 1/2 - 3 + 7 - 3 = 3/2
Hence, required remainder p(1/2) = 3/2
10. Factorise y² - 5y + 6 by using the Factor Theorem.
Sol. Let p(y) = y² - 5y + 6.
Now if p(y) = (y - a)(y - b) then (y - a)(y - b) = y² - (a + b)y + ab
So constant term is ab = 6.
The factors of 6 are 1, 2 and 3.
Now, p(2) = 2² - (5 × 2) + 6 = 0
∴ y - 2 is a factor of p(y).
Also, p(3) = 3² - (5 × 3) + 6 = 0
∴ y - 3 is also a factor of y² - 5y + 6
Therefore, y² - 5y + 6 = (y - 2)(y - 3)
11. Factorize: x² + 5x - 24.
Sol. The given expression is x² + 5x - 24.
We try to split 5 into two parts whose sum is 5 and product -24.
Clearly, 8 + (-3) = 5 and 8(-3) = -24.
So we write middle term 5x as (8x - 3x)
∴ x² + 5x - 24 = x² + 8x - 3x - 24
= x(x + 8) - 3(x + 8)
= (x + 8)(x - 3)
Hence, x² + 5x - 24 = (x + 8)(x - 3).
12. Factorize: x² + 3√3x + 6 by splitting the middle term.
Sol. In order to factorize x² + 3√3x + 6, we try to split 3√3 into two parts whose sum is 3√3 and product is 6.
Clearly, 2√3 + √3 = 3√3 and 2√3 × √3 = 6
So, we write the middle term 3√3x as 2√3x + √3x
∴ x² + 3√3x + 6
= x² + 2√3x + √3x + 6
= x(x + 2√3) + √3(x + 2√3) = (x + 2√3)(x + √3)
13. Factorize: 4√3x² + 5x - 2√3.
Sol. Here 4√3 × (-2√3) = -24
We try to split 5x into two parts whose sum is 5 and product is = -24
Clearly, 8 + (-3) = 5 and 8 × -3 = -24
∴ Middle term 5x can be written as 8x - 3x
∴ 4√3x² + 5x - 2√3
= 4√3x² + 8x - 3x - 2√3
= (4√3x² + 8x) - (3x + 2√3) = 4x(√3x + 2) - √3(√3x + 2) = (√3x + 2)(4x - √3)
ILLUSTRATIONS
1. Find the remainder when x⁴ + x³ - 2x² + x - 1 is divided by x - 1.
Sol. p(x) = x⁴ + x³ - 2x² + x + 1, and the zero of x - 1 is 1
∴ p(1) = (1)⁴ + (1)³ - 2(1)² + 1 + 1
= 1 + 1 - 2 + 1 + 1 = 2
Hence the remainder theorem, 2 is the remainder, when x⁴+x³ - 2x² + x + 1 is divided by x-1.
2. Show that (x - 3) is a factor of (x³ + x² - 17x + 15)
Sol. Applying factor theorem (x - 3) will be a factor of f(x) ; if f(3) = 0
Now, substituting x = 3 in (1), we get
f(3) = (3)³ + (3)² - 17 × (3) + 15
= 27 + 9 - 51 + 15 = 51 - 51 = 0
Hence (x - 3) is a factor of f(x).
3. Factorize 6x² + 17x + 5 by splitting the middle term, and by using the factor Theorem.
Sol. I : By splitting method
If we can find two numbers p and q such that p + q = 17 and pq = 30, then we can get the factors.
Now, let us see for pairs of factors of 30. Some are 1 and 30, 2 and 15, 3 and 10, 5 and 6 of these pairs, 2 and 15 will give up p + q = 17.
∴ 6x² + 17x + 5 = 6x² + (2 + 15)x + 5
= 6x² + 2x + 15x + 5
= 2x(3x + 1) + 5(3x + 1)
Hence, 6x² + 17x + 5 = (3x + 1)(2x + 5)
II : Using the factor theorem
6x² + 17x + 5
6p(x) = 6[x² + (17/6)x + 5/6]
If a and b are the zeroes of p(x), then
6x² + 17x + 5 = 6(x - a)(x - b)
∴ ab = 5/6
Let us look at some possibilities for a and b.
They could be ±1/2, ±1/3, ±5/3, ±5/2, ±1
Now, p(1/2) = 6x² + (17/6)x + 5/6
= (1/4) + (17/12) + 5/6 = (1 + 17 + 5)/4 - 12 + 6
= (1 + 17 + 5)/4 - 12 - 6 ≠ 0
But p(-1/3) = (1/9) + (17/18) + 5/6
= (2 - 17 + 15)/18 = (17 - 17)/18 = 0/18 = 0
∴ (x + 1/3) is a factor of p(x)
Similarly, by trial and error method, we can find that (x + 5/2) is a factor of p(x).
Hence, 6x² + 17x + 5 = 6(x + 1/3)(x + 5/2)
= 6[(3x + 1)/3][(2x + 5)/2] = (3x + 1)(2x + 5).
4. Expand:
(i) (2a + 3b + 4c)²
(ii) (3a - 4b - 5c)²
Sol. (i) (2a)² + (3b)² + (4c)² + 2.(2a).(3b) + 2.(3b).(4c) + 2.(4c).(2a)
= 4a² + 9b² + 16c² + 12ab + 24bc + 16ca
[(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca]
(ii) (3a)² + (4b)² + (5c)² - 2.(3a).(4b) + 2.(4b).(5c) - 2.(5c).(3a)
= 9a² + 16b² + 25c² - 24ab + 40bc - 30ca
[(a-b-c)² = a²+b² + c² - 2ab + 2bc - 2ca].
5. Factorize: 6x² + 7x - 3.
Sol. The given expression is 6x² + 7x - 3.
Here, 6 × (-3) = -18
So, we try to split 7 into two parts whose sum is 7 and product -18
Clearly, 9 + (-2) = 7 and 9 × (-2) = -18
∴ Middle term 7x can be written as 9x - 2x
∴ 6x² + 7x - 3 = 6x² + 9x - 2x - 3
= 3x(2x + 3) - 1(2x + 3)
= (2x + 3)(3x - 1)
Hence, (6x² + 7x - 3) = (2x + 3)(3x - 1).
6. Factorize: 4√3x² + 5x - 2√3.
Sol. Here 4√3 × (-2√3) = -24
We try to split 5x into two parts whose sum is 5 and product is = -24
Clearly, 8 + (-3) = 5 and 8 × -3 = -24
∴ Middle term 5x can be written as 8x - 3x
∴ 4√3x² + 5x - 2√3
= 4√3x² + 8x - 3x - 2√3
= (4√3x² + 8x) - (3x + 2√3) = 4x(√3x + 2) - √3(√3x + 2) = (√3x + 2)(4x - √3)
7. Find the value of a, if x - a is a factor of x³ - a²x + x + 3.
Sol. Let p(x) = x³ - a²x + x + 3 be the given polynomial. By factor theorem, (x - a) is a factor of p(x) if p(a) = 0.
Now,
Putting x = a in the given expression, we have
⇒ a³ - a²·a + a + 3 = 0 ⇒ a³ - a³ + a + 3 = 0 ⇒ a + 3 = 0 ⇒ a = -3
Hence, (x - a) is a factor of the given polynomial, if a = -3
8. Find the value of q, if x - 3 is a factor of 3x² + qx + 9.
Sol. Let p(x) = 3x² + qx + 9 be the given polynomial. According to question (x - 3) is a factor of p(x)
∴ we have p(3) = 0
⇒ 3(-3)² + q(-3) + 9 = 0 ⇒ 27 - 3q + 9 = 0 ⇒ 36 - 3q = 0 ⇒ q = 12
Hence, x - 3 is a factor of 3x² + qx + 9 if q = 12.
9. Factorise 6x² + 17x + 5 by splitting the middle term, and by using the Factor Theorem.
Sol. Splitting method: If we can find two numbers s and t such that s + t = 17 and st = 6×5 = 30, then we can get the factors.
We try to split 17x into two parts whose sum is 17 and product is 30
∴ 2 + 15 = 17 and 2 × 15 = 30
So, 6x² + 17x + 5 = 6x² + (2 + 15)x + 5
= 6x² + 2x + 15x + 5
= 2x(3x + 1) + 5(3x + 1)
= (3x + 1)(2x + 5)
10. Factorise x³ - 23x² + 142x - 120.
Sol. Let p(x) = x³ - 23x² + 142x - 120
Factors of the constant term i.e., -120 are ±1, ±2, ±3, ±4, ±5, ±6, ±8, ±10, ±12, ±15, ±20, ±24, ±30, ±60.
Now we conclude that p(1) = 0.
So x - 1 is a factor of p(x).
∴ x³ - 23x² + 142x - 120 = x³ - x² - 22x² + 22x + 120x - 120
= x²(x - 1) - 22x(x - 1) + 120(x - 1)
[Taking x² common from consecutive two terms]
= (x - 1)(x² - 22x + 120) [Taking (x - 1) common]
Note: It can also be done by dividing p(x) by x - 1.
Now x² - 22x + 120 can be factorised either by splitting the middle term or by using the Factor theorem. By splitting the middle term, we have:
x² - 22x + 120 = x² - 12x - 10x + 120 [∵ -12 - 10 = -22 and -12 × 10 = 120]
= x(x - 12) - 10(x - 12)
= (x - 12)(x - 10)
So, x³ - 23x² + 142x - 120 = (x - 1)(x - 10)(x - 12)
11. Factorize x³ - 3x² - 9x - 5
Sol. Let p(x) = x³ - 3x² - 9x - 5
Factors of constant term (-5) are ±1 and ±5.
We find that
p(-1) = (-1)³ - 3(-1)² - 9(-1) - 5 = 0
∴ (x + 1) is a factor of p(x).
[Using long division]
Thus we have,
p(x) = (x + 1)(x² - 4x - 5) = (x + 1)[x² - 5x + x - 5]
= (x + 1)[x(x - 5) + 1(x - 5)]
= (x + 1)(x + 1)(x - 5) = (x + 1)²(x - 5)
12. For what value of k is the polynomial (2x⁴ + 3x³ + 2kx² + 3x + 6) exactly divisible by (x + 2)?
Sol. Let f(x) = 2x⁴ + 3x³ + 2kx² + 3x + 6.
Now by taking x + 2 = 0 we have x = -2.
By the factor theorem, f(x) will be exactly divisible by (x + 2) if f(-2) = 0.
Now, f(-2) = 2(-2)⁴ + 3(-2)³ + 2k(-2)² + 3(-2) + 6
= 32 - 24 + 8k - 6 + 6 = 8k + 8.
∴ f(-2) = 0 ⇒ 8k + 8 = 0
⇒ 8k = -8
⇒ k = -1.
Hence, f(x) is exactly divisible by (x + 2), when k = -1.
13. Find the following products using appropriate identities:
(i) (x + 3)(x + 3)
(ii) (x + 3)(x - 5)
Sol. (i) Here we can use Identity (x + y)² = x² + 2xy + y².
Putting y = 3 in it, we get
(x + 3)(x + 3) = (x + 3)² = x² + 2(x)(3) + (3)²
= x² + 6x + 9
(ii) Using Identity (x + a)(x + b) = x² + (a + b)x + ab, we have
(x + 3)(x - 5) = x² + (-3 + 5)x + (-3)(5)
= x² + 2x - 15
14. Factorize:
(i) 49a² + 70ab + 25b²
(ii) (36/4)x² - y²/9
Sol. (i) Here we can see that
49a² + 70ab + 25b² = (7a)² + 2·7a·5b + (5b)²
Comparing the given expression with x² + 2xy + y², we observe that x = 7a and y = 5b.
Using Identity x² + 2xy + y² = (x + y)²
We get,
49a² + 70ab + 25b² = (7a + 5b)² = (7a + 5b)(7a + 5b)
(ii) We have (36/4)x² - y²/9 = (5x/2)² - (y/3)²
Now comparing it with Identity x² - y² = (x + y)(x - y)
We get, (5x/2)² - (y/3)² = (5x/2 + y/3)(5x/2 - y/3)
15. If x + y = 12 and xy = 32, find the value of x² + y².
Sol. We have,
(x + y)² = x² + y² + 2xy
⇒ 12² = x² + y² + 2×32 [Putting x + y = 12 and xy = 32]
⇒ 144 = x² + y² + 64
⇒ 144 - 64 = x² + y² ⇒ x² + y² = 80
16. If 3x + 2y = 12 and xy = 6, find the value of 9x² + 4y².
Sol. We have,
(3x + 2y)² = (3x)² + (2y)² + 2·3x·2y
⇒ (3x + 2y)² = 9x² + 4y² + 12xy
⇒ 12² = 9x² + 4y² + 12×6 [Putting 3x + 2y = 12 and xy = 6]
⇒ 144 = 9x² + 4y² + 72 ⇒ 144 - 72 = 9x² + 4y² ⇒ 9x² + 4y² = 72
17. Write the following in expanded form:
(i) (-x + 2y + z)²
(ii) (1/4·a - 1/2·b + 1)²
Sol. (i) We have,
(-x + 2y + z)²
= {(-x) + 2y + z}²
= (-x)² + (2y)² + z² + 2(-x)(2y) + 2·2y·z + 2(-x)·z
= x² + 4y² + z² - 4xy + 4yz - 2zx
(ii) We have,
(1/4·a - 1/2·b + 1)² = [(a/4) + (-b/2) + 1]²
= (a/4)² + (-b/2)² + 1² + 2(a/4)(-b/2) + 2(-b/2)(1) + 2(1)(a/4)
= a²/16 + b²/4 + 1 - ab/4 - b + a/2
18. Write the following cubes in the expanded form:
(i) (3a + 4b)³
(ii) (5p - 3q)³
Sol. (i) Comparing the given expression with (x + y)³, we find that x = 3a and y = 4b.
So using (x + y)³ = x³ + y³ + 3xy(x + y)
(3a + 4b)³ = (3a)³ + (4b)³ + 3(3a)(4b)(3a + 4b)
= 27a³ + 64b³ + 36ab(3a + 4b)
= 27a³ + 64b³ + 108a²b + 144ab²
(ii) Comparing the given expression with (x - y)³, we find that
x = 5p, y = 3q
So, using (x - y)³ = x³ - y³ - 3xy(x - y)
(5p - 3q)³ = (5p)³ - (3q)³ - 3(5p)(3q)(5p - 3q)
= 125p³ - 27q³ - 45pq(5p - 3q)
= 125p³ - 27q³ - 225p²q + 135pq²
19. Evaluate each of the following using suitable identities:
(i) (104)³
(ii) (999)³
Sol. (i) We have
(104)³ = (100 + 4)³
= (100)³ + (4)³ + 3(100)(4)(100 + 4)
= 1000000 + 64 + 1200(100 + 4)
[(x + y)³ = x³ + y³ + 3xy(x + y)]
= 1000000 + 64 + 120000 + 4800 = 1124864
(ii) We have
(999)³ = (1000 - 1)³
= (1000)³ - (1)³ - 3(1000)(1)(1000 - 1)
[(x - y)³ = x³ - y³ - 3xy(x - y)]
= 1000000000 - 1 - 3000(1000 - 1)
= 1000000000 - 1 - 3000000 + 3000
= 997002999
20. Find the following products:
(i) (x + y + 2z)(x² + y² + 4z² - xy - 2yz - 2zx)
(ii) (2x - y + 3z)(4x² + y² + 9z² + 2xy - 3yz - 6xz)
Sol. (i) We have,
(x + y + 2z)(x² + y² + 4z² - xy - 2yz - 2zx)
= (x + y + 2z)(x² + y² + (2z)² - x·y - y·(2z) - (2z)·x)
= (X + Y + Z)(X² + Y² + Z² - XY - YZ - ZX), where X = x, Y = y, Z = 2z
= X³ + Y³ + Z³ - 3XYZ
= x³ + y³ + (2z)³ - 3·x·y·2z
= x³ + y³ + 8z³ - 6xyz
(ii) We have,
(2x - y + 3z)(4x² + y² + 9z² + 2xy - 3yz - 6xz)
= (2x + (-y) + 3z){(2x)² + (-y)² + (3z)² - 2x·(-y) - (-y)·(3z) - 2x·3z}
= (X + Y + Z)(X² + Y² + Z² - XY - YZ - ZX), where X = 2x, Y = -y, Z = 3z
= X³ + Y³ + Z³ - 3XYZ
= (2x)³ + (-y)³ + (3z)³ - 3·2x·(-y)·3z
= 8x³ - y³ + 27z³ + 18xyz
21. If x + y + z = 0, show that x³ + y³ + z³ = 3xyz.
Sol. x + y + z = 0 [Given]
⇒ x + y = -z ...(i)
Taking cube of both sides, we get
x³ + y³ + 3xy(x + y) = (-z)³
⇒ x³ + y³ + 3xy(-z) = -z³ [From equation (i)]
⇒ x³ + y³ - 3xyz = -z³
⇒ x³ + y³ + z³ = 3xyz
22. Without actually calculating the cubes. Find the value of each of the following:
(i) (-12)³ + 7³ + 5³
(ii) (28)³ + (-15)³ + (-13)³
Sol. (i) Let x = -12, y = 7 and z = 5. Then,
x + y + z = -12 + 7 + 5 = 0
⇒ x³ + y³ + z³ = 3xyz [∵ x + y + z = 0]
⇒ (-12)³ + 7³ + 5³ = 3(-12)·7·5 = -1260
(ii) Let x = 28, y = -15 and z = -13. Then,
x + y + z = 28 - 15 - 13 = 0
⇒ x³ + y³ + z³ = 3xyz
⇒ (28)³ + (-15)³ + (-13)³ = 3×28×(-15)×(-13) = 16380
23. If a + b + c = 9 and ab + bc + ca = 40, find a² + b² + c².
Sol. We know that
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
⇒ 9² = a² + b² + c² + 2×40
⇒ 81 = a² + b² + c² + 80 ⇒ a² + b² + c² = 81 - 80 ⇒ a² + b² + c² = 1
24. If a² + b² + c² = 250 and ab + bc + ca = 3, find a + b + c.
Sol. We know that
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
⇒ (a + b + c)² = 250 + 2×3
⇒ (a + b + c)² = 256
⇒ (a + b + c)² = (±16)²
⇒ a + b + c = ±16 [Taking square root of both sides]
25. Using suitable identity, factorize 3a² + 3b² + 6ab - 25c².
Sol. Given expression is 3a² + 3b² + 6ab - 25c²
By rearranging the terms we have 3a² + 3b² + 6ab - 25c²
Above expression can be written as (√3a)² + (√3b)² + 2(√3a)(√3b) - 25c²
= (√3a + √3b)² - (5c)² [(a + b)² = a² + b² + 2ab]
= (√3a + √3b + 5c)(√3a + √3b - 5c) [a² - b² = (a + b)(a - b)]
26. Using suitable identity, factorize p⁴ - 81q⁴
Sol. p⁴ - 81q⁴ = (p²)² - (9q²)²
= (p² + 9q²)(p² - 9q²) [a² - b² = (a + b)(a - b)]
= (p² + 9q²)[(p)² - (3q)²]
= (p² + 9q²)(p + 3q)(p - 3q)
27. Using suitable identity, factorize (x² + y²)² - 10(x² + y²) + 25
Sol. The given polynomial is (x² + y²)² - 10(x² + y²) + 25 ..... (i)
Let x² + y² = a
Then from (i),
Given polynomial = a² - 10a + 25 = a² - 2·5·a + (5)²
= (a - 5)² = (x² + y² - 5)²
= (x² + y² - 5)(x² + y² - 5)
Hence (x² + y²)² - 10(x² + y²) + 25 = (x² + y² - 5)(x² + y² - 5).
28. If x² + 1/x² = 27, find the values of each of the following:
(i) x + 1/x
(ii) x - 1/x
Sol. (i) We have,
(x + 1/x)² = x² + 1/x² + 2·x·(1/x) [∵ (x + y)² = x² + 2xy + y²]
⇒ (x + 1/x)² = x² + 1/x² + 2
⇒ (x + 1/x)² = 27 + 2 [∵ x² + 1/x² = 27 (given)]
⇒ (x + 1/x)² = 29
⇒ x + 1/x = ±√29 [Taking square root of both sides]
(ii) We have,
(x - 1/x)² = x² + 1/x² - 2·x·(1/x)
⇒ (x - 1/x)² = x² + 1/x² - 2
⇒ (x - 1/x)² = 27 - 2 [∵ x² + 1/x² = 27(given)]
⇒ (x - 1/x)² = 25
⇒ (x - 1/x)² = (±5)²
⇒ x - 1/x = ±5
29. If x - 1/x = 4, then evaluate x² + 1/x² and x⁴ + 1/x⁴.
Sol. x - 1/x = 4
⇒ (x - 1/x)² = x² + 1/x² - 2·x·(1/x) [∵ a² + b² = a² + b² - 2ab]
⇒ (4)² = x² + 1/x² - 2
⇒ 16 + 2 = x² + 1/x²
⇒ 18 = x² + 1/x²
(x² + 1/x²)² = (x²)² + (1/x²)² + 2·x²·(1/x²)
⇒ x⁴ + 1/x⁴ + 2 = (18)²
⇒ x⁴ + 1/x⁴ = 324 - 2
⇒ x⁴ + 1/x⁴ = 322
30. If a + b + c = 15 and a² + b² + c² = 83, find the value of a³ + b³ + c³ - 3abc.
Sol. We know that
a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)
or a³ + b³ + c³ - 3abc = (a + b + c){(a² + b² + c²) - (ab + bc + ca)} ...(i)
The values of a + b + c and a² + b² + c² are given. So, we need the value of ab + bc + ca.
Now,
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
⇒ (15)² = 83 + 2(ab + bc + ca)
⇒ 225 = 83 + 2(ab + bc + ca) ⇒ 225 - 83 = 2(ab + bc + ca)
⇒ 142 = 2(ab + bc + ca)
⇒ ab + bc + ca = 142/2 = 71
Substituting the value of ab + bc + ca in (i), we get
a³ + b³ + c³ - 3abc = 15×(83 - 71) = 15×12 = 180
31. Give possible expressions for the length and breadth of the rectangle whose area is 25a² - 35a + 12.
Sol. We know that
Area of a rectangle = Length × Breadth
We have,
Area = 25a² - 35a + 12
By splitting the middle term
⇒ Area = 25a² - 20a - 15a + 12
[∵ -20 - 15 = -35 and -20 × -15 = 300]
⇒ Area = 5a(5a - 4) - 3(5a - 4) = (5a - 4)(5a - 3)
Hence, the possible expressions for the length and breadth are:
| Length | Breadth |
|---|---|
| (i) 5a - 4 | 5a - 3 |
| (ii) 5a - 3 | 5a - 4 etc. |
32. What are the possible expressions for the dimensions of a cuboid whose volume is 2ky² - 6ky - 20k.
Sol. We know that
Volume of a cuboid = Length × Breadth × Height
We have,
Volume of the cuboid = 2ky² - 6ky - 20k
= 2k(y² - 3y - 10) [Taking 2k common]
= 2k(y² - 5y + 2y - 10)
= 2k[y(y - 5) - 2(y - 5)] = 2k(y - 5)(y + 2)
Hence, the possible expressions for the dimensions of the cuboid are:
| Length | Breadth | Height |
|---|---|---|
| (i) 2k | y + 5 | y - 2 |
| (ii) k | y + 5 | 2(y - 2) |
| (iii) k | 2(y + 5) | y - 2 |
| (iv) y - 2 | y + 5 | 2k |
| (v) y + 5 | 2k | y - 2 |
| (vi) 2(y - 2) | y + 5 | k etc. |