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Samacheer Kalvi 10th Maths Exercise 3.3 Solutions

10th Maths Exercise 3.3 solutions (Chapter 3 Algebra) with step-by-step answers for all 9 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance6 answers · full steps below
  1. Q2 (i)LCM=(a+6)(a−2)(a−3)LCM=(a+6)(a-2)(a-3)
  2. Q2 (ii)LCM=x(x−3a)2(x2+3ax+9a2)LCM=x(x-3a)^2(x^2+3ax+9a^2)
  3. Q3 (i)GCD=4x2(x−1)GCD=4x^2(x-1)
  4. Q3 (ii)GCD=x2−xy+y2GCD=x^2-xy+y^2
  5. Q4 (i)q(x)=a2−5a−14q(x)=a^2-5a-14
  6. Q4 (ii)p(x)=(x2+xy+y2)(x2−y2)p(x)=(x^2+xy+y^2)(x^2-y^2)

Q1. Find the LCM and GCD for the following and verify that f(x)×g(x)=LCM×GCDf(x) \times g(x) = LCM \times GCD

(i)

21x2y, 35xy221x^2y,\ 35xy^2

Show steps (7)

  1. f(x)=21x2y=3×7×x2×yf(x)=21x^2y=3\times 7\times x^2\times y

  2. g(x)=35xy2=5×7×x×y2g(x)=35xy^2=5\times 7\times x\times y^2

  3. GCD = common factors with least powers =7xy=7xy

  4. LCM = all factors with highest powers =3×5×7×x2y2=105x2y2=3\times5\times7\times x^2y^2=105x^2y^2

  5. f(x)×g(x)=21x2y×35xy2=735x3y3f(x)\times g(x)=21x^2y\times 35xy^2=735x^3y^3

  6. LCM×GCD=105x2y2×7xy=735x3y3LCM\times GCD=105x^2y^2\times 7xy=735x^3y^3

  7. Both are equal.

Answer

LCM=105x2y2LCM=105x^2y^2, GCD=7xyGCD=7xy and f(x)×g(x)=LCM×GCD=735x3y3f(x)\times g(x)=LCM\times GCD=735x^3y^3

(ii)

(x3−1)(x+1), (x3+1)(x^3-1)(x+1),\ (x^3+1)

Show steps (7)

  1. f(x)=(x3−1)(x+1)=(x−1)(x2+x+1)(x+1)f(x)=(x^3-1)(x+1)=(x-1)(x^2+x+1)(x+1)

  2. g(x)=x3+1=(x+1)(x2−x+1)g(x)=x^3+1=(x+1)(x^2-x+1)

  3. GCD =x+1=x+1

  4. LCM =(x−1)(x2+x+1)(x+1)(x2−x+1)=(x3−1)(x3+1)=x6−1=(x-1)(x^2+x+1)(x+1)(x^2-x+1)=(x^3-1)(x^3+1)=x^6-1

  5. f(x)×g(x)=(x3−1)(x+1)(x3+1)=(x6−1)(x+1)f(x)\times g(x)=(x^3-1)(x+1)(x^3+1)=(x^6-1)(x+1)

  6. LCM×GCD=(x6−1)(x+1)LCM\times GCD=(x^6-1)(x+1)

  7. Both are equal.

Answer

LCM=x6−1LCM=x^6-1, GCD=x+1GCD=x+1 and f(x)×g(x)=LCM×GCD=(x6−1)(x+1)f(x)\times g(x)=LCM\times GCD=(x^6-1)(x+1)

(iii)

(x2y+xy2), (x2+xy)(x^2y+xy^2),\ (x^2+xy)

Show steps (7)

  1. f(x)=x2y+xy2=xy(x+y)f(x)=x^2y+xy^2=xy(x+y)

  2. g(x)=x2+xy=x(x+y)g(x)=x^2+xy=x(x+y)

  3. GCD =x(x+y)=x(x+y)

  4. LCM =xy(x+y)=xy(x+y)

  5. f(x)×g(x)=xy(x+y)×x(x+y)=x2y(x+y)2f(x)\times g(x)=xy(x+y)\times x(x+y)=x^2y(x+y)^2

  6. LCM×GCD=xy(x+y)×x(x+y)=x2y(x+y)2LCM\times GCD=xy(x+y)\times x(x+y)=x^2y(x+y)^2

  7. Both are equal.

Answer

LCM=xy(x+y)LCM=xy(x+y), GCD=x(x+y)GCD=x(x+y) and f(x)×g(x)=LCM×GCD=x2y(x+y)2f(x)\times g(x)=LCM\times GCD=x^2y(x+y)^2

Q2. Find the LCM of each pair of the following polynomials

(i)

a2+4a−12, a2−5a+6a^2+4a-12,\ a^2-5a+6 whose GCDGCD is a−2a-2

Show steps (6)

  1. LCM=f(x)×g(x)GCDLCM=\dfrac{f(x)\times g(x)}{GCD}

  2. f(x)×g(x)=(a2+4a−12)(a2−5a+6)f(x)\times g(x)=(a^2+4a-12)(a^2-5a+6)

  3. a2+4a−12=(a+6)(a−2)a^2+4a-12=(a+6)(a-2)

  4. a2−5a+6=(a−2)(a−3)a^2-5a+6=(a-2)(a-3)

  5. LCM=(a+6)(a−2)(a−2)(a−3)a−2LCM=\dfrac{(a+6)(a-2)(a-2)(a-3)}{a-2}

  6. =(a+6)(a−2)(a−3)=(a+6)(a-2)(a-3)

Answer

LCM=(a+6)(a−2)(a−3)LCM=(a+6)(a-2)(a-3)

(ii)

x4−27a3x, (x−3a)2x^4-27a^3x,\ (x-3a)^2 whose GCDGCD is (x−3a)(x-3a)

  1. LCM=f(x)×g(x)GCDLCM=\dfrac{f(x)\times g(x)}{GCD}

  2. x4−27a3x=x(x3−27a3)=x(x−3a)(x2+3ax+9a2)x^4-27a^3x=x(x^3-27a^3)=x(x-3a)(x^2+3ax+9a^2)

  3. LCM=x(x−3a)(x2+3ax+9a2) (x−3a)2(x−3a)LCM=\dfrac{x(x-3a)(x^2+3ax+9a^2)\,(x-3a)^2}{(x-3a)}

  4. =x(x−3a)2(x2+3ax+9a2)=x(x-3a)^2(x^2+3ax+9a^2)

Answer

LCM=x(x−3a)2(x2+3ax+9a2)LCM=x(x-3a)^2(x^2+3ax+9a^2)

Q3. Find the GCD of each pair of the following polynomials

(i)

12(x4−x3), 8(x4−3x3+2x2)12(x^4-x^3),\ 8(x^4-3x^3+2x^2) whose LCM is 24x3(x−1)(x−2)24x^3(x-1)(x-2)

Show steps (6)

  1. GCD=f(x)×g(x)LCMGCD=\dfrac{f(x)\times g(x)}{LCM}

  2. 12(x4−x3)=12x3(x−1)12(x^4-x^3)=12x^3(x-1)

  3. 8(x4−3x3+2x2)=8x2(x−1)(x−2)8(x^4-3x^3+2x^2)=8x^2(x-1)(x-2)

  4. f(x)×g(x)=96x5(x−1)2(x−2)f(x)\times g(x)=96x^5(x-1)^2(x-2)

  5. GCD=96x5(x−1)2(x−2)24x3(x−1)(x−2)GCD=\dfrac{96x^5(x-1)^2(x-2)}{24x^3(x-1)(x-2)}

  6. =4x2(x−1)=4x^2(x-1)

Answer

GCD=4x2(x−1)GCD=4x^2(x-1)

(ii)

(x3+y3), (x4+x2y2+y4)(x^3+y^3),\ (x^4+x^2y^2+y^4) whose LCM is (x3+y3)(x2+xy+y2)(x^3+y^3)(x^2+xy+y^2)

  1. GCD=f(x)×g(x)LCMGCD=\dfrac{f(x)\times g(x)}{LCM}

  2. GCD=(x3+y3)(x4+x2y2+y4)(x3+y3)(x2+xy+y2)GCD=\dfrac{(x^3+y^3)(x^4+x^2y^2+y^4)}{(x^3+y^3)(x^2+xy+y^2)}

  3. =x4+x2y2+y4x2+xy+y2=\dfrac{x^4+x^2y^2+y^4}{x^2+xy+y^2}

  4. x4+x2y2+y4=(x2+xy+y2)(x2−xy+y2)x^4+x^2y^2+y^4=(x^2+xy+y^2)(x^2-xy+y^2)

  5. GCD=x2−xy+y2GCD=x^2-xy+y^2

Answer

GCD=x2−xy+y2GCD=x^2-xy+y^2

Q4. Given the LCM and GCD of the two polynomials p(x)p(x) and q(x)q(x) …

Given the LCM and GCD of the two polynomials p(x)p(x) and q(x)q(x) find the unknown polynomial in the following table

(i)

LCM =a3−10a2+11a+70= a^3-10a^2+11a+70, GCD =a−7= a-7, p(x)=a2−12a+35p(x) = a^2-12a+35, q(x)=?q(x) = ?

S.No. LCM GCD p(x)p(x) q(x)q(x)
(i) a3−10a2+11a+70a^3-10a^2+11a+70 a−7a-7 a2−12a+35a^2-12a+35
(ii) (x4−y4)(x4+x2y2+y4)(x^4-y^4)(x^4+x^2y^2+y^4) (x2−y2)(x^2-y^2) (x4−y4)(x2+y2−xy)(x^4-y^4)(x^2+y^2-xy)
Show steps (6)

  1. p(x)×q(x)=LCM×GCDp(x)\times q(x)=LCM\times GCD

  2. q(x)=LCM×GCDp(x)=(a3−10a2+11a+70)(a−7)a2−12a+35q(x)=\dfrac{LCM\times GCD}{p(x)}=\dfrac{(a^3-10a^2+11a+70)(a-7)}{a^2-12a+35}

  3. a3−10a2+11a+70=(a+2)(a−5)(a−7)a^3-10a^2+11a+70=(a+2)(a-5)(a-7)

  4. a2−12a+35=(a−5)(a−7)a^2-12a+35=(a-5)(a-7)

  5. q(x)=(a+2)(a−5)(a−7)(a−7)(a−5)(a−7)q(x)=\dfrac{(a+2)(a-5)(a-7)(a-7)}{(a-5)(a-7)}

  6. =(a+2)(a−7)=a2−5a−14=(a+2)(a-7)=a^2-5a-14

Answer

q(x)=a2−5a−14q(x)=a^2-5a-14

(ii)

LCM =(x4−y4)(x4+x2y2+y4)= (x^4-y^4)(x^4+x^2y^2+y^4), GCD =(x2−y2)= (x^2-y^2), q(x)=(x4−y4)(x2+y2−xy)q(x) = (x^4-y^4)(x^2+y^2-xy), p(x)=?p(x) = ?

S.No. LCM GCD p(x)p(x) q(x)q(x)
(i) a3−10a2+11a+70a^3-10a^2+11a+70 a−7a-7 a2−12a+35a^2-12a+35
(ii) (x4−y4)(x4+x2y2+y4)(x^4-y^4)(x^4+x^2y^2+y^4) (x2−y2)(x^2-y^2) (x4−y4)(x2+y2−xy)(x^4-y^4)(x^2+y^2-xy)

  1. p(x)=LCM×GCDq(x)p(x)=\dfrac{LCM\times GCD}{q(x)}

  2. p(x)=(x4−y4)(x4+x2y2+y4)(x2−y2)(x4−y4)(x2+y2−xy)p(x)=\dfrac{(x^4-y^4)(x^4+x^2y^2+y^4)(x^2-y^2)}{(x^4-y^4)(x^2+y^2-xy)}

  3. =(x4+x2y2+y4)(x2−y2)x2−xy+y2=\dfrac{(x^4+x^2y^2+y^4)(x^2-y^2)}{x^2-xy+y^2}

  4. x4+x2y2+y4=(x2+xy+y2)(x2−xy+y2)x^4+x^2y^2+y^4=(x^2+xy+y^2)(x^2-xy+y^2)

  5. p(x)=(x2+xy+y2)(x2−y2)p(x)=(x^2+xy+y^2)(x^2-y^2)

Answer

p(x)=(x2+xy+y2)(x2−y2)p(x)=(x^2+xy+y^2)(x^2-y^2)

Quick check

  1. Graph of a linear equation is a _______

    1. (1) straight line
    2. (2) circle
    3. (3) parabola
    4. (4) hyperbola
    Show answer

    Answer: (1) straight line

    1. A linear equation in two variables has degree one, and its graph is a straight line.

    Question 1 of 5

  2. 3y−3y÷7y−73y2\dfrac{3y - 3}{y} \div \dfrac{7y - 7}{3y^2} is

    1. (1) 9y7\dfrac{9y}{7}
    2. (2) 9y3(21y−21)\dfrac{9y^3}{(21y - 21)}
    3. (3) 21y2−42y+213y3\dfrac{21y^2 - 42y + 21}{3y^3}
    4. (4) 7(y2−2y+1)y2\dfrac{7(y^2 - 2y + 1)}{y^2}
    Show answer

    Answer: (1) 9y7\dfrac{9y}{7}

    1. 3y−3y÷7y−73y2=3(y−1)y×3y27(y−1)\dfrac{3y-3}{y} \div \dfrac{7y-7}{3y^2} = \dfrac{3(y-1)}{y} \times \dfrac{3y^2}{7(y-1)}

    2. Cancel (y−1)(y-1) and one yy.

    3. =9y7= \dfrac{9y}{7}

    Question 2 of 5

  3. Transpose of a column matrix is

    1. (1) unit matrix
    2. (2) diagonal matrix
    3. (3) column matrix
    4. (4) row matrix
    Show answer

    Answer: (4) row matrix

    1. A column matrix has order m×1m \times 1.

    2. Its transpose has order 1×m1 \times m, which is a row matrix.

    Question 3 of 5

  4. The solution of (2x−1)2=9\left(2x - 1\right)^2 = 9 is equal to

    1. (1) −1-1
    2. (2) 22
    3. (3) −1, 2-1,\ 2
    4. (4) None of these
    Show answer

    Answer: (3) −1, 2-1,\ 2

    1. (2x−1)2=9⇒2x−1=±3(2x-1)^2 = 9 \Rightarrow 2x - 1 = \pm 3.

    2. 2x−1=3⇒x=22x - 1 = 3 \Rightarrow x = 2.

    3. 2x−1=−3⇒x=−12x - 1 = -3 \Rightarrow x = -1.

    Question 4 of 5

  5. The values of aa and bb if 4x4−24x3+76x2+ax+b4x^4 - 24x^3 + 76x^2 + ax + b is a perfect square are

    1. (1) 100,120100,120
    2. (2) 10,1210,12
    3. (3) −120,100-120,100
    4. (4) 12,1012,10
    Show answer

    Answer: (3) −120,100-120,100

    1. Let 4x4−24x3+76x2+ax+b=(2x2−6x+c)24x^4 - 24x^3 + 76x^2 + ax + b = (2x^2 - 6x + c)^2.

    2. Expanding: 4x4−24x3+(36+4c)x2−12cx+c24x^4 - 24x^3 + (36 + 4c)x^2 - 12cx + c^2.

    3. Compare x2x^2 terms: 36+4c=76⇒c=1036 + 4c = 76 \Rightarrow c = 10.

    4. So a=−12(10)=−120a = -12(10) = -120 and b=102=100b = 10^2 = 100.

    Question 5 of 5
Chapter 3 formulas
  • Roots of a quadratic equation

    ax2+bx+c=0,  (a≠0)  ⇒  x=−b±b2−4ac2aax^2 + bx + c = 0,\; (a \neq 0) \;\Rightarrow\; x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

  • Sum of the roots

    α+β=−ba=−Co-efficient of xCo-efficient of x2\alpha + \beta = \dfrac{-b}{a} = \dfrac{-\text{Co-efficient of } x}{\text{Co-efficient of } x^2}

  • Product of the roots

    αβ=ca=Constant termCo-efficient of x2\alpha\beta = \dfrac{c}{a} = \dfrac{\text{Constant term}}{\text{Co-efficient of } x^2}

  • Quadratic equation formed from its roots

    x2−(α+β)x+αβ=0x^2 - (\alpha + \beta)x + \alpha\beta = 0

  • Discriminant

    Δ=b2−4ac\Delta = b^2 - 4ac

  • Nature of roots: real and unequal

    Δ>0  ⇒  the roots are real and unequal\Delta > 0 \;\Rightarrow\; \text{the roots are real and unequal}

  • Nature of roots: real and equal

    Δ=0  ⇒  the roots are real and equal\Delta = 0 \;\Rightarrow\; \text{the roots are real and equal}

  • Nature of roots: no real roots

    Δ<0  ⇒  there are no real roots\Delta < 0 \;\Rightarrow\; \text{there are no real roots}

  • Cases for a system of linear equations in three variables

    (i) Unique solution(ii) Infinitely many solutions(iii) No solution\text{(i) Unique solution}\quad \text{(ii) Infinitely many solutions}\quad \text{(iii) No solution}

  • LCM of algebraic expressions

    LCM=expression of lowest degree that each given expression exactly divides\text{LCM} = \text{expression of lowest degree that each given expression exactly divides}

  • Quadratic polynomial and its zeroes

    p(x)=ax2+bx+c  (a≠0)  ⇒  p(x) has atmost two zeroes, which lie on the x-axisp(x) = ax^2 + bx + c \;(a \neq 0) \;\Rightarrow\; p(x) \text{ has atmost two zeroes, which lie on the } x\text{-axis}

  • Order of a matrix

    Order of A=(Number of rows)×(Number of columns)=m×n\text{Order of } A = (\text{Number of rows}) \times (\text{Number of columns}) = m \times n

  • Row matrix (row vector)

    A=(a11a12⋯a1n)1×nA = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \end{pmatrix}_{1 \times n}

  • Column matrix (column vector)

    A=(a11a21⋮am1)m×1A = \begin{pmatrix} a_{11} \\ a_{21} \\ \vdots \\ a_{m1} \end{pmatrix}_{m \times 1}

  • Square matrix

    Number of rows=Number of columns,  order n×n\text{Number of rows} = \text{Number of columns},\; \text{order } n \times n

  • Zero matrix (null matrix)

    aij=0 for all i, ja_{ij} = 0 \text{ for all } i,\, j

  • Transpose of a matrix

    (AT)ij=aji(rows and columns of A interchanged)\left(A^{T}\right)_{ij} = a_{ji} \quad (\text{rows and columns of } A \text{ interchanged})

  • Diagonal matrix

    aij=0 for i≠j(square matrix)a_{ij} = 0 \text{ for } i \neq j \quad (\text{square matrix})

  • Scalar matrix

    aij={k,i=j0,i≠ja_{ij} = \begin{cases} k, & i = j \\ 0, & i \neq j \end{cases}

  • Identity matrix (unit matrix)

    aij={1,i=j0,i≠ja_{ij} = \begin{cases} 1, & i = j \\ 0, & i \neq j \end{cases}

  • Lower triangular matrix

    aij=0 for i<j(all entries above the leading diagonal are zero)a_{ij} = 0 \text{ for } i < j \quad (\text{all entries above the leading diagonal are zero})

  • Upper triangular matrix

    aij=0 for i>j(all entries below the leading diagonal are zero)a_{ij} = 0 \text{ for } i > j \quad (\text{all entries below the leading diagonal are zero})

  • Equality of matrices

    A=B  ⟺  same order and aij=bij for all i, jA = B \iff \text{same order and } a_{ij} = b_{ij} \text{ for all } i,\, j

Answers checked against the Samacheer Kalvi textbook · Updated

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