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Samacheer Kalvi 10th Maths Exercise 3.2 Solutions

10th Maths Exercise 3.2 solutions (Chapter 3 Algebra) with step-by-step answers for all 10 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance10 answers · full steps below
  1. Q1 (i)GCD=(x−1)(x+3)=x2+2x−3\text{GCD}=(x-1)(x+3)=x^2+2x-3
  2. Q1 (ii)GCD=x2+1\text{GCD}=x^2+1
  3. Q1 (iii)GCD=x(x+2)2\text{GCD}=x(x+2)^2
  4. Q1 (iv)GCD=3(x2+1)\text{GCD}=3(x^2+1)
  5. Q2 (i)8x3y28x^3y^2
  6. Q2 (ii)36a3b2c36a^3b^2c
  7. Q2 (iii)48m2n248m^2n^2
  8. Q2 (iv)LCM=(p−1)(p−2)(p+2)\text{LCM}=(p-1)(p-2)(p+2)
  9. Q2 (v)LCM=4(2x+1)(x+3)(x−3)\text{LCM}=4(2x+1)(x+3)(x-3)
  10. Q2 (vi)LCM=8x2(2x−3y)3(4x2+6xy+9y2)\text{LCM}=8x^2(2x-3y)^3(4x^2+6xy+9y^2)

Question 1

(i)

Find the GCD of the given polynomials: x4+3x3−x−3x^4 + 3x^3 - x - 3, x3+x2−5x+3x^3 + x^2 - 5x + 3

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

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

  3. For x3+x2−5x+3x^3+x^2-5x+3, put x=1x=1: 1+1−5+3=01+1-5+3=0, so (x−1)(x-1) is a factor.

  4. x3+x2−5x+3=(x−1)(x2+2x−3)=(x−1)(x−1)(x+3)x^3+x^2-5x+3=(x-1)(x^2+2x-3)=(x-1)(x-1)(x+3)

  5. Common factors: (x−1)(x-1) and (x+3)(x+3).

Answer

GCD=(x−1)(x+3)=x2+2x−3\text{GCD}=(x-1)(x+3)=x^2+2x-3

(ii)

Find the GCD of the given polynomials: x4−1x^4 - 1, x3−11x2+x−11x^3 - 11x^2 + x - 11

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

  2. x3−11x2+x−11=x2(x−11)+1(x−11)=(x−11)(x2+1)x^3-11x^2+x-11=x^2(x-11)+1(x-11)=(x-11)(x^2+1)

  3. The only common factor is x2+1x^2+1.

Answer

GCD=x2+1\text{GCD}=x^2+1

(iii)

Find the GCD of the given polynomials: 3x4+6x3−12x2−24x3x^4 + 6x^3 - 12x^2 - 24x, 4x4+14x3+8x2−8x4x^4 + 14x^3 + 8x^2 - 8x

Show steps (6)

  1. 3x4+6x3−12x2−24x=3x(x3+2x2−4x−8)3x^4+6x^3-12x^2-24x=3x(x^3+2x^2-4x-8)

  2. =3x[x2(x+2)−4(x+2)]=3x(x+2)(x2−4)=3x(x+2)2(x−2)=3x[x^2(x+2)-4(x+2)]=3x(x+2)(x^2-4)=3x(x+2)^2(x-2)

  3. 4x4+14x3+8x2−8x=2x(2x3+7x2+4x−4)4x^4+14x^3+8x^2-8x=2x(2x^3+7x^2+4x-4)

  4. Put x=−2x=-2: −16+28−8−4=0-16+28-8-4=0, so (x+2)(x+2) is a factor.

  5. =2x(x+2)(2x2+3x−2)=2x(x+2)(x+2)(2x−1)=2x(x+2)2(2x−1)=2x(x+2)(2x^2+3x-2)=2x(x+2)(x+2)(2x-1)=2x(x+2)^2(2x-1)

  6. Common: xx and (x+2)2(x+2)^2 (numerical GCD of 33 and 22 is 11).

Answer

GCD=x(x+2)2\text{GCD}=x(x+2)^2

(iv)

Find the GCD of the given polynomials: 3x3+3x2+3x+33x^3 + 3x^2 + 3x + 3, 6x3+12x2+6x+126x^3 + 12x^2 + 6x + 12

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

  2. =3[x2(x+1)+1(x+1)]=3(x+1)(x2+1)=3[x^2(x+1)+1(x+1)]=3(x+1)(x^2+1)

  3. 6x3+12x2+6x+12=6(x3+2x2+x+2)6x^3+12x^2+6x+12=6(x^3+2x^2+x+2)

  4. =6[x2(x+2)+1(x+2)]=6(x+2)(x2+1)=6[x^2(x+2)+1(x+2)]=6(x+2)(x^2+1)

  5. Common numerical factor 33, common algebraic factor (x2+1)(x^2+1).

Answer

GCD=3(x2+1)\text{GCD}=3(x^2+1)

Question 2

(i)

Find the LCM of the given expressions: 4x2y4x^2y, 8x3y28x^3y^2

  1. LCM of coefficients: LCM(4,8)=8(4,8)=8

  2. LCM of variables: LCM(x2y,  x3y2)=x3y2(x^2y,\;x^3y^2)=x^3y^2

  3. Multiply both parts.

Answer

8x3y28x^3y^2

(ii)

Find the LCM of the given expressions: 9a3b29a^3b^2, 12a2b2c12a^2b^2c

  1. LCM of coefficients: LCM(9,12)=36(9,12)=36

  2. LCM of variables: LCM(a3b2,  a2b2c)=a3b2c(a^3b^2,\;a^2b^2c)=a^3b^2c

  3. Multiply both parts.

Answer

36a3b2c36a^3b^2c

(iii)

Find the LCM of the given expressions: 16m16m, 12m2n212m^2n^2, 8n28n^2

  1. LCM of coefficients: LCM(16,12,8)=48(16,12,8)=48

  2. LCM of variables: LCM(m,  m2n2,  n2)=m2n2(m,\;m^2n^2,\;n^2)=m^2n^2

  3. Multiply both parts.

Answer

48m2n248m^2n^2

(iv)

Find the LCM of the given expressions: p2−3p+2p^2 - 3p + 2, p2−4p^2 - 4

  1. p2−3p+2=(p−1)(p−2)p^2-3p+2=(p-1)(p-2)

  2. p2−4=(p+2)(p−2)p^2-4=(p+2)(p-2)

  3. Take each factor the greatest number of times it occurs.

Answer

LCM=(p−1)(p−2)(p+2)\text{LCM}=(p-1)(p-2)(p+2)

(v)

Find the LCM of the given expressions: 2x2−5x−32x^2 - 5x - 3, 4x2−364x^2 - 36

  1. 2x2−5x−3=2x2−6x+x−3=(x−3)(2x+1)2x^2-5x-3=2x^2-6x+x-3=(x-3)(2x+1)

  2. 4x2−36=4(x2−9)=4(x+3)(x−3)4x^2-36=4(x^2-9)=4(x+3)(x-3)

  3. Take all factors with highest powers.

Answer

LCM=4(2x+1)(x+3)(x−3)\text{LCM}=4(2x+1)(x+3)(x-3)

(vi)

Find the LCM of the given expressions: (2x2−3xy)2(2x^2 - 3xy)^2, (4x−6y)3(4x - 6y)^3, 8x3−27y38x^3 - 27y^3

  1. (2x2−3xy)2=[x(2x−3y)]2=x2(2x−3y)2(2x^2-3xy)^2=[x(2x-3y)]^2=x^2(2x-3y)^2

  2. (4x−6y)3=[2(2x−3y)]3=8(2x−3y)3(4x-6y)^3=[2(2x-3y)]^3=8(2x-3y)^3

  3. 8x3−27y3=(2x−3y)(4x2+6xy+9y2)8x^3-27y^3=(2x-3y)(4x^2+6xy+9y^2)

  4. Highest powers: 88, x2x^2, (2x−3y)3(2x-3y)^3 and (4x2+6xy+9y2)(4x^2+6xy+9y^2).

Answer

LCM=8x2(2x−3y)3(4x2+6xy+9y2)\text{LCM}=8x^2(2x-3y)^3(4x^2+6xy+9y^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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