{"id":1084,"date":"2023-11-13T07:28:35","date_gmt":"2023-11-13T01:58:35","guid":{"rendered":"https:\/\/wordpress-505192-1602719.cloudwaysapps.com\/?p=1084"},"modified":"2023-11-14T15:24:26","modified_gmt":"2023-11-14T09:54:26","slug":"samacheer-kalvi-8th-maths-term-1-chapter-3-intext-questions","status":"publish","type":"post","link":"https:\/\/samacheerkalviguru.com\/samacheer-kalvi-8th-maths-term-1-chapter-3-intext-questions\/","title":{"rendered":"Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Intext Questions"},"content":{"rendered":"

Students can Download Maths Chapter 3 Algebra Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions<\/a> Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.<\/p>\n

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Intext Questions<\/h2>\n

Exercise 3.1<\/span><\/p>\n

Recap Page No. 66 and 67<\/span><\/p>\n

Question 1.
\nWrite the numbers of terms in the following expressions.
\n(i) x + y + z – xyz
\nSolution:
\n4 terms<\/p>\n

(ii) m2<\/sup>n2<\/sup>c
\nSolution:
\n1 term<\/p>\n

(iii) a2<\/sup>b2<\/sup>c – ab2<\/sup>c2<\/sup> + a2<\/sup>bc2<\/sup> + 3abc
\nSolution:
\n4 terms<\/p>\n

(iv) 8x2<\/sup> – 4xy + 7xy2<\/sup>
\nSolution:
\n3 terms
\n\"Samacheer<\/p>\n

Question 2.
\nIdentify the numerical co-efficient of each term in the following expressions.
\nQuestion 1.
\n2x2<\/sup> – 5xy + 6y2<\/sup> + 7x – 10y + 9
\nSolution:
\nNumerical co efficient in 2x2<\/sup> is 2
\nNumerical co efficient in -5xy is -5
\nNumerical co efficient in 6y2<\/sup> is 6
\nNumerical co efficient in 7x is 7
\nNumerical co efficient in -10y is – 10
\nNumerical co-efficient in 9 is 9<\/p>\n

\"Samacheer<\/p>\n

Question 2.
\n\\(\\frac{x}{3}+\\frac{2 y}{5}-x y+7\\)
\nSolution:
\n\"Samacheer
\nNumerical co efficient in -xy is -1
\nNumerical co efficient in 7 is 7<\/p>\n

\"Samacheer<\/p>\n

Question 3.
\nPick out the like terms from the following.
\n\"Samacheer
\nSolution:
\n\"Samacheer<\/p>\n

Question 4.
\nAdd : 2x, 6y, 9x – 2y
\nSolution:
\n2x + 6y + 9x – 2y = 2x + 9x + 6y – 2y = (2 + 9)x + (6 – 2)y = 11x + 4y<\/p>\n

Question 5.
\nSimplify : (5x3<\/sup> y3<\/sup> – 3x2<\/sup> y2<\/sup> + xy + 7) + (2xy + x3<\/sup>y3<\/sup> – 5 + 2x2<\/sup>y2<\/sup>)
\nSolution:
\n(5x3<\/sup>y3<\/sup> – 3x2<\/sup>y2<\/sup> + xy + 7) + (2xy + x3<\/sup>y3<\/sup> – 5 + 2x2<\/sup>y2<\/sup>)
\n= 5x3<\/sup>y3<\/sup> + x3<\/sup>y3<\/sup> – 3x2<\/sup>y2<\/sup> + 2x2<\/sup>y2<\/sup> + xy + 2xy + 7 – 5
\n= (5 + 1)x3<\/sup>y3<\/sup> + (-3 + 2)x2<\/sup>y2<\/sup> +(1 +2)xy + 2
\n= 6x3<\/sup>y3<\/sup> – x2<\/sup>y2<\/sup> + 3xy + 2<\/p>\n

\"Samacheer<\/p>\n

Question 6.
\nThe sides of a triangle are 2x – 5y + 9, 3y + 6x – 7 and -4x + y +10 . Find perimeter of the triangle.
\nSolution:
\nPerimeter of the triangle = Sum of three sides
\n= (2x – 5y + 9) + (3y + 6x – 7) + (-4x + y + 10)
\n= 2x – 5y + 9 + 3y + 6x – 7 – 4x + y + 10
\n= 2x + 6x – 4x – 5y + 3y + y + 9 – 7 + 10
\n= (2 + 6 – 4)x + (-5 + 3 + 1)y + (9 – 7 + 10)
\n= 4x – y + 12
\n\u2234 Perimeter of the triangle = 4x – y + 12 units.<\/p>\n

Question 7.
\nSubtract -2mn from 6mn.
\nSolution:
\n6 mn – (-2mn) = 6mn + (+2mn) = (6 + 2) mn = 8mn<\/p>\n

\"Samacheer<\/p>\n

Question 8.
\nSubtract 6a2<\/sup> – 5ab + 3b2<\/sup> from 4a2<\/sup> – 3ab + b2<\/sup>.
\nSolution:
\n(4a2<\/sup> – 3ab+ b2<\/sup>) – (6a2<\/sup>– 5ab + 3b2<\/sup>)
\n= (4a2<\/sup> – 6a2<\/sup>) + (- 3ab -(-5 ab)] + (b2<\/sup>– 3b2<\/sup>)
\n= (4 – 6) a2<\/sup> + [-3ab + (+ 5ab)] + (1 – 3) b2<\/sup>
\n= [4 + (- 6)] a2<\/sup> + (-3 + 5) ab + [1+ (-3)]b2<\/sup>
\n= -2a2<\/sup> + 2ab – 2b2<\/sup><\/p>\n

\"Samacheer<\/p>\n

Question 9.
\nThe length of a log is 3a + 4b – 2 and a piece (2a – b) is remove from it. What is the length of the remaining log?
\nSolution:
\nLength of the log = 3a + 4b – 2
\n\"Samacheer
\nLength of the piece removed = 2a – b
\nRemaining length of the log = (3a + 4b – 2) – (2a – b)
\n= (3a – 2a) + [4b – (-b)] – 2
\n= (3 – 2)a + (4 + 1)b – 2
\n= a + 5b – 2<\/p>\n

\"Samacheer<\/p>\n

Question 10.
\nA tin had ‘x’ litre oil. Another tin had (3x2<\/sup> + 6x – 5) litre of oil. The shopkeeper added (x + 7) litre more to the second tin. Later he sold (x2<\/sup> + 6) litres of oil from the second tin. How much oil was left In the second tin?
\nSolution:
\nQuantity of oil in the second tin = 3x2<\/sup> + 6x – 5 litres.
\nQuantity of oil added = x + 7 litres
\n\u2234 Total quantity of oil in the second tin
\n= (3x2<\/sup> + 6x – 5) + (x + 7) litres
\n= 3x2<\/sup> + (6x + x) + (-5 + 7)
\n= 3x2<\/sup> + (6 + 1)x + 2
\n= 3x2<\/sup> + 7x + 2 litres
\nQuantity of oil sold = x + 6 litres
\n\u2234 Quantity of oil left in the second tin = (3x2<\/sup> + 7x + 2) – (x2<\/sup> + 6)(3x2<\/sup> – x2<\/sup> ) + 7x + (2 – 6)
\n= (3 – 1)x2<\/sup> + 7x + (-4) = 2x2<\/sup> + 7x – 4
\nQuantity of oil left = 2x2<\/sup> + 7x – 4 litres<\/p>\n

Try this Page No. 70<\/span><\/p>\n

Question 1.
\nEvery algebraic expression is a polynomial. Is this statement true? Why?
\nSolution:
\nNo, This statement is not true. Because Polynomials contain only whole numbers as the powers of their variables. But an algebraic expression may contains fractions and negative powers on their variables.
\nEg. 2y2<\/sup> + 5y-1<\/sup> – 3 is a an algebraic expression. But not a polynomial.<\/p>\n

Try this Page No. 71<\/span><\/p>\n

Question 2.
\n-(5y2<\/sup> + 2y – 6) Is this correct? If not, correct the mistake.
\nSolution:
\nTaking -(5y2<\/sup> + 2y – 6) = 5y2<\/sup> + [(-)(+) 2y] + [(-) \u00d7 (-)6]
\n= -5y2<\/sup> – 2y + 6
\n\u2260 -5y2<\/sup> – 2y + 6
\n\u2234 Correct answer is -5y2<\/sup> + 2y – 6 = -(5y2<\/sup> + 2y + 6)<\/p>\n

\"Samacheer<\/p>\n

Try this Page No 71<\/span><\/p>\n

(i) 3ab2<\/sup>, -2a2<\/sup>b3<\/sup>
\n(ii) 4xy, 5y2<\/sup>x, (-x2<\/sup>)
\n(iii) 2m, -5n, -3p
\nSolution:
\n(i) (3ab2<\/sup>) \u00d7 (-2a2<\/sup>b2<\/sup>) = (+) \u00d7 (-) \u00d7 (3 \u00d7 2) \u00d7 (a \u00d7 a2<\/sup>) \u00d7 (b2<\/sup> \u00d7 b3<\/sup>) = -6a3<\/sup> b5<\/sup><\/p>\n

(ii) (4xy) \u00d7 (5y2<\/sup>x) \u00d7 (-x2<\/sup>)
\n= (+) \u00d7 (+) \u00d7 (-) \u00d7 (4 \u00d7 5 \u00d7 1) \u00d7 (x \u00d7 x \u00d7 x2<\/sup>) \u00d7 (y \u00d7 y2<\/sup>)
\n= -20x4<\/sup>y3<\/sup><\/p>\n

(iii) (2m) \u00d7 (-5n) \u00d7 (-3p) = (+) \u00d7 (-) \u00d7 (-) \u00d7 (2 \u00d7 5 \u00d7 3) \u00d7 m \u00d7 n \u00d7 p
\n= + 30mnp = 30 mnp<\/p>\n

Try this Page No. 71<\/span><\/p>\n

Question 1.
\nWhy 3 + (4x – 7y) \u2260 12x – 21y?
\nSolution:
\nAddition and multiplication are different
\n3 + (4x – 7y) = 3 + 4x – 7y
\nWe can add only like terms.<\/p>\n

\"Samacheer<\/p>\n

Try this Page No. 72<\/span><\/p>\n

Question 1.
\nWhich is corrcet? (3a)2<\/sup> is equal to
\n(i) 3a2<\/sup>
\n(ii) 32<\/sup>a
\n(iii) 6a2<\/sup>
\n(iv) 9a2<\/sup>
\nSolution:
\n(3a) =32<\/sup>a2 <\/sup>= 9a2<\/sup>
\n(iv) 9a2<\/sup> is the correct answer<\/p>\n

Try These Page No.72<\/span><\/p>\n

Question 1.
\nMultiply
\n(i) (5x2<\/sup> + 7x – 3) by 4x2<\/sup>
\nSolution:
\n(5x2<\/sup> + 7x – 3) \u00d7 4x2<\/sup>
\n= 4x2<\/sup>(5x2<\/sup> + 7x – 3) Multiplication is commutative
\n= 4x2<\/sup> (5x2<\/sup> + 4x2<\/sup> (7x) + 4x2<\/sup> (-3)
\n= (4 \u00d7 5)(x2<\/sup> \u00d7 x2<\/sup>) + (4 \u00d7 7)(x2<\/sup> \u00d7 x) + (4 \u00d7 -3)(x2<\/sup>)
\n= 20x4<\/sup> + 28x3<\/sup> – 12x2<\/sup><\/p>\n

(ii) (10x – 7y + 5z) by 6xyz
\nSolution:
\n(10x – 7y + 5z) by 6xyz
\n(10x – 7y + 5z) \u00d7 6xyz = 6xyz (10x – 7y + 5z) [\u2235 Multiplication is commutative]
\n= 6xy (10x) + 6xyz (-7y) + 6xyz (5z)
\n= (6 \u00d7 10)(x \u00d7 x \u00d7 y \u00d7 z) + (6 \u00d7 -7) + (x \u00d7 y \u00d7 y \u00d7 z) + (6 \u00d7 5)(x \u00d7 y \u00d7 z \u00d7 z)
\n= 60x2<\/sup>yz + (-42xy2<\/sup>z) + 30xyz2<\/sup>
\n= 60x2<\/sup>yz – 42x2<\/sup>z + 30xyz2<\/sup><\/p>\n

(iii) (ab + 3bc – 5ca) by – 3abc
\nSolution:
\n(ab + 3bc – 5ca) \u00d7 (- 3abc) = (-3abc) (ab + 3bc – 5ca)
\n[\u2235 Multiplication is commutativel
\n= (-3abc) (ab) + (-3abc) (3bc) + (-3abc) (5ca)
\n= (-3)(a \u00d7 a \u00d7 b \u00d7 b \u00d7 c) + (- 3 \u00d7 3) + (a \u00d7 b \u00d7 b \u00d7 c \u00d7 c)
\n= -3a2<\/sup>b2<\/sup>c – 9ab2<\/sup>c2<\/sup> – 30a2<\/sup>bc2<\/sup><\/p>\n

\"Samacheer<\/p>\n

Try these Page No. 74<\/span><\/p>\n

Question 1.
\nMultiply
\n(i) (a – 5) and (a + 4)
\nSolution:
\n(a – 5) (a + 4) = a(a + 4) – 5 (a + 4)
\n= (a \u00d7 a) + (a \u00d7 4) + (-5 \u00d7 a) + (-5 \u00d7 4)
\n= a2<\/sup> + 4a – 5a – 20 = a2<\/sup> – a – 20<\/p>\n

(ii) (a + b) and (a – b)
\nSolution:
\n(a + b) (a – b) = a(a – b) + b (a – b)
\n= (a \u00d7 a) + (a \u00d7 -b)+(b \u00d7 a) + b(-b)
\n= a2<\/sup> – ab + ab – b2<\/sup> = a2<\/sup> – b2<\/sup><\/p>\n

(iii) (m4<\/sup> + n4<\/sup>) and (m – n)
\nSolution:
\n(m4<\/sup> + n4<\/sup>)(m – n) = m4<\/sup>(m – n) + n4<\/sup>(m – n)
\n= (m4<\/sup> \u00d7 m) + (m4<\/sup> \u00d7 (-n)) + (n4<\/sup> \u00d7 m (n4<\/sup> \u00d7 (-n))
\n= m5<\/sup> – m4<\/sup>n + mn4<\/sup> – n5<\/sup><\/p>\n

(iv) (2x + 3)(x – 4)
\nSolution:
\n(2x + 3)(x – 4) = 2x(x – 4) + 3(x – 4)
\n= (2x2<\/sup> \u00d7 x) – (2x \u00d7 4) + (3 \u00d7 x) – (3 \u00d7 4)
\n= 2x2<\/sup> – 8x + 3x – 12 = 2x2<\/sup> – 5x – 12<\/p>\n

(v) (x – 5)(3x + 7)
\nSolution:
\n(x – 5)(3x + 7) = x(3x + 7) – 5(3x + 7)
\n= (x \u00d7 3x) + (x \u00d7 7) + (-5 \u00d7 3x) + (-5 \u00d7 7)
\n= 3x2<\/sup> + 7x – 15x – 35
\n= 3x2<\/sup> – 8x – 35<\/p>\n

(vi) (x – 2)(6x – 3)
\nSolution:
\n(x – 2)(6x – 3) \u00d7 (6x – 3) – 2(6x – 3)
\n= (x \u00d7 6x)+(x \u00d7 (-3) \u00d7 (2 \u00d7 6x) – (2 \u00d7 3)
\n= 6x2<\/sup> – 3x – 12x + 6
\n= 6x2<\/sup> – 15x + 6<\/p>\n

Try this Page No. 74<\/span><\/p>\n

Question 2.
\n3x2<\/sup> (x4<\/sup> – 7x3<\/sup> + 2), what is the highest power in the expression.
\nSolution:
\n3x2<\/sup>(x4<\/sup> – 7x3<\/sup> + 2) = (3x2<\/sup>) (x4<\/sup>) + 3x2<\/sup> (-7x3<\/sup>)+ (3x2<\/sup>)2<\/sup>
\n= 3x6<\/sup> – 21x5<\/sup> + 6x2<\/sup>
\nHighest power is 6 in x6<\/sup>.<\/p>\n

\"Samacheer<\/p>\n

Exercise 3.2<\/span><\/p>\n

Try this Page No. 77<\/span><\/p>\n

Question 1.
\nAre the following correct?
\n(i) \\(\\frac{x^{3}}{x^{8}}=x^{8-3}=x^{5}\\)
\n(ii) \\(\\frac{10 m^{4}}{10 m^{4}}=0\\)
\n(iii) When a monomial is divided by itself, we will get I?
\nSolution:
\n\"Samacheer<\/p>\n

Try this Page No. 77<\/span><\/p>\n

Question 1.
\nDivide
\n\"Samacheer
\nSolution:
\n\"Samacheer<\/p>\n

Try this Page No. 78<\/span><\/p>\n

Question 1.
\nAre the following divisions correct ?
\n\"Samacheer
\nSolution:
\n\"Samacheer<\/p>\n

Try this Page No. 78<\/span><\/p>\n

Question 1.
\n\"Samacheer
\nSolution:
\n\"Samacheer
\n\"Samacheer<\/p>\n

Exercise 3.3<\/span><\/p>\n

Try these Page No. 81<\/span><\/p>\n

Question 1.
\n1. (p + 2)2<\/sup> = …….
\n2. (3 – a)2<\/sup> = …….
\n3. (62<\/sup> – x2<\/sup>) = ………
\n4. (a + b)2<\/sup> – (a – b)2<\/sup> = …….
\n= a2<\/sup> + 2ab + b2<\/sup> – a2<\/sup> – 2ab – b2<\/sup>
\n= (1 – 1)a2<\/sup> + (2 + 2)ab + (+1 – 1 )b2<\/sup> = 4ab
\n5. (a + b)2<\/sup> = (a + b) \u00d7 (a + b)
\n6. (m + n)( m – n) = m2<\/sup> – n2<\/sup>
\n7. (m + 7)2<\/sup> = m2<\/sup> + 14m + 49
\n8. (k2<\/sup> – 36) \u2261 k2<\/sup> – 62<\/sup> = (k + 6)(k – 6)
\n9. m2<\/sup> – 6m + 9 = (m – 3)2<\/sup>
\n10. (m – 10)(m + 5) = m2<\/sup> + (-10 + 5)m + (-10)(5) = m2<\/sup> – 5m – 50
\nSolution:
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Try these page No. 83<\/span><\/p>\n

Question 1.
\nExpand using appropriate identities
\nQuestion 1.
\n(3p + 2q)2
\n<\/sup>Solution:
\n(3p + 2q)2<\/sup>
\nComparing (3p + 2q)2<\/sup> with (a + b)2<\/sup>, we get a = 3p and b = 2q.
\n(a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup>
\n(3p + 2q)2<\/sup> = (3p)2<\/sup>+ 2(3p) (2q) + (2q)2<\/sup>
\n= 9p2<\/sup> + 12pq + 4q2<\/sup><\/p>\n

\"Samacheer<\/p>\n

Question 2.
\n(105)2<\/sup>
\nSolution:
\n(105)2<\/sup> = (100 + 5)2<\/sup>
\nComparing (100 + 5)2<\/sup> with (a + b)2<\/sup>, we get a = 100 and b = 5.
\n(a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup>
\n(100 + 5)2<\/sup> = (100)2<\/sup> + 2(100)(5) + 52<\/sup> = 1oooo + 1000 + 25
\n1052<\/sup> = 11,025<\/p>\n

Question 3.
\n( 2x – 5d)2<\/sup>
\nSolution:
\n(2x – 5d)2<\/sup>
\nComparing with (a – b)2<\/sup>, we get a = 2x b = 5d.
\n(a – b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup>
\n(2x – 5d)2<\/sup> = (2x)2<\/sup> – 2(2x)(5d) + (5d)2<\/sup>
\n= 2x2<\/sup> – 20 xd + 52<\/sup>d2<\/sup> = 4x2<\/sup> – 20 xd + 25d2<\/sup><\/p>\n

\"Samacheer<\/p>\n

Question 4.
\n(98)2<\/sup>
\nSolution:
\n(98)2<\/sup> = (100 – 2)2<\/sup>
\nComparing (100 – 2)2<\/sup> with (a – b)2<\/sup> we get
\na = 100, b = 2
\n(a – b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup>
\n(100 – 2)2<\/sup> = 1002<\/sup> – 2(100)(2) + 22<\/sup>
\n= 10000 – 400 + 4 = 9600 + 4 = 9604<\/p>\n

Question 5.
\n(y – 5)(y + 5)
\nSolution:
\n(y – 5)(y + 5)
\nComparing (y – 5) (y + 5) with (a – b) (a + b) we get
\na = y; b = 5
\n(a – b)(a + b) = a2<\/sup> – b2<\/sup>
\n(y – 5)(y + 5) = y2<\/sup> – 52<\/sup> = y2<\/sup> – 25<\/p>\n

\"Samacheer<\/p>\n

Question 6.
\n(3x)2<\/sup> – 52<\/sup>
\nSolution:
\n(3x)2<\/sup> – 52<\/sup>
\nComparing (3x)2<\/sup> – 52<\/sup> with a2<\/sup> – b2<\/sup> we have
\na = 3x; b = 5
\n(a2<\/sup> – b2<\/sup>) = (a + b)(a – b)
\n(3x)2<\/sup> – 52<\/sup> = (3x + 5)(3x – 5) = 3x(3x – 5) + 5(3x – 5)
\n= (3x) (3x) – (3x)(5) + 5(3x) – 5(5)
\n= 9x2<\/sup> – 15x + 15x – 25 = 9x2<\/sup> – 25<\/p>\n

Question 7.
\n(2m + n)(2m +p)
\nSolution:
\n(2m + n) (2m + p)
\nComparing (2m + n) (2m + p) with (x + a) (x + b) we have
\nx = 2n; a = n ;b = p
\n(x – a)(x + b) = x2<\/sup> + (a + b)x + ab
\n(2m +n) (2m +p) = (2m2<\/sup>) + (n +p)(2m) + (n) (p)
\n= 22<\/sup>m2<\/sup> + n(2m) + p(2m) + np
\n= 4m2<\/sup> + 2mn + 2mp + np<\/p>\n

\"Samacheer<\/p>\n

Question 8.
\n203 \u00d7 197
\nSolution:
\n203 \u00d7 197 = (200 + 3)(200 – 3)
\nComparing (a + b) (a – b) we have
\na = 200, b = 3
\n(a + b)(a – b) = a2<\/sup> – b2<\/sup>
\n(200 + 3)(200 – 3) = 2002<\/sup> – 32<\/sup>
\n203 \u00d7 197 = 40000 – 9
\n203 \u00d7 197 = 39991<\/p>\n

Question 9.
\nFind the area of the square whose side is (x – 2)
\nSolution:
\nSide of a square = x – 2
\n\u2234 Area = Side \u00d7 Side
\n= (x – 2) (x – 2) = x(x – 2) – 2(x – 2)
\n= x(x) + (x)(-2) + (-2)(x) + (-2)(-2)
\n= x – 2x – 2x + 4x2<\/sup> – 4x + 4<\/p>\n

\"Samacheer<\/p>\n

Question 10.
\nFind the area of the rectangle whose length and breadth are (y + 4) and (y – 3).
\nSolution:
\nLength of the rectangle = y+ 4
\nbreadth of the rectangle = y – 3
\nArea of the rectangle = length \u00d7 breadth
\n= (y + 4)(y – 3) = y2<\/sup> + (4 +(-3))y + (4)(-3)
\n= y2<\/sup> + y – 12<\/p>\n

Try these Page No. 88<\/span><\/p>\n

Question 1.
\nExpand :
\nQuestion 1.
\n(x + 4)3<\/sup>
\nSolution:
\nComparing (x + 4)3<\/sup> with (a + b)3<\/sup>, we have a = x and b = 4.
\n(a + b)3<\/sup> = a3<\/sup> + 3a2<\/sup>b + 3ab2<\/sup> + b3<\/sup>
\n(x + 4)3<\/sup> = x3<\/sup> + 3x2<\/sup>(4) + 3(x)(4)2<\/sup> + 43<\/sup>
\n= x3<\/sup> + 12x2<\/sup> + 48x + 64<\/p>\n

\"Samacheer<\/p>\n

Question 2.
\n( y – 2)2<\/sup>
\nSolution:
\nComparing (y – 2) with (a – b)3<\/sup> we have a = y b = z
\n(a – b)3<\/sup> = a3<\/sup> – 3a2<\/sup>b + 3ab2<\/sup> – b3<\/sup>
\n(y – 2)2<\/sup> = y3<\/sup> – 3y(2) + 3y(2)2<\/sup> + 23<\/sup>
\n= y3<\/sup> – 6y2<\/sup> + 12y + 8<\/p>\n

Question 3.
\n(x + 1)(x + 3)(x + 5)
\nSolution:
\nComparing (x + 1) (x + 3) (x + 5) with (x + a) (x + b) (x + c) we have
\na = 1
\nb = 3
\nand c = 5
\n\"Samacheer<\/p>\n

Exercise 3.4<\/span><\/p>\n

Try These Page No.92<\/span><\/p>\n

Question 1.
\nFactorize the following:
\nQuestion 1.
\n3y + 6
\nSolution:
\n3y + 6
\n3y + 6 = 3 \u00d7 y + 2 \u00d7 3
\nTaking out the common factor 3 from each term we get 3 (y + 2)
\n\u2234 3y + 6 = 3(y + 2)<\/p>\n

\"Samacheer<\/p>\n

Question 2.
\n10x2<\/sup> + 15y
\nSolution:
\n10x2<\/sup> + 15y2<\/sup>
\n10x2<\/sup> + 15y2<\/sup> = (2 \u00d7 5 \u00d7 x \u00d7 x) + (3 \u00d7 5 \u00d7 y \u00d7 y)
\nTaking out the common factor 5 we have
\n10x2<\/sup> + 15y2<\/sup> = 5(2x2<\/sup> + 3y2<\/sup>)<\/p>\n

Question 3.
\n7m(m – 5) + 1(5 – m)
\nSolution:
\n7m(m – 5) + 1(5 – m)
\n7m(m – 5) + 1(5 – m) = 7m(m – 5) + (-1)(-5 + m)
\n= 7m(m – 5) – 1 (m – 5)
\nTaking out the common binomial factor (m – 5) = (m – 5)(7m – 1)<\/p>\n

\"Samacheer<\/p>\n

Question 4.
\n64 – x2<\/sup>
\nSolution:
\n64 – x2<\/sup>
\n64 – x2<\/sup> = 82<\/sup> – x2<\/sup>
\nThis is of the form a2<\/sup> – b2<\/sup>
\nComparing with a2<\/sup> – b2<\/sup> we have a = 8, b = x
\na2<\/sup> – b2<\/sup> = (a + b)(a – b)
\n64 – x2<\/sup> = (8 + x)(8 – x)<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 3 Algebra Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations. Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Intext Questions Exercise 3.1 Recap …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":"","footnotes":""},"categories":[4],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/1084"}],"collection":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/comments?post=1084"}],"version-history":[{"count":1,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/1084\/revisions"}],"predecessor-version":[{"id":59904,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/1084\/revisions\/59904"}],"wp:attachment":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/media?parent=1084"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/categories?post=1084"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/tags?post=1084"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}