Class-8 Maths- Factorisation Ex 12.3
Keeping the examination point of view in mind the mathematicsandinformationtechnology.com team has prepared NCERT Solutions for Class-8 Maths- Factorisation Ex 12.1. The NCERT Solutions for chapter Factorisation solutions explains the easy and simple way to solve the problems. By understanding these ways in NCERT Solutions for Class 8, students will be confident while solving such problems found in Class-8 Maths- Factorisation Ex 12.1.
1. Carry out the following divisions. (i) \(28x^4 ÷ 56x\)
(ii) \(–36y^3 ÷ 9y^2\)
(iii) \(66pq^2r^3 ÷ 11qr^2\)
(iv) \(34x^3y^3z^3 ÷ 51xy^2z^3\)
(v) \(12a^8b^8 ÷ (– 6a^6b^4)\)
Solution:
(i) \(28x^4 ÷ 56x \)
\( 28x^4 = 2 × 2 × 7 × x × x × x × x \)
\(56x =2 × 2 × 2 × 7 × x \)
\[ 28x^4 ÷ 56x = {(2 × 2 × 7 × x × x × x × x) \over (2 × 2 × 2 × 7 × x)} \]
= \(x^3\over 2\)
(ii) \(- 36y3 ÷ 9y2 \)
\( -36y3 = - 2 × 2 × 3 × 3 × y × y × y \)
\( 9y2= 3 × 3 × y × y \)
\[-36y^3 ÷ 9y^2 = {(-2 × 2 × 3 × 3 × y × y × y)\over (3 × 3 × y × y)} \]
\( = -4y \)
(iii) \(66 pq^2r^3 ÷ 11qr^2 \)
\( 66pq^2r^3 =2 × 3 × 11 × p × q × q × r × r × r \)
\(11qr^2=11 × q × r × r \)
\[66 pq^2r^3 ÷ 11qr^2 = {(2 × 3 × 11 × p × q × q × r × r × r) \over (11 × q × r × r)}\]
\( = 6pqr \)
(iv) \(34x^3y^3z^3 ÷ 51xy^2z^3 \)
\(34x^3y^3z^3 =2 × 17 × x × x × x × y × y × y × z × z × z \)
\(51xy^2z^3=3 × 17 × x × y × y × z × z × z \)
\[34x^3y^3z^3 ÷ 51xy^2z^3 = {(2 × 17 × x × x × x × y × y × y × z × z × z) \over (3 × 17 × x × y × y × z × z × z)} \]
\(= 2x^2y \over 3 \)
(v) 12a^8b^8 ÷ (-6a^6b^4)
12a^8b^8 = 2 × 2 × 3 × a^8 × b^8
-6a^6b^4 = -2 × 3 × a^6 × b^4
\[12a^8b^8 ÷ (-6a^6b^4) = {(2 × 2 × 3 × a8 × b8) \over (-2 × 3 × a6 × b4)}\]
\(= -2a^2b^4\)
3.Divide the given polynomial by the given monomial.
(i) \((5x^2 - 6x) ÷ 3x\)
(ii) \((3y^8 - 4y^6 + 5y^4) ÷ y^4\)
(iii) \(8(x^3y^2z^2 + x^2y^3z^2 + x^2y^2z^3) ÷ 4x2y^2z^2\)
(iv) \((x^3 + 2x^2 + 3x) ÷ 2x\)
(v) \((p^3q^6 - p^6q^3) ÷ p^3q^3\)
Solution:
(i) \((5x^2 - 6x) ÷ 3x\)
\((5x^2 - 6x) = x(5x - 6)\)
Then, \[(5x^2 - 6x) ÷ 3x = {x(5x - 6) \over 3x}\]
\(= {(5x - 6) \over 3}\)
(ii) \((3y^8 - 4y^6 + 5y^4) ÷ y^4\)
\((3y^8 - 4y^6 + 5y^4) = y^4(3y^4 - 4y^2 + 5)\)
Then, \[(3y^8 - 4y^6 + 5y^4) ÷ y^4 = {y^4(3y^4 - 4y^2 + 5) \over y^4}\]
\(= 3y^4 - 4y^2 + 5\)
(iii) \(8(x^3y^2z^2 + x^2y^3z^2 + x^2y^2z^3) ÷ 4x^2y^2z^2\)
\(8(x^3y^2z^2 + x^2y^3z^2 + x^2y^2z^3) = 8x^2y^2z^2(x + y + z)\)
Then, \[8(x^3y^2z^2 + x^2y^3z^2 + x^2y^2z^3) ÷ 4x^2y^2z^2= {8x^2y^2z^2(x + y + z) \over 4x^2y^2z^2}\]
\(= 2(x + y + z)\)
(iv) \((x^3 + 2x^2 + 3x) ÷ 2x\)
\((x^3 + 2x^2 + 3x) = x(x^2 + 2x + 3)\)
Then, \[(x^3 + 2x^2 + 3x) ÷ 2x = {x(x^2 + 2x + 3) \over 2x}\]
\(= {(x^2 + 2x + 3) \over 2}\)
(v) \((p^3q^6 - p^6q^3) ÷ p^3q^3\)
\((p^3q^6 - p^6q^3) = p^3q^3 (q^3 - p^3)\)
Then, \[(p^3q^6 - p^6q^3) ÷ p^3q^3= {p^3q^3 (q^3 - p^3) \over p^3q^3}\]
\(= q^3 - p^3\)
(i) \(5(2x +1)(3x + 5) ÷ (2x +1) \)
(ii) \(26xy(x + 5)(y - 4) ÷ 13x( y - 4)\)
(iii) \(52 pqr(p + q)(q + r)(r + p) ÷ 104pq(q + r)(r + p)\)
(iv) \(20(y + 4) (y^2 + 5y + 3) ÷ 5(y + 4)\)
(v) \(x(x +1)(x + 2)(x + 3) ÷ x(x +1)\)
Solution:
(i) \(5(2x +1)(3x + 5) ÷ (2x +1)\)
\(= {5(2x +1)(3x + 5) \over (2x +1)}\)
\(= 5(3x + 5)\)
(ii) \(26xy(x + 5)(y - 4) ÷ 13x(y - 4)\)
\(= 2 × 13 × xy(x + 5)(y - 4) \over 13x(y - 4)\)
\(= 2y(x + 5)\)
(iii) \(52pqr (p + q)(q + r)(r + p) ÷ 104pq(q + r)(r + p)\)
\(= 2 × 2 × 13 × p × q × r × ( p + q) × (q + r) × (r + p) \over 2 × 2 × 2 × 13 × p × q × (q + r) × (r + p)\)
\(= r(p + q) \over 2\)
(iv) \(20(y + 4) (y^2 + 5y + 3) ÷ 5(y + 4)\)
\(= {2 × 2 × 5 × (y + 4) × (y^2 + 5 y + 3) \over 5 × (y + 4)}\)
\(= 4(y^2 + 5 y + 3)\)
(v) \(x(x +1)(x + 2)(x + 3) ÷ x(x +1)\)
\(= (x + 2)(x + 3)\)
5.Factorize the expressions and divide them as directed.
(i) \((y^2 + 7y + 10) ÷ (y + 5) \)
Solution:
(i) \[(y^2 + 7y + 10) ÷ (y + 5)\\ (y^2 + 7y + 10) = y^2 + 2 y + 5 y +10 \\= y(y + 2) + 5(y + 2) \\= (y + 2)(y + 5)\]
Thus, \[(y^2 + 7y + 10) ÷ (y + 5) \\= {(y + 2)(y + 5) \over (y + 5) = y + 2} \]
(ii) \((m^2 -14m - 32) ÷ (m + 2) \)
Solutions:
\[(m^2 -14m - 32) ÷ (m + 2)\\ (m^2 -14m - 32)\\ = m^2 + 2m -16m - 32 \\= m(m + 2) -16(m + 2)\\= (m + 2)(m - 16)\]
Thus, \[(m^2 -14m - 32) ÷ (m + 2) = {(m + 2)(m - 16) \over (m + 2)}\\= m - 16 \]
(iii)\( (5p^2 - 25p + 20) ÷ (p -1) \)
Solutions:
\[(5p^2 - 25p + 20) ÷ (p -1)\\ (5p^2 - 25p + 20) = \\5(p^2 - 5p + 4) \\= 5(p^2 - p - 4p + 4) \\= 5[p(p - 1) - 4(p - 1)] \\= 5(p - 1)(p - 4)\]
Thus, \[(5p^2 - 25p + 20) ÷ (p - 1) \\= {5(p - 1)(p - 4) \over (p -1) = 5(p - 4)} \]
(iv) \(4yz(z^2 + 6z -16) ÷ 2y(z + 8) \)
Solutions:
\(4yz(z^2 + 6z -16) ÷ 2y(z + 8) \)
\( 4yz(z^2 + 6z -16) = 4 yz (z^2 - 2z + 8z -16) \)
\(= 4 yz [z(z - 2) + 8(z - 2)] \)
\(= 4 yz(z - 2)(z + 8) \)
Thus, \[4yz(z^2 + 6z -16) ÷ 2y(z + 8)\\= {4 yz(z - 2)(z + 8) \over 2y(z + 8)}\\ = 2z(z - 2) \]
(v) \(5pq(p^2 - q^2) ÷ 2p(p + q) \)
Solutions:
\(5pq(p^2 - q^2) ÷ 2p(p + q)\)
\( 5pq(p^2 - q^2) = 5pq(p - q)(p + q) [ a^2 - b2 = (a + b)(a - b)] \)
Thus, \[5pq(p^2 - q^2) ÷ 2p(p + q) \\= {5pq(p - q)(p + q) \over 2p(p + q)}\\ = {5q(p - q) \over 2} \]
(vi) \(12xy(9x^2 -16y^2) ÷ 4xy(3x + 4y) \)
Solutions:
\(12xy(9x^2 -16y^2) ÷ 4xy(3x + 4y) \)
\( 12xy(9x^2 -16y^2) = 12xy[(3x)^2 - (4y)^2] \)
\(= 12xy(3x - 4y)(3x + 4y) [ a^2 - b^2 = (a + b)(a - b)] \)
\(= 2 × 2 × 3 × x × y × (3x - 4y) × (3x + 4y)\)
Thus, \(12xy(9x^2 -16y^2) ÷ 4xy(3x + 4y)\)
\(= {2 × 2× 3 × x × y × (3x - 4y) × (3x + 4y) \over 4xy(3x + 4y)}\\ = 3(3x - 4y) \)
(vii) \(39y^3(50y^2 - 98) ÷ 26y^2(5y + 7) \)
Solutions:
\(39y^3(50y^2 - 98) ÷ 26y^2(5y + 7) \)
\(39y^3(50y^2 - 98) = 3 × 13 × y × y × y × [2 × (25y^2 - 49)] \)
\(= 3 × 13 × 2 × y × y × y × [(5y)^2 - (7)^2] \)
\[= 3 × 13 × 2 × y × y × y(5y - 7)(5y + 7) [ a^2 - b^2 = (a + b)(a - b)]\]
\(26y^2(5y + 7) \\= 2 × 13 × y × y × (5y + 7) \)
Thus, \(39y^3(50y^2 - 98) ÷ 26y^2(5y + 7) \)
\[= 3 × 13 × 2 × y × y × y(5y - 7)(5y + 7) \over 2 × 13 × y × y × (5y + 7) \\ = 3y(5y - 7)\]
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