{"id":581105,"date":"2022-02-23T04:23:07","date_gmt":"2022-02-23T04:23:07","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=581105"},"modified":"2022-02-25T05:21:24","modified_gmt":"2022-02-25T05:21:24","slug":"maharashtra-board-solutions-class-9-maths-part-1-chapter-3-3-polynomials","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-9-maths-part-1-chapter-3-3-polynomials\/","title":{"rendered":"Maharashtra Board Solutions Class 9-Maths (Part 1): Chapter 3.3- Polynomials"},"content":{"rendered":"\n

Class 9: Maths Chapter 3.3 solutions. Complete Class 9 Maths Chapter 3.3 Notes.<\/p>\n\n\n\n

Maharashtra Board Solutions Class 9-Maths (Part 1): Chapter 3.3- Polynomials<\/h2>\n\n\n\n

Maharashtra Board 9th Maths Chapter 3.3, Class 9 Maths Chapter 3.3 solutions<\/p>\n\n\n\n

Question 1.
Divide each of the following polynomials by synthetic division method and also by linear division method. Write the quotient and the remainder.<\/strong>
i. (2m2<\/sup> \u2013 3m + 10) \u00f7 (m \u2013 5)
ii. (x4<\/sup> + 2x3<\/sup> + 3x2<\/sup> + 4x + 5) \u00f7 (x + 2)
iii. (y3<\/sup> \u2013 216) \u00f7 (y \u2013 6)
iv. (2x4<\/sup> + 3x3<\/sup> + 4x \u2013 2x2<\/sup>) \u00f7 (x + 3)
v. (x4<\/sup> \u2013 3x2<\/sup> \u2013 8) \u00f7 (x + 4)
vi. (y3<\/sup> \u2013 3y2<\/sup> + 5y \u2013 1) \u00f7 (y \u2013 1)
Solution:
<\/strong>i. Synthetic division:
(2m2<\/sup> \u2013 3m + 10) \u00f7 (m \u2013 5)
Dividend = 2m\u00b2 \u2013 3m + 10
\u2234 Coefficient form of dividend = (2, -3, 10)
Divisor = m \u2013 5
\u2234 Opposite of -5 is 5.
\"Maharashtra
Coefficient form of quotient = (2, 7)
\u2234 Quotient = 2m + 7,
Remainder = 45
Linear division method:
2m2<\/sup> \u2013 3m + 10
To get the term 2m2<\/sup>, multiply (m \u2013 5) by 2m and add 10m,
= 2m(m \u2013 5) + 10m- 3m + 10
= 2m(m \u2013 5) + 7m + 10
To get the term 7m, multiply (m \u2013 5) by 7 and add 35
= 2m(m \u2013 5) + 7(m- 5) + 35+ 10
= (m \u2013 5) (2m + 7) + 45
\u2234 Quotient = 2m + 7,
Remainder = 45<\/p>\n\n\n\n

ii. Synthetic division:
(x4<\/sup> + 2x3<\/sup> + 3x2<\/sup> + 4x + 5) \u00f7 (x + 2)
Dividend = x4<\/sup> + 2x3<\/sup> + 3x2<\/sup> + 4x + 5
\u2234 Coefficient form of dividend = (1, 2, 3, 4, 5)
Divisor = x + 2
\u2234 Opposite of + 2 is -2.
\"Maharashtra
Coefficient form of quotient = (1, 0, 3, -2)
\u2234 Quotient = x3<\/sup> + 3x \u2013 2,
Remainder = 9<\/p>\n\n\n\n

Linear division method:
x4<\/sup> + 2x3<\/sup> + 3x2<\/sup> + 4x + 5
To get the term x4<\/sup>, multiply (x + 2) by x3<\/sup> and subtract 2x3<\/sup>,
= x3<\/sup>(x + 2) \u2013 2x3<\/sup> + 2x3<\/sup> + 3x2<\/sup> + 4x + 5
= x3<\/sup>(x + 2) + 3x2<\/sup> + 4x + 5
To get the term 3x2<\/sup>, multiply (x + 2) by 3x and subtract 6x,
= x3<\/sup>(x + 2) + 3x(x + 2) \u2013 6x + 4x + 5
= x3<\/sup>(x + 2) + 3x(x + 2) \u2013 2x + 5
To get the term -2x, multiply (x + 2) by -2 and add 4,
= x3<\/sup>(x + 2) + 3x(x + 2) \u2013 2(x + 2) + 4 + 5
= (x + 2) (x3 + 3x \u2013 2) + 9
\u2234 Quotient = x3<\/sup> + 3x \u2013 2,
Remainder \u2013 9<\/p>\n\n\n\n

iii. Synthetic division:
(y3<\/sup> \u2013 216) \u00f7 (y \u2013 6)
Dividend = y3<\/sup> \u2013 216
\u2234 Index form = y3<\/sup> + 0y3<\/sup> + 0y \u2013 216
\u2234 Coefficient form of dividend = (1, 0, 0, -216)
Divisor = y \u2013 6
\u2234 Opposite of \u2013 6 is 6.
\"Maharashtra
Coefficient form of quotient = (1, 6, 36)
\u2234 Quotient = y2<\/sup> + 6y + 36,
Remainder = 0<\/p>\n\n\n\n

Linear division method:
y3<\/sup> \u2013 216
To get the term y3<\/sup>, multiply (y \u2013 6) by y2<\/sup> and add 6y2<\/sup>,
= y2<\/sup>(y \u2013 6) + 6y2<\/sup> \u2013 216
= y2<\/sup>(y \u2013 6) + 6ysup>2 \u2013 216
To get the, term 6 y2<\/sup> multiply (y \u2013 6) by 6y and add 36y,
= y2<\/sup>(y \u2013 6) + 6y(y \u2013 6) + 36y \u2013 216
= y2<\/sup>(y \u2013 6) + 6y(y \u2013 6) + 36y \u2013 216
To get the term 36y, multiply (y- 6) by 36 and add 216,
= y2<\/sup> (y \u2013 6) + 6y(y \u2013 6) + 36(y \u2013 6) + 216 \u2013 216
= (y \u2013 6) (y2<\/sup> + 6y + 36) + 0
Quotient = y2<\/sup> + 6y + 36
Remainder = 0<\/p>\n\n\n\n

iv. Synthetic division:
(2x4<\/sup> + 3x3<\/sup> + 4x \u2013 2x2<\/sup>) \u00f7 (x + 3)
Dividend = 2x4<\/sup> + 3x3<\/sup> + 4x \u2013 2x2<\/sup>
\u2234 Index form = 2x4<\/sup> + 3x3<\/sup> \u2013 2x2<\/sup> + 4x + 0
\u2234 Coefficient form of the dividend = (2,3, -2,4,0)
Divisor = x + 3
\u2234 Opposite of + 3 is -3
\"Maharashtra
Coefficient form of quotient = (2, -3, 7, -17)
\u2234 Quotient = 2x3<\/sup> \u2013 3x2<\/sup> + 7x \u2013 17,
Remainder = 51<\/p>\n\n\n\n

Linear division method:
2x4<\/sup> + 3x3<\/sup> + 4x \u2013 2x2<\/sup> = 2x2<\/sup> + 3x3<\/sup> \u2013 2x2<\/sup> + 4x
To get the term 2x4<\/sup>, multiply (x + 3) by 2x3<\/sup> and subtract 6x3<\/sup>,
= 2x3<\/sup>(x + 31 \u2013 6x3<\/sup> + 3x3<\/sup> \u2013 2x2<\/sup> + 4x
= 2x3<\/sup>(x + 3) \u2013 3x3<\/sup> \u2013 2x2<\/sup> + 4x<\/p>\n\n\n\n

To get the term \u2013 3x3<\/sup>, multiply (x + 3) by -3x2<\/sup> and add 9x2<\/sup>,
= 2x3<\/sup>(x + 3) \u2013 3x2<\/sup>(x + 3) + 9x2<\/sup> \u2013 2x2<\/sup> + 4x
= 2x3<\/sup>(x + 3) \u2013 3x2<\/sup>(x + 3) + 7x2<\/sup> + 4x<\/p>\n\n\n\n

To get the term 7x2<\/sup>, multiply (x + 3) by 7x and subtract 21x,
= 2x3<\/sup>(x + 3) \u2013 3x2<\/sup>(x + 3) + 7x(x + 3) \u2013 21x + 4x
= 2x3<\/sup>(x + 3) \u2013 3x2<\/sup>(x + 3) + 7x(x + 3) \u2013 17x<\/p>\n\n\n\n

To get the term -17x, multiply (x + 3) by -17 and add 51,
= 2x3<\/sup>(x + 3) \u2013 3x2<\/sup>(x + 3) + 7x(x+3) \u2013 17(x + 3) + 51
= (x + 3) (2x3<\/sup> \u2013 3x2<\/sup> + 7x- 17) + 51
\u2234 Quotient = 2x3<\/sup> \u2013 3x2<\/sup> + 7x \u2013 17,
Remainder = 51<\/p>\n\n\n\n

v. Synthetic division:
(x4<\/sup> \u2013 3x2<\/sup> \u2013 8) + (x + 4)
Dividend = x4<\/sup> \u2013 3x2<\/sup> \u2013 8
\u2234 Index form = x4<\/sup> + 0x3<\/sup> \u2013 3x2<\/sup> + 0x \u2013 8
\u2234 Coefficient form of the dividend = (1,0, -3,0, -8)
Divisor = x + 4
\u2234 Opposite of + 4 is -4
\"Maharashtra
\u2234 Coefficient form of quotient = (1, -4, 13, -52)
\u2234 Quotient = x3<\/sup> \u2013 4x2<\/sup> + 13x \u2013 52,
Remainder = 200<\/p>\n\n\n\n

Linear division method:
x4<\/sup> \u2013 3x2<\/sup> \u2013 8
To get the term x4<\/sup>, multiply (x + 4) by x3<\/sup> and subtract 4x3<\/sup>,
= x3<\/sup>(x + 4) \u2013 4x3<\/sup> \u2013 3x2<\/sup> \u2013 8
= x3<\/sup>(x + 4) \u2013 4x3<\/sup> \u2013 3x2<\/sup> \u2013 8
To get the term \u2013 4x3<\/sup>, multiply (x + 4) by -4x2<\/sup> and add 16x2<\/sup>,
= x3<\/sup>(x + 4) \u2013 4x2<\/sup> (x + 4) + 16x2<\/sup> \u2013 3x2<\/sup> \u2013 8
= x3<\/sup>(x + 4) \u2013 4x2<\/sup> (x + 4) + 13x2<\/sup> \u2013 8
To get the term 13x2<\/sup>, multiply (x + 4) by 13x and subtract 52x,
= x3<\/sup>(x + 4) \u2013 4x2<\/sup>(x + 4) + 13x(x + 4) \u2013 52x \u2013 8
= x3<\/sup>(x + 4) \u2013 4x2<\/sup>(x + 4) + 13x(x + 4) \u2013 52x \u2013 8
To get the term -52x, multiply (x + 4) by \u2013 52 and add 208,
= x3<\/sup>(x + 4) \u2013 4x2<\/sup>(x + 4) + 13x(x + 4) \u2013 52(x + 4) + 208 \u2013 8
= (x + 4) (x3<\/sup> \u2013 4x2<\/sup> + 13x \u2013 52) + 200
\u2234 Quotient = x3<\/sup> \u2013 4x2<\/sup> + 13x \u2013 52,
Remainder 200<\/p>\n\n\n\n

vi. Synthetic division:
(y3<\/sup> \u2013 3y2<\/sup> + 5y \u2013 1) \u00f7 (y \u2013 1)
Dividend = y3<\/sup> \u2013 3y2<\/sup> + 5y \u2013 1
Coefficient form of the dividend = (1, -3, 5, -1)
Divisor = y \u2013 1
\u2234Opposite of -1 is 1.
\"Maharashtra
\u2234 Coefficient form of quotient = (1, -2, 3)
\u2234 Quotient = y2<\/sup> \u2013 2y + 3,
Remainder = 2<\/p>\n\n\n\n

Linear division method:
y3<\/sup> -3y2<\/sup> + 5y \u2013 1
To get the term y3<\/sup> , multiply (y \u2013 1) by y2<\/sup> and add y2<\/sup>
= y2<\/sup> (y \u2013 1) + y2<\/sup> \u2013 3y2<\/sup> + 5y \u2013 1
= y2<\/sup> (y \u2013 1) \u2013 2y2<\/sup> + 5y \u2013 1
To get the term -2y2<\/sup>, multiply (y \u2013 1) by -2y and subtract 2y,
= y2<\/sup> (y \u2013 1) \u2013 2y(y \u2013 1) \u2013 2y + 5y \u2013 1
= y2<\/sup> (y \u2013 1) \u2013 2y(y \u2013 1) + 3y \u2013 1
To get the term 3y, multiply (y \u2013 1) by 3 and add 3,
= y2<\/sup> (y \u2013 1) \u2013 2y(y \u2013 1) + 3(y- 1) + 3 \u2013 1
= (y \u2013 1)(y2<\/sup> \u2013 2y + 3) + 2
\u2234 Quotient = y2<\/sup> \u2013 2y + 3,
Remainder = 2.<\/p>\n\n\n\n

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Maharashtra Board Solutions Class 9-Maths (Part 1): Chapter 3.3- Polynomials<\/p>\n\n\n\n

Download PDF<\/strong>: Maharashtra Board Solutions Class 9-Maths (Part 1): Chapter 3.3- Polynomials PDF<\/a><\/p>\n\n\n\n

Chapterwise Maharashtra Board Solutions Class 9 Maths :<\/strong><\/h2>\n\n\n\n

Part 1<\/strong><\/p>\n\n\n\n