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In the division by synthetic method the divisor polynomial is a whose degree is 1. in the form x + a or X Practice set 3.3 1. Divide each of the following polynomials by synthetic division method and linear division method. Write the quotient and the remainder. (i) (2m2 - 3m + 10) = (m - 5) (ii) (x4 + 2x3 + 3x2 + 4x + 5) = (x + 2) + (iii) (13 - 216 ; ( 22 23 1 17​

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Answered by ushamisha5923
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Step-by-step explanation:

. Synthetic division:

(2m2 – 3m + 10) ÷ (m – 5)

Dividend = 2m2 – 3m + 10

∴ Coefficient form of dividend = (2, -3, 10)

Divisor = m – 5

∴ Opposite of -5 is 5.

Coefficient form of quotient = (2, 7)

∴ Quotient = 2m + 7,

Remainder = 45

Linear division method:

2m2 – 3m + 10 To get the term 2m2,

multiply (m – 5) by 2m and

add 10m, = 2m(m – 5) + 10m- 3m + 10 = 2m(m –

5) + 7m + 10 To get the term 7m,

multiply (m – 5) by 7

and add 35 = 2m(m – 5) + 7(m- 5) + 35+ 10

= (m – 5) (2m + 7) + 45

∴ Quotient = 2m + 7,

Remainder = 45

ii. Synthetic division:

(x4 + 2x3 + 3x2 + 4x + 5) ÷ (x + 2)

Dividend = x4 + 2x3 + 3x2 + 4x + 5

∴ Coefficient form of dividend = (1, 2, 3, 4, 5)

Divisor = x + 2

∴ Opposite of + 2 is -2

Coefficient form of quotient = (1, 0, 3, -2)

∴ Quotient = x3 + 3x – 2,

Remainder = 9

Linear division method:

x4 + 2x3 + 3x2 + 4x + 5

To get the term x4,

multiply (x + 2) by x3

and subtract 2x3,

= x3(x + 2) – 2x3 + 2x3 + 3x2 + 4x + 5

= x3(x + 2) + 3x2 + 4x + 5

To get the term 3x2,

multiply (x + 2) by 3x and subtract 6x,

= x3(x + 2) + 3x(x + 2) – 6x + 4x + 5

= x3(x + 2) + 3x(x + 2) – 2x + 5

To get the term -2x,

multiply (x + 2) by -2

and add 4,

= x3(x + 2) + 3x(x + 2) – 2(x + 2) + 4 + 5

= (x + 2) (x3 + 3x – 2) + 9

∴ Quotient = x3 + 3x – 2,

Remainder – 9

iii. Synthetic division:

(y3 – 216) ÷ (y – 6)

Dividend = y3 – 216

∴ Index form = y3 + 0y3 + 0y – 216

∴ Coefficient form of dividend = (1, 0, 0, -216)

Divisor = y – 6

∴ Opposite of – 6 is 6.

Coefficient form of quotient = (1, 6, 36)

∴ Quotient = y2 + 6y + 36,

Remainder = 0

Linear division method:

y3 – 216

To get the term y3,

multiply (y – 6) by y2

and add 6y2, = y2(y – 6) + 6y2 – 216

= y2(y – 6) + 6ysup>2 – 216

To get the, term 6y2

multiply (y – 6) by 6y

and add 36y,

= y2(y – 6) + 6y(y – 6) + 36y – 216 = y2(y – 6)

+ 6y(y – 6) + 36y – 216

To get the term 36y, multiply (y- 6) by 36

and add 216, = y2(y – 6) + 6y(y – 6) + 36(y – 6)

+ 216 – 216 = (y – 6) (y2 + 6y + 36) + 0

Quotient = y2 + 6y + 36

Remainder = 0

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