DIVISION PROCESS IN THE ATTACHMENT .

SOLUTION :

i) Given : x³ - 2 x² - 5 x + 6

Let f(x) = x³ - 2 x² - 5 x + 6

The factors of the constant term 6 is ±1,±2,±3.

Put x = 1

f(1) = 1³ - 2 (1)² - 5 (1) + 6

= 1 - 2 - 5 + 6

= 7 - 7 = 0

Hence, (x - 1) is a factor of f(x).

To find other two factors we divide f(x) by (x-1).

x³ - 2 x² - 5 x + 6 / (x-1)

= (x² - x - 6)

x³ - 2 x² - 5 x + 6 = (x² - x - 6) (x-1)

x³ - 2 x² - 5 x + 6 = (x² - 3x +2x - 6) (x-1)

= [x(x -3) +2(x-3)] (x-1)

[by splitting the middle term]

x³ - 2 x² - 5 x + 6 = (x - 3) (x + 2) (x -1)

Hence, the required factors of the cubic Polynomial are (x - 1) (x - 3) (x + 2)

ii)

Let f (x) = 4 x³ - 7 x + 3

The factors of the constant term 3 is ±1,±3.

Put, x = 1in f(x)

f(1) = 4 (1)³ -7 (1) + 3

= 4 - 7 + 3 = 7 - 7 = 0

Hence, (x - 1) is a factor of f(x).

To find other two factors we divide f(x) by x-1.

4 x³ - 7 x + 3 / x-1 = 4 x² + 4 x - 3

4 x³ - 7 x + 3 = ( 4 x² + 4 x - 3) (x -1)

4 x³ - 7 x + 3 = ( 4 x² + 6 x - 2x - 3) (x -1)

[by splitting the middle term]

= 2x(2x + 3) -1(2x +3) (x -1)

4 x³ - 7 x + 3 = (2x +3) (2 x - 1) (x -1)

Hence, the required factors of the cubic Polynomial are (2x +3) (2 x - 1) (x -1)

iii) Let f (x) = x³ - 23 x² + 142 x - 120

The factors of the constant term 120 is ±1,±2,±3,±4,±5,±6,±,10,±12...

Put, x = 1in f(x)

f (1) = 1³ - 23 (1)² + 142 (1) - 120

= 1 - 23 + 142 - 120 = 143 - 143 = 0

Hence, (x - 1) is a factor of f(x).

To find other two factors we divide f(x) by x-1.

x³ - 23 x² + 142 x - 120 / (x - 1) = x² - 22 x + 120

x³ - 23 x² + 142 x - 120 =( x² - 22 x + 120)(x -1)

x³ - 23 x² + 142 x - 120 =[( x² - 12x -10 x + 120)] (x -1)

[ by middle term splitting]

x³ - 23 x² + 142 x - 120 = x(x - 12) -10(x -12) (x -1)

x³ - 23 x² + 142 x - 120 = (x - 12) (x -10) (x -1)

Hence, the required factors of the cubic Polynomial are (x - 1) (x - 12) (x - 10)

iv)

Let f (x) = 4 x³ - 5 x² + 7 x - 6

The factors of the constant term 6 is ±1,±2,±3.

Put x = 1,in f(x)

f (1) = 4 (1)³ - 5 (1)² + 7 (1) - 6

= 4 - 5 + 7 - 6 = 11 - 11 = 0

Hence, (x - 1) is a factor of f(x).

To find other two factors we divide f(x) by x-1.

4 x³ - 5 x² + 7 x - 6 / (x -1) = 4 x² - x + 6

This quadratic equation (4 x² - x + 6) can not be further factorise.

Hence, the required factors of the cubic Polynomial are (x -1) (4 x² - x + 6)

v)

Let f (x) = x³ - 7 x + 6

The factors of the constant term 6 is ±1,±2,±3.

Put x = 1,in f(x)

f (1) = (1)³ - 7 (1) + 6

= 1 - 7 + 6 = 7 - 7 = 0

Hence, (x - 1) is a factor of f(x).

To find other two factors we divide f(x) by x-1.

x³ - 7 x + 6 / (x -1) = x² + x - 6

x³ - 7 x + 6 = (x - 1) (x² + x - 6)

[factorising (x² + x - 6) by middle term splitting]

x³ - 7 x + 6 = (x - 1) (x² + 3x -2x - 6)

x³ - 7 x + 6 = (x - 1) (x(x +3) - 2(x + 3))

x³ - 7 x + 6 = (x - 1) (x-2) (x+3)

Hence, the required factors of the cubic Polynomial are (x - 1) (x-2) (x+3)

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