Question

solve Q no 1 and 2 of example 6 without using -8 i want a new method?
^_^
irrelevant or wrong answers will be reported

Answer 1

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1.

= x^3 + 3x^2 + 3x -7

The factors of 7 are 1 , 7 , -1 and -7.

By substituting x = 1.

= ( 1 )^3 + 3(1)^2 + 3 ( 1 ) -7

= 1 + 3 + 3 - 7

= 7 - 7

= 0.

Since, by substituting the value of x = 1 it resulted in 0.

So, ( x - 1 ) is one of its factor.

= x^3 + 3x^2 + 3× - 7

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

= x^2 ( x - 1 ) + 4x ( x - 1 ) + 7x - 7

= x^2 ( x - 1 ) + 4x ( x - 1 ) + 7 ( x - 1 )

= ( x - 1 ) ( x^2 + 4x + 7 ).

Note: It is long division method in short.We cam apply this to even find remainder.

2.

= x^3 - 3x^2 + 3x + 7

Factors of 7 = 1 , 7 , -1 , -7.

By substituting x = -1.

= ( -1 )^3 - 3 ( -1 )^2 + 3 ( -1 ) + 7

= -1 -3 - 3 + 7

= -7 + 7

= 0.

Since when x = -1 then it resulted in 0.

So, ( x + 1 ) is one of its factor.

= x^3 - 3x^2 + 3x + 7

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

= x^2 ( x + 1 ) - 4x ( x + 1 ) + 7x + 7

= x^2 ( x + 1 ) - 4x ( x + 1 ) + 7 ( x + 1 ).

= ( x + 1 ) ( x^2 - 4x + 7 ).

Answer 2

= x^3 + 3x^2 + 3× - 7

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

= x^2 ( x - 1 ) + 4x ( x - 1 ) + 7x - 7

= x^2 ( x - 1 ) + 4x ( x - 1 ) + 7 ( x - 1 )

= ( x - 1 ) ( x^2 + 4x + 7 ).



2.

= x^3 - 3x^2 + 3x + 7

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

= x^2 ( x + 1 ) - 4x ( x + 1 ) + 7x + 7

= x^2 ( x + 1 ) - 4x ( x + 1 ) + 7 ( x + 1 ).

= ( x + 1 ) ( x^2 - 4x + 7 ).