Math, asked by palakyadav760, 5 hours ago

(x³ - 7x² + 6x) ÷ (x - 6) = *


x³ - x²

x² - x

x - 1

x² - 6x

Answers

Answered by kswasti71
2

Answer:

=x+1

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Step-by-step explanation:

SOLUTION

TO DETERMINE

\displaystyle \sf{ \frac{ {x}^{3} + 7 {x}^{2} - x - 7 }{ {x}^{2} + 6x - 7 } }

x

2

+6x−7

x

3

+7x

2

−x−7

EVALUATION

Here the given expression is

\displaystyle \sf{ \frac{ {x}^{3} + 7 {x}^{2} - x - 7 }{ {x}^{2} + 6x - 7 } }

x

2

+6x−7

x

3

+7x

2

−x−7

Numerator

= \displaystyle \sf{ {x}^{3} + 7 {x}^{2} - x - 7 }=x

3

+7x

2

−x−7

For x = - 1 the value of the numerator is zero

∴ x + 1 is a factor of the numerator

\therefore \: \: \displaystyle \sf{ {x}^{3} + 7 {x}^{2} - x - 7 }∴x

3

+7x

2

−x−7

= \displaystyle \sf{ {x}^{3} + {x}^{2} + 6 {x}^{2} + 6x -7 x - 7 }=x

3

+x

2

+6x

2

+6x−7x−7

= \displaystyle \sf{ {x}^{2}(x + 1) + 6x(x + 1) - 7(x + 1) }=x

2

(x+1)+6x(x+1)−7(x+1)

\displaystyle \sf{ = (x + 1)({x}^{2} + 6x - 7) }=(x+1)(x

2

+6x−7)

Hence

\displaystyle \sf{ \frac{ {x}^{3} + 7 {x}^{2} - x - 7 }{ {x}^{2} + 6x - 7 } }

x

2

+6x−7

x

3

+7x

2

−x−7

= \displaystyle \sf{ \frac{ (x + 1)({x}^{2} + 6x - 7 ) }{ {x}^{2} + 6x - 7 } }=

x

2

+6x−7

(x+1)(x

2

+6x−7)

\sf{ = x + 1}=x+1

FINAL ANSWER

\boxed{ \: \: \displaystyle \sf{ \frac{ {x}^{3} + 7 {x}^{2} - x - 7 }{ {x}^{2} + 6x - 7 } = x + 1 } \: \: }

x

2

+6x−7

x

3

+7x

2

Answered by at046082
2

Step-by-step explanation:

x³-7x²+6x) ÷(x-6)

taking out x common from x³-7x+6x

x(x²-7x+6) ÷(x-6)

now factorizing

x{x²-(1+6)x +6} ÷ (x-6)

x(x²-x-6x +6) ÷(x-6)

x{x(x-1) -6(x-1)} ÷(x-6)

x(x-1) (x-6) /(x-6)

now we can see that x-6 can be cancel so we will cancel x-6

so now we will get

x(x-1)

the ans is x(x-1)

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