Question

(x³+7x²-x-7)÷(x²+6x-7)​

Answer 1

SOLUTION

TO DETERMINE

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

EVALUATION

Here the given expression is

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

Numerator

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

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

∴ x + 1 is a factor of the numerator

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

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

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

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

Hence

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

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

$\sf{ = x + 1}$

FINAL ANSWER

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

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