Question

If x^3+4x^2- 6x -5 =b(x+5) , then find the value of b​

Answer 1

Answer:

x² - x - 1

Step-by-step explanation:

b = (x³ + 4x² - 6x - 5) ÷ (x - 5)

Answer 2

$Given \:x^3+4x^2- 6x -5 =b(x+5)$

$\implies b(x+5) = x^3+4x^2- 6x -5$

$\implies b = \frac{x^3+4x^2- 6x -5}{(x+5)}$

* _____________________*

x+5)x³ + 4x²-6x-5(x²-x-1

*****x³ + 5x²

______________

******** - x² - 6x

******** - x² - 5x

______________

************** - x - 5

************** - x - 5

______________

Remainder (0)

________________________**/

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

$\red{b} = \frac{ (x+5)(x^{2} - x - 1 )}{(x+5)} \\= (x^{2} - x - 1 )$

Therefore.,

$\red { Value \:of \:b } \green { =(x^{2} - x - 1 )}$

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