Question

On dividing x3 + x2+x-2 by a polynomial g(x), the quotient and remainder were
x2 + 2x + 1 and 2x – 1 respectively. Find g(x). *-​

Answer 1

Answer:

g(x) = x - 1

Step-by-step explanation:

p(x) = x³ + x² + x - 2

g(x) = ?

quotient = x² + 2x + 1

reminder = 2x - 1

divident = divisor × quotient + reminder

(x³ + x² + x - 2) = g(x) × (x²+2x+1)+(2x - 1)

(x³+x²+x-2) -(2x-1) = g(x) × (x² + 2x + 1)

x³ + x² + x - 2 - 2x + 1 = g(x)×(x² + 2x + 1)

x³ + x² - x -1 = g(x) × (x² + 2x + 1)

(x³ + x² - x - 1)/(x² + 2x + 1) = g(x)

|______________|

x² + 2x + 1 | x³ + x² - x - 1 | x - 1

x³ + 2x² + x

- - -

_____________________

-x² - 2x - 1

-x² - 2x - 1

+ + +

_________________

0

g(x) = x - 1

Answer 2

Answer:

Required value of g(x) is (x-1)

Step-by-step explanation:

Given,x3 + x2+x-2 by a polynomial g(x), the quotient and remainder were

x2 + 2x + 1 and 2x – 1 respectively

So, Divident=${x}^{3} + {x}^{2} + x - 2$

Quotient $= {x}^{2} + 2x + 1$

remainder$= 2x - 1$

Now we know

$Divident =divisor \times quotient+ remainder$

${x}^{3} + {x}^{2} + x - 2 = g(x) \times ( {x}^{2} + 2x + 1) + (2x - 1)$

$({x}^{2} + 2x + 1)g(x) = {x}^{3} + {x}^{2} + x - 2 - 2x + 1 = {x}^{3} + {x}^{2} - x - 1$

So, $g(x) = \frac{ {x}^{3} + {x}^{2} - x - 1 }{ {x}^{2} + 2x + 1 }$

$g(x) = \frac{ {x}^{2}(x + 1) - (x + 1) }{ {(x + 1)}^{2} } = \frac{( {x + 1)}( {x}^{2} - 1) }{ {(x + 1)}^{2} } = \frac{ {x}^{2} - 1 }{x + 1} = \frac{(x + 1)(x - 1)}{x + 1} = x - 1$

So value of g(x) is (x-1).

Here applied formula is,

${a}^{2} - {b}^{2} = (a + b)(a - b)$