Question

using remainder theorem find the remainder when: (a) ( x3 -6x2+9x+3) is divided by (x-1)​

Answer 1

$\textbf{Given:}$

$\text{Dividend= x^3 -6x^2+9x+3 }$

$\text{Divisor= x-1 }$

$\textbf{To find:}$

$\text{Remainder}$

$\textbf{Solution:}$

$\textbf{Remainder theorem:}$

$\text{The remainer when P(x) is divided by (x-a) is P(a)}$

$\text{Let}\;P(x)=x^3 -6x^2+9x+3$

$\text{Using remainder theorem,}$

$\text{The remainder when P(x) is divided by x-1 }$

$=P(1)$

$=1^3 -6(1)^2+9(1)+3$

$=1-6+9+3$

$=13-6$

$=7$

$\textbf{Answer:}$

$\textbf{The remainder when \bf\,x^3 -6x^2+9x+3 is divided by \bf\,x-1 is 7}$

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Answer 2

Using remainder theorem we have to find the remainder when

x³ - 6x² + 9x + 3 is divided by (x - 1)

solution : remainder theorem : if a polynomial f(x) is divided by (x - a) then remainder must be f(a).

here let f(x) = x³ - 6x² + 9x + 3

(x - a) = (x - 1) ⇒a = 1 [on comparing ]

then remainder = f(a) = f(1)

= (1)³ - 6(1)³ + 9(1) + 3

= 1 -6 + 9 + 3

= 13 - 6 = 7

Therefore the remainder will be 7 after dividing (x³ - 6x² + 9x + 3) by (x - 1).

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