if p (x) = x cube -9x-3 is divided by x, then the remainder is
Answers
Answer:
iven:
\text{Dividend=$x^3 -6x^2+9x+3$}Dividend=x
3
−6x
2
+9x+3
\text{Divisor=$x-1$}Divisor=x−1
\textbf{To find:}To find:
\text{Remainder}Remainder
\textbf{Solution:}Solution:
\textbf{Remainder theorem:}Remainder theorem:
\text{The remainer when P(x) is divided by (x-a) is P(a)}The remainer when P(x) is divided by (x-a) is P(a)
\text{Let}\;P(x)=x^3 -6x^2+9x+3LetP(x)=x
3
−6x
2
+9x+3
\text{Using remainder theorem,}Using remainder theorem,
\text{The remainder when $P(x)$ is divided by $x-1$}The remainder when P(x) is divided by x−1
=P(1)=P(1)
=1^3 -6(1)^2+9(1)+3=1
3
−6(1)
2
+9(1)+3
=1-6+9+3=1−6+9+3
=13-6=13−6
=7=7
\textbf{Answer:}Answer:
\textbf{The remainder when $\bf\,x^3 -6x^2+9x+3$ is divided by $\bf\,x-1$ is 7}The remainder when x
3
−6x
2
+9x+3 is divided by x−1 is 7
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If a quadratic equation of the form ax^2 + c when divided by x and (x + 1) leaves remainder 2 and 4 respectively, then the value of a^2 + c^2 is
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