Math, asked by sohansmulik, 9 months ago

Using remainder theorem, find the value of ‘a’ if the division of x³ +5x² -ax+6 by (x-1) leaves the remainder 2.

Answers

Answered by Kartikempire
4

Answer:

10

Step-by-step explanation:

Let P(x)=x³ +5x²-ax+6

Zero of x-1 =0

=>x =1

P(1)=(1)³+5(1)² -a(1)+6

=>2 =1+5-a+6

=>2 =12-a

=>-10=-a

=>10 =a

Answered by ItzAditt007
12

AnswEr:-

Your answer is 10.

ExplanaTion:-

Given:-

  • Cubic polynomial, \tt x^3+5x^2-ax+6 is divided by (x-1) and leaves Remainder 2.

To Find:-

  • The value of 'a' by using remainder theorem.

So Here,

It is given that when \tt x^3+5x^2-ax+6 is divided by (x-1) then it leaves 2 as Remainder.

So If we subtract 2 from the given cubic polynomial then the resulted polynomial should be completely divisible by (x-1).

So lets subtract:-

\\ \tt\longrightarrow ({x}^{3}   + 5 {x}^{2}  - ax + 6) - 2. \\  \\ \tt =  {x}^{3}  + 5 {x}^{2}  - ax + 6 - 2. \\  \\ \tt =  {x}^{3}  - 5 {x}^{2}  - ax + 4.\\

\therefore x³+5x²-ax+4 is completely divisible by (x-1).

So we can say that \tt(x-1) is a factor of \tt x^3+5x^2-ax+6.

Thus by remainder theorem we get,

\\ \tt\longrightarrow(x - 1) = 0. \\  \\ \tt\longrightarrow x - 1 = 0 \\  \\ \tt\longrightarrow x = 1.\\

\tt \because (x-1) is a factor of \tt x^3+5x-ax+4, so when we put the value of x in the above polynomial then it must be equal to 0.

So by putting the value of x we get,

\\ \tt\mapsto {x}^{3}  + 5 {x}^{2}  - ax + 4 = 0. \\  \\ \tt\mapsto {1}^{3}  + 5(1) {}^{2}  - a(1) + 4 = 0. \\  \\ \tt\mapsto 1 + 5 - a + 4 = 0. \\  \\ \tt\mapsto 6 - a + 4 = 0. \\  \\ \tt\mapsto  10 - a = 0-10. \\  \\  \tt\mapsto \cancel{-}a = \cancel{-}10.\\ \\ \tt\mapsto a=10.\\

\tt\therefore The required value of a is 10.

Similar questions