Question

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

Answer 1

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

Answer 2

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.