Question

x^2-ax^3+bx^2-cx+8=0 divided by [x-1] leaves a remainder of 4 divided by [x+1] leaves a remainder of 3 then b=

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{x^4-ax^3+bx^2-cx+8\;is\;divided\;by\;(x-1)\;leaves\;a}$

$\textsf{remainder of 4 and divided by (x+1) leaves a remainder of 3}$

$\underline{\textbf{To find:}}$

$\textsf{The value of 'b'}$

$\underline{\textbf{Solution:}}$

$\underline{\textbf{Remainder theorem:}}$

$\textsf{The remainder when P(x) is divided by (x-a) is P(a)}$

$\textsf{The remainder when P(x) is divided by (x+a) is P(-a)}$

$\mathsf{Let\;f(x)=x^4-ax^3+bx^2-cx+8}$

$\textsf{The remainder when f(x) is divided by (x-1) = 4}$

$\implies\mathsf{f(1)=4}$

$\implies\mathsf{1^4-a(1)^3+b(1)^2-c(1)+8=4}$

$\implies\mathsf{1-a+b-c+8=4}$

$\implies\mathsf{-a+b-c+9=4}$

$\implies\mathsf{-a+b-c=4-9}$

$\implies\mathsf{-a+b-c=-5}$-------(1)

$\textsf{The remainder when f(x) is divided by (x+1) = 3}$

$\implies\mathsf{f(-1)=3}$

$\implies\mathsf{(-1)^4-a(-1)^3+b(-1)^2-c(-1)+8=3}$

$\implies\mathsf{1+a+b+c+8=3}$

$\implies\mathsf{a+b+c+9=3}$

$\implies\mathsf{a+b+c=3-9}$

$\implies\mathsf{a+b+c=-6}$-------(2)

$\implies\mathsf{-a+b-c=-5}$-------(1)

$\implies\mathsf{a+b+c=-6}$-------(2)

$\textsf{Adding (1) and (2), we get}$

$\mathsf{2b=-11}$

$\implies\boxed{\mathsf{b=\dfrac{-11}{2}}}$

$\therefore\mathsf{The\;value\;of\;b\;is\;\dfrac{-11}{2}}$

Answer 2

Step-by-step explanation:

By remainder theorem ,

x⁴-ax³+bx²-cx+8=p(x).(x-1)+4

Where p(x) be a cubic polynomial.

Since we do not know p(x) to eliminate it put x=1 which

makes p(x).(x-1)=0

Then,

1-a+b-c+8=4

So -a+b-c=-5 …. let it be equation 1

Similarly

x⁴-ax³+bx²-cx+8=q(x).(x+1)+3

Put x=-1

1+a+b+c+8=3

a+b+c=-6 .let it be equation 2

Adding equations 1 and 2

We get

2b=-11

b=-11/2

therefore when we divide -11/2 we get -5.5

hope it is helpful