Question

Let R₁ and R₂ be the remainders when the polynomials x³-6x²+2x-k and kx³+12x²+14x-3 are divided by (1-2x) and (2x+1) respectively. If R₁ - R₂ = 25/8, find the value of k.​

Answer 1

$\large\underline{\sf{Given \: info-}}$

When, p(x) = x^3 - 6x^2 + 2x - k, is divided by (1 - 2x), Remainder = R1

When, g(x) = kx^3 + 12x^2 + 14x - 3, is divided by (2x + 1), Remainder = R2

R1 - R2 = 25/8

• Find the value of k.

$\large\underline{\sf{Solution-}}$

We are given that,

$\sf\longmapsto x^3-6x^2+2x-k\div(1-2x)=R_1$

And,

$\sf\longmapsto kx^3+12x^2+14x-3\div(2x+1)=R_2$

Taking p(x) first,

We have,

$\sf\longmapsto x^3-6x^2+2x-k\div(1-2x)=R_1$

That is,

$\sf\longmapsto p(x)\div(1-2x)=R_1$

Let, 1 - 2x = 0.

So, x = 1/2

As,

$\sf\longmapsto p(x)\div(1-2x)=R_1$

So, by Factor Theorem,

$\sf\longmapsto p\left(\dfrac{1}{2}\right)=R_1$

So,

$\sf\longmapsto\left(\dfrac{1}{2}\right)^3-6\left(\dfrac{1}{2}\right)^2+2\left(\dfrac{1}{2}\right)-k=R_1$

$\sf\longmapsto\dfrac{1}{8}-6\!\!\!/^3\left(\dfrac{1}{4\!\!\!/_2}\right)+2\!\!\!/\left(\dfrac{1}{2\!\!\!/}\right)-k=R_1$

So,

$\sf\longmapsto \dfrac{1}{8}-\dfrac{3}{2}+1-k=R_1$

Taking LCM,

$\sf\longmapsto \dfrac{1-3(4)+1(8)-k(8)}{8}=R_1$

$\sf\longmapsto \dfrac{1-12+8+8k}{8}=R_1$

So,

$\sf\longmapsto \dfrac{-3-8k}{8}=R_1 - - - -(1)$

Taking g(x) now,

We have,

$\sf\longmapsto kx^3+12x^2+14x-3\div(2x+1)=R_2$

That is,

$\sf\longmapsto g(x)\div(2x+1)=R_2$

Let, 2x + 1 = 0.

So, x = -1/2

As,

$\sf\longmapsto g(x)\div(2x+1)=R_2$

So, by Factor Theorem,

$\sf\longmapsto g\left(\dfrac{-1}{2}\right)= R_2$

So,

$\sf\longmapsto k\left(\dfrac{-1}{2}\right)^3+12\left(\dfrac{-1}{2}\right)^2+14\left(\dfrac{-1}{2}\right)-3= R_2$

$\sf\longmapsto k\dfrac{-1}{8}+12\!\!\!/^3\left(\dfrac{1}{4\!\!\!/}\right)+14\!\!\!\!\!/^7\left(\dfrac{-1}{2\!\!\!/}\right)-3= R_2$

So,

$\sf\longmapsto \dfrac{-k}{8}+3\!\!\!/-7-3\!\!\!/= R_2$

$\sf\longmapsto \dfrac{-k}{8}-7= R_2$

Taking LCM,

$\sf\longmapsto \dfrac{-k-7(8)}{8}= R_2$

So,

$\sf\longmapsto \dfrac{-k-56}{8}= R_2 - - - -(2)$

Now, we are also given that,

$\sf\longmapsto R_1-R_2=\dfrac{25}{8}$

From (1) & (2),

$\sf\longmapsto R_1-R_2=\dfrac{25}{8}$

$\sf\longmapsto \dfrac{-3-8k}{8} - \dfrac{-k-56}{8} =\dfrac{25}{8}$

Combining fractions,

$\sf\longmapsto \dfrac{-3-8k+k+56}{8\!\!\!/}=\dfrac{25}{8\!\!\!}$

$\sf\longmapsto 53-7k=25$

$\sf\longmapsto 53-25=7k$

$\sf\longmapsto 7k=28$

So,

$\sf\longmapsto k=\dfrac{28\!\!\!\!\!/^{\:\:4}}{7\!\!\!/}$

Hence,

$\longmapsto\bf k=4$

Therefore, the value of k is 4.

Answer 2

Answer:

this is your correct answer