Question

A polynomial f(x) with rational coefficients leaves remainder 15, when divided by(x – 3) and remainder 2x + 1, when divided by (x – 1)2. If ‘p’ is coefficient of x^2 of its remainder which will come out if f(x) is divided by (x – 3)(x – 1)^2 then find p.

Answer 1

$\large\text{\underline{Let's begin:-}}$

Let us consider the identity,

$\hookrightarrow f(x)=(x-3)(x-1)^{2}Q(x)+R(x)$

which can be obtained by division. $\large\text{[1]}$

If an n-th degree polynomial divides another polynomial, the maximum degree of the remainder is n-1. $\large\text{[2]}$

$\large\text{\underline{Solution:-}}$

By remainder theorem, the first condition is $f(3)=15$.

(By $\large\text{[1]}$ and $\large\text{[2]}$)

The remainder of $f(x)$ by $(x-1)^{2}$ is equivalent to the remainder of $R(x)$ by $(x-1)^{2}$.

Since $2x+1$ is the remainder if $f(x)$ is divided by $(x-1)^{2}$, and as the maximum degree of $R(x)$ is 2,

$\hookrightarrow R(x)=p(x-1)^{2}+2x+1$

Now, according to $f(3)=15$, we get,

$\hookrightarrow f(3)=R(3)$

$\hookrightarrow R(3)=15$

$\hookrightarrow p\times(3-1)^{2}+2\times3+1=15$

$\hookrightarrow 4p+7=15\implies\therefore p=2$

$\large\text{\underline{Result:-}}$

The required value of $p$ is 2.

Answer 2

Given :-

A  polynomial f(x) with rational coefficients leaves remainder 15, when divided by(x – 3) and remainder 2x + 1, when divided by (x – 1)2.

To Find :-

If ‘p’ is coefficient of x^2 of its remainder which will come out if f(x) is divided by (x – 3)(x – 1)^2 then find p.

Solution :-

x - 3 = 0

x = 3

f(3) = 15 (..1)

(x - 3)(x - 1)²[q(x)][r(x)]

Let, remainder be the standard form of a quadratic polynomial i.e ax² + bx + c

(a - b)² = a² - 2ab + b²

(x - 3)[x² - 2(1)(x) + (1)²][q(x)][ax² + bx + c]

(x - 3)(x² - 2x + 1)[q(x)](ax² + bx + c)

Now,

15 = ax² + bx + c

15 = a(3)² + b(3) + c

15 = a(9) + 3b + c

15 = 9a + 3b + c (2)

x² - 2x + 1)ax² + bx + c(

We get

a(x² - 2x + 1) + (b + 2a)x + c - a

Remainder is 2x + 1

(b + 2a) = 2

b + 2a = 2

b = 2 - 2a (3)

And

(c - a) = 1

c - a = 1

c = 1 + a (4)

Putting value in eq 2

9a + 3b + c = 15

9a + 3(2 - 2a) + 1 + a = 15

9a + 6 - 6a + 1 + a = 15

4a + 7 = 15

4a = 15 - 7

4a = 8

a = 8/4

a = 2

As a is the coefficient of x² and p is also the coefficient of x². So,

a = p

a = 2

p = 2