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)? If 'p' is coefficient of xof its remainder which will
come out if f(x) is divided by (x - 3)(x - 1) then find p.​

Answer 1

Answer:

$p=2$

Step-by-step explanation:

Since on dividing the polynomial $f(x)$, by $(x-3)$, the remainder is 15

Therefore, by remainder theorem

$f(3)=15$

When the polynomial is divided by $(x-1)^2(x-3)$, the remainder will be a polynomial of degree 2

Let the remainder be

$ax^2+bx+c$

Therefore, by Euclid's division lemma, the polynomial can be written as

$f(x)=(x-3)(x-1)^2.q(x)+(ax^2+bx+c)$ ....(a)

$f(3)=15$

Therefore,

$15=0+a\times 3^2+b\times 3+c$

$\implies 9a+3b+c=15$ ........ (1)

Again if (a) is divided by $(x-1)^2$, the remainder will be the remainder obtained by dividing $ax^2+bx+c$ by $(x-1)^2$ or $x^2-2x+1$

Dividing $ax^2+bx+c$ by $x^2-2x+1$ and writing it in the form of Euclid's division lemma we get

$ax^2+bx+c=a(x^2-2x+1)+(b+2a)x+(c-a)$

But given that the remainder is $2x+1$

Therefore,

$b+2a=2$  ..........(2)

And

$c-a=1$  ............(3)

Putting the value of c from eq (3) and value of b from eq (2), in eq (1)

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

$\implies 9a+6-6a+a+1=15$

$\implies 4a=15-6-1$

$\implies 4a=8$

$\implies a=2$

Therefore the coefficient of $x^2$ in the remainder is 2

Thus, $p=2$

Hope this helps.