Question

let p(x) be a polynomial which when divided by x-1,x-2,x-3 leaves remainder 3,5and 2 respectively. Determine the remainder when px is divided by (x-1)(x-2)(x-3)​

Answer 1

$r(x) = - \frac{5}{2} {x}^{2} + \frac{19}{2} x - 4$

EXPLAINATION

ɢɪᴠᴇɴ

p(x) leaves 3,5and 2 as remainder when divided by x-1,x-2 and x-3 respectively. Therefore, by remainder theorem

p(1) = 3

p(2) = 5

p(3) = 2

ᴄᴏɴᴄᴇᴘᴛs

⇒ In polynomial division, degree of the remainder is always less that the degree of divisor.

⇒ If a polynomial p(x) leaves remainder r(x) when divided by a divisor d(x), then p(x) and r(x) leaves a same remainder when divided by a factor of d(x).

Suppose (x-a) is a factor of d(x), then

p(a) = r(a)

ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴ

Let r(x) = ax² + bx + c be the remainder when p(x) is divided by (x-1)(x-2)(x-3).

ᴇǫᴜᴀᴛɪᴏɴs

By the above concept

p(x) and r(x) will leave a same remainder when divided by (x-1), so

r(1) = p(1)

or a + b + c = 3--------------(1)

Similarly

r(2) = p(2)

or 4a + 2b + c = 5----------------(2)

r(3) = p(3)

or 9a + 3b + c = 2 ----------------(3)

sᴏʟᴜᴛɪᴏɴ ʙʏ ᴇʟɪᴍɪɴᴀᴛɪᴏɴ

First, let us eliminate c

Substracting (1) from (2), we get

3a + b = 2 ---------------(4)

Substracting (1) from (3) we get

8a + 2b = - 1 -------------(5)

Now, let us eliminate b

Multiplying (4) by 2 and substracting from (5)

(8a + 2b) - (6a + 2b) = -1 - 4

or 2a = -5

$or \: a = - \frac{5}{2}$

Substituing value of a in (4), we get

$3( \frac{ - 5}{2} ) + b = 2$

$b = 2 + \frac{15}{2} = \frac{4 + 15}{2}$

$b = \frac{19}{2}$

Substituting value of a and b in (1), we get

$- \frac{5}{2} + \frac{19}{2} + c = 3$

$c = 3 + \frac{5}{2} - \frac{19}{2}$

$c = \frac{6 + 5 - 19}{2}$

$c = \frac{ - 8}{2}$

$c = - 4$

ʀᴇᴍᴀɪɴᴅᴇʀ

The required remainder is

$r(x) = a {x}^{2} + bx + c$

$r(x) = - \frac{5}{2} {x}^{2} + \frac{19}{2} x - 4$