Question

if quadratic polynomial divided by(x-1),(x-2) gives the remainder 6,30 respectively and it is divisible by (x-3) ,then find polynomial​

Answer 1

Answer:

3x²+15x-12

Step-by-step explanation:

let f(x) be the polynomial,

ax²+bx+c

given f(x) leaves remainder 6 when divided by (x-1),

=> f(1) = 6

=> a(1²) + b(1) + c = 6

=> a+b+c = 6    ---------------------- 1

also given f(x) leaves remainder 30 when divided by(x-2),

=> f(2) = 30

=>a(2²)+b(2)+c = 30

=> 4a+2b+c = 30    ---------------------- 2    

also it is divisible by (x-3)

=> f(3) = 0

=> a(3²)+b(3)+c=0

=> 9a+3b+c = 0    ---------------------- 3

2 - 1  => 4a - a + 2b - b+c -c = 30-6

        => 3a+b=24    ---------------------- 4

3 - 2 => 9a-4a+3b-2b+c-c = 0-30

        => 5a+b = 30    ---------------------- 5

5 - 4 => 5a-3a+b-b = 30-24

       => 2a = 6   => a=3

substitute a=3 in 4, we get

3(3)+b = 24

=> 9+b=24  => b = 24-9 = 15  => b = 15

substitute a=3 and b=15 in 1,

3+15+c = 6

=> c = 6-18 = -12

=> c = -12

=> f(x) = 3x²+15x-12

Answer 2

Given:

A quadratic polynomial, f(x) leaves remainder 6 when divided by (x-1) and 30 when divided by (x-2). f(x) is divisible by (x-3).

To find:

The polynomial f(x).

Solution:

Let f(x) = ax² + bx + c.

According to the question,

When, x - 1 = 0, x =1; f(1) = 6

• a(1²) + b(1) + c = 6  ⇒ a+b+c = 6 ..(1) and,

When, x - 2 = 0, x = 2; f(2) = 30

• a(2²)+b(2)+c = 30 ⇒ 4a + 2b + c = 30..(2) and,

f(x) is divisible by (x-3) so, f(3) = 0

• a(3²)+b(3)+c=0  => 9a+3b+c = 0..(3)

On solving (1), (2) and (3), we get:

a = 3; b = 15; and c = -12.

Therefore, the quadratic polynomial is f(x) = 3x²+15x-12.