Question

The polynomial f(x)= x⁴-2x³+3x²-ax-b when divided by(x-1) and (x+1) leaves the remainders 5 and 9 respectively. find the value of a and b .Hence, find the remainder 2hen f(x) is divided by (x-3)​

Answer 1

GIVEN :–

• Polynomial f(x)= x⁴-2x³+3x²-ax-b when divided by (x-1) and (x+1) leaves the remainders 5 and 9 respectively.

TO FIND :–

• Value of a and b = ?

• Remainder when f(x) is divided by (x-3) = ?

SOLUTION :–

• If (x-1) is a factor of polynomial f(x)= x⁴-2x³+3x²-ax-b then f(1) = 0.

$\\ \bf \implies f(x)= {x}^{4} -2 {x}^{3} +3 {x}^{2} -ax-b$

$\\ \bf \: \: \because \: \: f(1)= 0$

$\\ \bf \implies f(1)= {(1)}^{4} -2 {(1)}^{3} +3 {(1)}^{2} -a(1)-b$

$\\ \bf \implies 0=1-2+3 -a-b$

$\\ \bf \implies a+b=2 \: \: \: \: \: \: \: \: - - - - eq.(1)$

• If (x+1) is a factor of polynomial f(x)= x⁴-2x³+3x²-ax-b then f(-1) = 0.

$\\ \bf \implies f(x)= {x}^{4} -2 {x}^{3} +3 {x}^{2} -ax-b$

$\\ \bf \: \: \because \: \: f( - 1)= 0$

$\\ \bf \implies f( - 1)= {( - 1)}^{4} -2 {( - 1)}^{3} +3 {( - 1)}^{2} -a( - 1)-b$

$\\ \bf \implies 0=1+2+3+a-b$

$\\ \bf \implies b - a=6 \: \: \: \: \: \: \: \: - - - - eq.(2)$

• Add eq.(1) and eq.(2) –

$\\ \bf \implies (a+b) + (b - a)=2+6$

$\\ \bf \implies 2b=8$

$\\ \large\implies{\boxed{ \bf b=4}}$

• By eq.(1) –

$\\ \bf \implies a+4=2$

$\\ \bf \implies a=2-4$

$\\ \large\implies{\boxed{ \bf a=-2}}$

• Hence the equation is f(x)= x⁴-2x³+3x²+2x-4 .

$\\ \rule{400}{2}$

☛ Remainder when f(x)= x⁴-2x³+3x²+2x-4 is divided by (x-3) :–

$\\ \bf \implies f(x)= {x}^{4} -2 {x}^{3} +3 {x}^{2} + 2x-4$

• Now put x = 3 –

$\\ \bf \implies f(3)={(3)}^{4} -2 {(3)}^{3} +3 {(3)}^{2} + 2(3)-4$

$\\ \bf \implies f(3)=81-54 +27+ 6-4$

$\\ \bf \implies f(3)=81-27+2$

$\\ \large\implies { \boxed{ \bf f(3)=56}}$

Answer 2

Answer:

Answer :

Given :

Expression is f(x) = 2x + 3x - ax + b

(x-1) when divided leaves remainder 5

(x+1) when divided leaves remainder 19

Required to find :

Values of " a " and " b "

Solution :

Let's consider the given expression

f(x) = 2x + 3x - ax + b

(x-1) when divided leaves remainder 5

So, let

x - 1 = 0

x = 1

Hence,

f (1) = 2(1) + 3(1) - a(1) + b = 5

2 + 3 - a + b - 5 = 0

5 - a + b - 5 = 0

0 - a + b = 0

Then ,

b = a. -----------> equation 1

Similarly,

(x+1) when divided leaves remainder 19

So,

Let x + 1 = 0

x = - 1

Hence,

f(- 1) = 2(-1) + 3(-1) - a (-1) + b = 19

- 2 - 3 + a + b = 19

- 5 + a + b - 19 = 0 ( substitute the value of " b " from equation 1 )

- 5 + a + a - 19 = 0

- 24 + 2a = 0

2a = 24

a = 24/2

a = 12

hence ,

Value of a = 12

value of b = ?

but b = a ( from equation 1 )

So, b = 12

Therefore,

Value of a = 12

Value of b = 12

This states that ,

a = b = 12