Question

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

Answer 1

Answer:

they golden ruls are;

Step-by-step explanation:

Real Account :

debit what's comes in

credit what's goes out

Personal Account :

debit receiver

credit giver

Nominal Account :

debit the expense of loss

credit the income and gains.

Answer 2

Answer:

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 –

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

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

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

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