Question

If f(x)=xpower4-2x³+3x²-ax+b is a polynomial such that when it is divided by (x-1) and (x+1) the remainders are 5 and 19 respectively. Determine the remainder when f(x) is divided by (x-2)

Answer 1

f(x)=4-2x³+3x²-ax+b

f(1)=4-2(1)³+3(1)²-a(1)+b=5

4-2+3-a+b=5

-a+b=0...(i)

f(-1)=4-2(-1)³+3(-1)²-a(-1)+b=19

4+2+3+a+b=19

a+b-10=0...(ii)

From (i) and (ii):

a+b-10=-a+b

2a=10

a=5

putting the value of 'a' in (i):

-a+b=0

-5+b=0

b=5

Now :

f(2)=4-2(2)³+3(2)²-5(2)+5

=4-16+12-10+5

=21-26

=-5

Hence the required remainder is -5.

Answer 2

AnswEr:

When f(x) is divided by x-1 and x+1, the remainders are 5 and 19 respectively.

$\therefore$ $\qquad\sf\green{f(1)=5\:and\:f(-1)=19}$

$\tt{1-2+3-a+b=5\:and\:1+2+3+a+b=19}$

=> -a + b = 3 and a + b = 13

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Adding these two, we get : 2b = 16 $\implies$ b=8

Putting b=8 in a + b = 3, we get

$\qquad\tt{-a+8=3=a=5}$

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Substituting the values of a and in f(x) = x⁴ - 2x³ + 3x² - ax + b, we get

$\qquad\tt{f(x)=x^4-2x^3+3x^2-5x+8}$

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The remainder when f(x) is divided by x-2 is equal to f(2).

$\therefore$ Remainder = f(2) = 2⁴ - 2 × 2³ + 3 × 2² - 5 × 2 +8 = 16 - 16 + 12 - 10 + 8 = 10.

#BAL.

#Answerwithquality