Question

If a polynomial x^4+x^3-3x^2+ax-b is divided by x+a and 2x-a successively , then the remainder is 9 in each case. Show that a and b are given by the equations 5a^2-6a=20 and b =a^4-a^3-4a^2-9. Hence, find the remainder when 5x^2-6x+89 is divided by x-a.​

Answer 1

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Answer:

47

Step-by-step explanation:

Given that the equation

f(x) = x4 – 2x3 + 3x2 – ax +b

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

∴ f(-1) = 19 and f(1) = 5

(-1)4 – 2 (-1)3 + 3(-1)2 – a (-1) + b = 19

⇒ 1 +2 + 3 + a + b = 19

∴ a + b = 13 ——- (1)

According to given condition f(1) = 5

f(x) = x4 – 2x3 + 3x2 – ax

⇒ 14 – 2 3 + 3 2 – a (1) b = 5

⇒ 1 – 2 + 3 – a + b = 5

∴ b – a = 3 —— (2)

solving equations (1) and (2)

a = 5 and b = 8

Now substituting the values of a and b in f(x) , we get

∴ f(x) = x4 – 2x3 + 3x2 – 5x + 8

Also f(x) is divided by (x-3) so remainder will be f(3)

∴ f(x)= x4 – 2x3 + 3x2 – 5x + 8

⇒ f(3) = 34 – 2 × 33 + 3 × 32 – 5 × 3 + 8

= 81 – 54 + 27 – 15 + 8

= 47

Therefore, f(x) = x4 – 2x3 + 3x2 – ax +b when a=3 and b= 8 is 47

Answer 2

Answer:

Correct option is

A

10

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

∴f(1)=5 and f(−1)=19

⇒(1)

4

−2×(1)

3

+3×(1)

2

−a×1+b=5

and (−1)

4

−2×(−1)

3

+3×(−1)

2

−a×(−1)+b=19

⇒1−2+3−a+b=5

and 1+2+3+a+b=19

⇒2−a+b=5 and 6+a+b=19

⇒−a+b=3 and a+b=13

Adding these two equations, we get

(−a+b)+(a+b)=3+13

⇒2b=16⇒b=8

Putting b=8 and −a+b=3, we get

−a+8=3⇒a=−5⇒a=5

Putting the values of a and b in

f(x)=x

4

−2x

3

+3x

2

−5x+8

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

So, Remainder =f(2)=(2)

4

−2×(2)

3

+3×(2)

2

−5×2+8=16−16+12−10+8=10

Step-by-step explanation:

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