What must be subtracted from x 4 2x 3 2x 2 4x 6 so that the result is exactly divisible by x 2 2x 3 in remainder Theorem?

Question 25 Polynomials Exercise 2D

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Answer:

Consider,

p(x) = 𝑥^4 + 2𝑥^3 − 2𝑥^2 + 4𝑥 + 6

q(x) = 𝑥^2+2x – 3

Assume r(x) = ax + b

Let p(x) be subtracted from r(x) and divided by q(x)

f(x)

= p(x) – r(x)

By substituting (ax + b) in the place of r we get

= p(x) – (ax + b)

On further calculation

= (𝑥^4 + 2𝑥^3 − 2𝑥^2 + 4𝑥 + 6) – (ax + b)

= 𝑥^4 + 2𝑥^3 − 2𝑥^2 + (4 − 𝑎)𝑥 + 6 - b

q(x)

= 𝑥^2+2x – 3

We can further write it as

= 𝑥^2+ 3x - x – 3

By taking the common terms out

= x(x+3) – 1(x+3)

So we get

= (x-1) (x+3)

(x-1) and (x+3) are the factors of f(x) if f(1) and f(-3) = 0

f(1) = 0

By substituting 1 in the place of x

1^4 + 2(1)^3 − 2(1)^2 + (4 − 𝑎)(1) + 6 - b= 0

On further calculation

1 + 2 – 2 + 4 – a + 6 – b = 0

So we get

11 – a – b = 0

-a - b = -11 ……. (1)

f(-3) = 0

By substituting -3 in the place of x

(−3)^4 + 2(−3)^3 − 2(−3)^2 + (4 − 𝑎)(−3) + 6 – b = 0

So we get

81 – 54 – 18 – 12 + 3a + 6 – b = 0

3 + 3a – b = 0

3a – b = -3 …….. (2)

By subtracting both the equations (1) and (2)

-a – 3a – b + b = -11 + 3

-4a = - 8

Dividing both sides by -4

a = 2

Substitute a = 2 in (2)

3(2) – b = -3

6 – b = -3

On further calculation

b = 6 + 3

So we get

b = 9

Substituting a and b values in r(x) = ax + b

r(x) = 2x + 9

Therefore, (𝑥^4 + 2𝑥^3 − 2𝑥^2 + 4𝑥 + 6) is divisible by 𝑥^2+2x – 3 if 2x + 9 is subtracting from it.

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Text Solution

2x+99x+2`-x^2+5`None of these

Answer : A

Solution : Let g (x) `=(x^(2)+2x-3)=(x^(2)+3x-x-3)=x(x+3)-(x+3)=(x+3)(x-1)`. <br> When a polynomial is divided by a quadratic polynomial , then remainder is a linear expression ,say ax + b . <br> Let `p(x)=(x^(4)+2x^(3)-2x^(2)+4x+6)-(ax+b)` . Then, <br> `p(x)=x^(4)+2x^(3)-2x^(2)+(4-a)x+(6-b)`. <br> `thereforep(-3)=0 and p (1) = 0 rArr 3a-b=- 3 and a +b = 11`. <br> On solving these equations , we get a =2 and b =9 . <br> Hence , the required expression is ( 2x +9).