Question

if f(x)=2x^3+ax^2+bx-6 where a and b are constants ,when f(x) is divided by (2x-1) the remainder is -5 when f(x) is divided by (x+2) there is no remainder, then the value of a and b are

Answer 1

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Answer 2

Answer:

Value of a = 5 and b = -1

Step-by-step explanation:

The given function is f(x) = 2x³ + ax² + bx - 6

If f(x) is divided by (2x - 1) the remainder is (-5).

That means if we subtract remainder (-5) from the function then the given function f(x) will be divisible by the (2x - 1).

Or (2x - 1) will be the zero root of the function.

(2x - 1) = 0

x = $\frac{1}{2}$

$f(\frac{1}{2})-(-5)=2(\frac{1}{2})^{3}+a(\frac{1}{2})^{2}+b(\frac{1}{2})-6-(-5) = 0$

$\frac{1}{4}+a(\frac{1}{4})+b(\frac{1}{2})-1=0$

$1+a+2b-4=0$

a + 2b - 3 = 0

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

When f(x) is divided by (x+2) remainder is 0.

Therefore for x = -2, f(x) = 0

f(-2) = 2(-2)³+ a(-2)² + b(-2) - 6 = 0

-16 + 4a - 2b - 6 = 0

4a - 2b = 22

2a - b = 11 -------(2)

Equation (2)×2 - equation (1)

2(2a - b) + (a + 2b) = 22 + 3

4a - 2b + a + 2b = 25

5a = 25

a = 5

From equation (2),

2(5) - b = 11

-b = 11 - 10

b = -1

Therefore, value of a = 5 and b = -1.

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