Question

I am confused in this can someone help?

Answer 1

Answer:

$\huge\boxed{\sf a = -12 , b = 8}$

Step-by-step explanation:

Given polynomial is:

$f(x) = ax^3+ b^2 + 3x + 4$

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First factor is (x-1)

Let x - 1 = 0  => x = 1

Putting this in the above polynomial

$f(1) = a(1)^3 + b(1)^2 + 3(1) + 4$

Given that it's remainder is 3

$3 = a + b + 3 + 4$

$3 = a + b + 7$

Subtracting 7 to both sides

$-4 = a + b\\OR$

a + b = -4 -------------------------- (1)

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Second Factor is 2x + 1

Let 2x + 1 = 0 => x = - 1 / 2

Putting in the above polynomial

$f(-\frac{1}{2} ) = a (-\frac{1}{2} )^3 + b (-\frac{1}{2} )^2 + 3 (-\frac{1}{2} ) + 4$

The remainder is 6

$6 = -\frac{a}{8} + \frac{b}{4} - \frac{3}{2} + 4\\6 - 4 = -\frac{a}{8} + \frac{b}{4} - \frac{3}{2}\\2 = \frac{-a + 2b -12 }{8} \\-a + 2b - 12 = 16\\-a + 2b = 16 + 12$

-a + 2b = 28  --------------------------------(2)

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Adding both equations

a + b -  a + 2b = -4 + 28

3b = 24

Dividing both sides by 3

b = 8

$\rule[225]{225}{2}$

Putting b = 8 in Equation (1)

a + 8 = -4

a = - 4 - 8

a = -12

$\rule[225]{225}{2}$