Given that f(x)=x^3+4x^2+bx+c. When *f* is divided by (x-3), the remainder is 110. In addition, when *f* is divided by (x+2), the remainder is 150. The sum of b+c=?

Oct 1, 2017

Given that $f \left(x\right) = {x}^{3} + 4 {x}^{2} + b x + c$. When f is divided by $\left(x - 3\right)$, the remainder is 110.
So we have $f \left(3\right) = 110$

$\implies {3}^{3} + 4 \cdot {3}^{2} + 3 b + c = 110$

$\implies 3 b + c = 47. \ldots . . \left[1\right]$

In addition, when f is divided by $\left(x + 2\right)$, the remainder is 150.

This means

$f \left(- 2\right) = 150$

$\implies {\left(- 2\right)}^{3} + 4 \cdot {\left(- 2\right)}^{2} - 2 b + c = 150$

$\implies - 2 b + c = 142. \ldots \ldots \left[2\right]$

Subtracting [2] from [1] we get

$5 b = - 95$

$\implies b = - 19$

Inserting $b = - 19$ in [1] we get

$3 \cdot \left(- 19\right) + c = 47$

$\implies c = 104$

So the sum of $b + c = 104 - 19 = 85$