If x^3+2x^2-17x-17=(x-4)(x+5)(x+A)+Bx+C, how do you find the values of A, B and C?

Jun 29, 2017

A = 1, B = 2, C = 3

Explanation:

1. First, let's expand the right side of the equation.

$\left(x - 4\right) \left(x + 5\right) \left(x + A\right) + B x + C$

$\textcolor{b l u e}{{x}^{2} + x - 20} \textcolor{red}{\left(x + A\right)} + B x + C$

$\textcolor{v i o \le t}{{x}^{3} + A {x}^{2} + {x}^{2} + A x - 20 x - 20 A} + B x + C$

2. Rearrange the expression, so that like terms are next to each other (this makes it easier to visualize).

$\textcolor{v i o \le t}{{x}^{3}} \textcolor{b l u e}{+ A {x}^{2} + {x}^{2}} \textcolor{g r e e n}{+ A x - 20 x + B x} \textcolor{red}{- 20 A + C}$

3. Compare the right side of the equation to the left side. Is there any variable that we can find?

$\textcolor{v i o \le t}{{x}^{3}} \textcolor{b l u e}{+ 2 {x}^{2}} \textcolor{g r e e n}{- 17 x} \textcolor{red}{- 17} = \textcolor{v i o \le t}{{x}^{3}} \textcolor{b l u e}{+ A {x}^{2} + {x}^{2}} \textcolor{g r e e n}{+ A x - 20 x + B x} \textcolor{red}{- 20 A + C}$

We can find A!

$\textcolor{b l u e}{2 {x}^{2} = A {x}^{2} + {x}^{2}}$

Treat the ${x}^{2}$ just like any variable. In fact, let's remove it altogether to make solving for $A$ easier.

$\textcolor{b l u e}{2 = A + 1}$

Now, it's clear that A = 1. Since we know A, we can plug its value into our equation and solve for another variable.

The right side of the equation is now:

${x}^{3} + \textcolor{b l u e}{A} {x}^{2} + {x}^{2} + \textcolor{b l u e}{A} x - 20 x + B x - 20 \textcolor{b l u e}{A} + C$

${x}^{3} + \textcolor{b l u e}{1} {x}^{2} + {x}^{2} + \textcolor{b l u e}{1} x - 20 x + B x - 20 \textcolor{b l u e}{\cdot 1} + C$

${x}^{3} + 2 {x}^{2} - 19 x + B x - 20 + C$

${x}^{3} + 2 {x}^{2} - 17 x - 17 = {x}^{3} + 2 {x}^{2} - 19 x + B x - 20 + C$

4. We have enough information to solve for both B and C.

$- 19 x + B x = - 17 x$
$B x = 2 x$
$B = 2$

$- 20 + C = - 17$
$C = 3$

Hope this helps!