If x3+2x2−17x−17=(x−4)(x+5)(x+A)+Bx+C, how do you find the values of A,B and C?

1 Answer
Michelle
Jun 29, 2017

A = 1, B = 2, C = 3

Explanation:

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

(x−4)(x+5)(x+A)+Bx+C

x2+x−20(x+A)+Bx+C

x3+Ax2+x2+Ax−20x−20A+Bx+C

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

x3+Ax2+x2+Ax−20x+Bx−20A+C

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

x3+2x2−17x−17=x3+Ax2+x2+Ax−20x+Bx−20A+C

We can find A!

2x2=Ax2+x2

Treat the x2 just like any variable. In fact, let's remove it altogether to make solving for A easier.

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:

x3+Ax2+x2+Ax−20x+Bx−20A+C

x3+1x2+x2+1x−20x+Bx−20⋅1+C

x3+2x2−19x+Bx−20+C

Add the left side:

x3+2x2−17x−17=x3+2x2−19x+Bx−20+C

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

−19x+Bx=−17x
Bx=2x
B=2

−20+C=−17
C=3

Hope this helps!

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