What is the least common factor in equation $\frac{5}{x+4}=4+\frac{3}{x-2}$

Question

I am attempting to solve for x $\frac{5}{x+4}=4+\frac{3}{x-2}$

I know that I need to find the least common denominator. In this case, since I cannot see a clear relationship among them all I think it's just the product of all 3 denominators:

$\frac{5}{x+4}=4+\frac{3}{x-2}$ = $\frac{5}{x+4}=\frac{4}{1}+\frac{3}{x-2}$

LCD: $(x+4)(1)(x-2)$ = $(x+4)(x-2)$

Is this the LCD?

Because I tried to use this in solving my equation but I arrived at a quadratic. I don't think that my textbook wants me to use quadratics in this section but I'm not sure. Here's how I arrived at that:

$\frac{5}{x+4}=\frac{4}{1}+\frac{3}{x-2}$

$(x+4)(x-2)\frac{5}{x+4}=(x+4)(x-2)\frac{4}{1}+(x+4)(x-2)\frac{3}{x-2}$

Then cancel out common factors:

$(x-2)5=(x+4)(x-2)(4)+(x+4)(3)$

$5x-10=(x+4)(x-2)(4)+3x+12$

And if I multiple out the middle term I'll get a polynomial, which is unexpected so I'm not sure I'm on the right path here... am I?

$5x-10=4x^2-4x-8+3x+12$

$5x-10=4x^2-x+4$

Answer 1

Yes, you are on the right path. Yes, you do get a quadratic. The quadratic that you should get is $10-6x-4x^2$ and its roots are $1$ and $-\frac52$.

Answer 2

You have a mistake in this step:

$$5x - 10 = (x+4)(x-2)(4) + 3x+12$$ $$5x-10 = (x^2+2x-8)(4) + 3x+12$$ $$5x-10 = 4x^2+\color{red}{8x}-\color{red}{32}+3x+12$$ $$5x-10 = 4x^2+11x-20$$ $$4x^2-6x-10 = 0$$

Alternatively, you can save a step by grouping under the $x+4$ term: $$(x-2)(5)=(x+4)(x-2)(4)+(x+4)(3)$$ $$5x-10 = (x+4) \left(4(x-2) + 3 \right)$$

and you should get the same answer as before.