Compare your answer:

$x=5$, $x=-4$, and $x=\frac{9}{7}$

Rewriting each quadratic expression in factored form gives:

$(x-5)(x+4)(x-1)(x+3)=(x+4)(x-2)(x-5)(x-3)$

Step 1 - Writing the equation so that it equals zero yields:

$(x-5)(x+4)(x-1)(x+3)-(x+4)(x-2)(x-5)(x-3)=0$

Step 2 - Notice that the factors $(x-5)$ and $(x+4)$ appear in each of the terms. This means we can factor those binomials to the front of our expression. This means our expression is a list of factors and not terms.

$(x-5)(x+4)[(x-1)(x+3)-(x-2)(x-3)]=0$

Step 3 - Now, using the zero product property, we set each of the factors equal to zero and solve.

$(x-5)=0$, so $x=5$.

$(x+4)=0$, so $x=-4$.

$[(x-1)(x+3)-(x-2)(x-3)]=0$

Solving this equation will take a little more work!

$x^2+2x-3-(x^2-5x+6)=0$

$x^2+2x-3-x^2+5x-6=0$

$7x-9=0$

$7x=9$

$x=\frac{9}{7}$

Solving the linear equation gives $x=\frac{9}{7}$ .