we have the expression
[tex]\frac{4}{3x^2-23x+40}[/tex]
Rewrite as equivalent rational expressions with denominator (3x-8)(x-5)(x-3)
In this problem
3x^2-23x+40=3(x-5)(3x-8)
so
[tex]\frac{4}{3x^2-23x+40}=\frac{4}{3\mleft(x-5\mright)\mleft(3x-8\mright)}[/tex]
Multiply by (x-3)/(x-3)
[tex]\frac{4}{3(x-5)(3x-8)}\cdot\frac{(x-3)}{(x-3)}=\frac{4(x-3)}{3(x-5)(3x-8)(x-3)}[/tex]
Part 2
we have the expression
[tex]\frac{9x}{3x^2-17x+24}[/tex]
we have that
3x^2-17x+24=3(3x-8)(x-3)
so
[tex]\frac{9x}{3x^2-17x+24}=\frac{9x}{3\mleft(3x-8\mright)\mleft(x-3\mright)}=\frac{3x}{(3x-8)(x-3)}[/tex]
Multiply the expression by (x-5)/(x-5)
[tex]\frac{3x}{(3x-8)(x-3)}\cdot\frac{(x-5)}{(x-5)}=\frac{3x(x-5)}{(3x-8)(x-3)(x-5)}[/tex]
