Answer:
[tex]\frac{3(x-2)}{2(x-3)}[/tex]
Step-by-step explanation:
The given expression is
[tex]\frac{6x^{2} -54x+84}{8x^{2}-40x+48 } \div \frac{x^{2} +x-56}{2x^{2}+12x-32}[/tex]
We need to factor each quadratic expression
[tex]6x^{2} -54x+84[/tex]
First, we extract the GCF 6:
[tex]6(x^{2}-9x+14 )[/tex]
Then, we look for two numbers which product is 14 and which sum is 9, thos numbers are 7 and 2, so the factored expression is
[tex]6(x-7)(x-2)[/tex]
[tex]8x^{2} -40x+48[/tex]
We do the same process,
[tex]8(x^{2} -5x+6)[/tex]
We need to find two numbers which product is 6 and which sum is 5, those numbers are 3 and 2
[tex]8(x-3)(x-2)[/tex]
[tex]x^{2} +x-56[/tex]
We have to find two numbers which product is 56, and which difference is 1, those numbers are 8 and 7
[tex]x^{2} +x-56=(x+8)(x-7)[/tex]
[tex]2x^{2} +12x-32=2(x^{2}+6x-16 )=2(x+8)(x-2)[/tex]
Replacing all factors in the given expression, we have
[tex]\frac{6(x-7)(x-2)}{8(x-3)(x-2)} \div \frac{(x+8)(x-7)}{2(x+8)(x-2)} \\\frac{3(x-7)}{4(x-3)} \div \frac{(x-7)}{2(x-2)} \\\frac{3(x-7)}{4(x-3)} \times \frac{2(x-2)}{(x-7)} =\frac{3(x-2)}{2(x-3)}[/tex]
Therefore, the answer is
[tex]\frac{3(x-2)}{2(x-3)}[/tex]
