Question

2}-5x+6}{2x^{2}-7x+6 }" alt="\frac{x^{2}-5x+6}{2x^{2}-7x+6 }" align="absmiddle" class="latex-formula">
whats the simplest form and the exclusions aka holes

Answer 1

Answer:

$\frac{x-3}{2x-3}$. hole or removable discontinuity at x=2

Step-by-step explanation:

Well generally if you want the simplest form, you factor each the denominator and numerator and then see if you can cancel any of the factors out (because they're in the denominator and numerator)

So let's start by factoring the first equation:

$x^2-5x+6$

Now let's find what ac is (it's just c since a=1...)

$AC= 6$

List factors of -6

$\pm1, \pm2, \pm3, \pm6$.

Now we have to look for two numbers that add up to -5. It's a bit obvious here since there isn't many factors, but it's -2 and -3, and they're both negative since 6 is positive, and -5 is negative...

So using these two factors we get

$(x-2)(x-3)$

Ok now let's factor the second equation:

$2x^2-7x+6$

Multiply a and c

$AC = 12$

List factors of 12:

$\pm1, \pm2, \pm3, \pm4, \pm6, \pm12$.

Factors that add up to -7 and multiply to 12:

$-3\ and\ -4$

Rewrite equation:

$2x^2-4x-3x+6$

Group terms:

$(2x^2-4x)+(-3x+6)$

Factor out GCF:

$2x(x-2)-3(x-2)$

Rewrite:

$(2x-3)(x-2)$

Now let's write out the equation using these factors:

$\frac{(x-2)(x-3)}{(2x-3)(x-2)}$.

Here we can factor out the x-2 and the simplified form is:

$\frac{x-3}{2x-3}$

So we can "technically" define f(2) using the most simplified form, but it's removable discontinuity, so it has a hole as x=2. since it makes (x-2) equal to 0 (2-2) = 0.