Question

Simplify x^2-9 over x^2-3x

Answer 1

Answer:

$x^{2} -3x-\frac{9}{x^{2} }$  

Step-by-step explanation:

$x^{2} -9$ ÷ $x^{2} -3x$

To add or subtract expressions, expand them to make their denominators the same. Multiply $x^{2} -3x$ times $\frac{x^{2} }{x^{2} }$

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$\frac{(x^{2} -3x)x^{2} }{x^{2} } - \frac{9}{x^{2} }$

Since $\frac{(x^{2} -3x)x^{2} }{x^{2} }$ and $\frac{9}{x^{2} }$  have the same denominator, subtract them by subtracting their numerators.

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$\frac{(x^{2} -3x)x^{2} -9}{x^{2} }$

Do the multiplications in $(x^{2} -3x)x^{2} -9$

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$\frac{x^{4}-3x^{3}-9 }{x^{2} }$

Answer 2

Answer:

$\frac{x+3}{x}$ or x+3/x

Step-by-step explanation:

Using D.O.T.S (difference of two squares) formula :

$a^{2} -b^{2} = (a+b)(a-b)$

$x^{2} -9= (x+3)(x-3)$

Factorise the second bit:

$x^{2} -3x = x(x-3)$

Now write out this fraction:

$\frac{(x+3)(x-3)}{x(x-3)}$

Now we can cancel out repeated terms (x-3):

$\frac{(x+3)}{x}$
This is our final answer:

$\frac{x+3}{x}$