2 precal question help please?

Question

Solve and simplify if possible: (x^2/x - 2) - (4/x - 2)

for f(x) = 1/x^2-3, find f(2 + h)

llaffer · Accepted Answer

Presuming the first one is:

x² / (x - 2) - 4 / (x - 2)

You already have a common denominator, so you can subtract the numerators:

(x² - 4) / (x - 2)

You now have a difference of two squares, so factor, then cancel out the common factor with the denominator:

(x + 2)(x - 2) / (x - 2)

x + 2

---

f(x) = 1 / (x² - 3)

So:

f(2 + h) = 1 / [(2 + h)² - 3]

First, simplify the binomial square, then simplify the constant terms in the denominator:

f(2 + h) = 1 / (4 + 4h + h² - 3)
f(2 + h) = 1 / (1 + 4h + h²)

Rearranging into standard form:

f(2 + h) = 1 / (h² + 4h + 1)

Hammed Sanusi · Answer

x²/(x - 2) - 4/(x - 2)

Find the L.C.M to get. ..

(x² - 4)/(x - 2)

= (x² - 2²)/(x - 2)

Using difference of two square in the numerator. 
a² - b² = (a + b)(a - b)

= [(x + 2)(x - 2)]/(x - 2)

Noticed that (x - 2) will council leaving ..

(x + 2)

∴ The answer = (x + 2)

(2)

f(x) = 1/(x² - 3)

f(2 + h) = 1/[(2+h)² - 3)]
f(2 + h) = 1/[(2+h)(2+h) - 3]
f(2 + h) = 1/[4+2h+2h+h² - 3)]
f(2 + h) = 1/(4 + 4h + h² - 3)
f(x + h) = 1/(h² + 4h + 1)

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