Question

1 divide by cancelling out the common factor:
a) \\(\frac{x^2 + 2x + 3}{x - 1}\\)
b) \\(\frac{8h^5 - 32h^4 + 16h^3}{-8h^4}\\)
c) \\(\frac{a^2 + 8a + 15}{a^2 + a - 20}\\)

Explanation:

Part a)

Step1: Check numerator for factors

The quadratic $x^2+2x+3$ has discriminant $\Delta = 2^2 - 4(1)(3) = 4 - 12 = -8 < 0$, so it cannot be factored over real numbers.

Step2: Check for common factors

There are no common factors between $x^2+2x+3$ and $x-1$.

Part b)

Step1: Factor numerator

Factor out $8h^3$ from numerator:
$8h^5 - 32h^4 + 16h^3 = 8h^3(h^2 - 4h + 2)$

Step2: Rewrite the fraction

Substitute factored numerator into the fraction:
$\frac{8h^3(h^2 - 4h + 2)}{-8h^4}$

Step3: Cancel common factors

Cancel $8h^3$ from numerator and denominator:
$\frac{h^2 - 4h + 2}{-h} = -\frac{h^2 - 4h + 2}{h} = -h + 4 - \frac{2}{h}$

Part c)

Step1: Factor numerator

Factor the quadratic $a^2+8a+15$:
$a^2+8a+15 = (a+3)(a+5)$

Step2: Factor denominator

Factor the quadratic $a^2+a-20$:
$a^2+a-20 = (a+5)(a-4)$

Step3: Cancel common factors

Cancel the common factor $(a+5)$ from numerator and denominator:
$\frac{(a+3)(a+5)}{(a+5)(a-4)} = \frac{a+3}{a-4}$ (where $a
eq -5$)

Answer:

a) $\frac{x^2+2x+3}{x-1}$ (cannot be simplified further by cancelling common factors)
b) $-h + 4 - \frac{2}{h}$ or $\frac{-h^2 + 4h - 2}{h}$
c) $\frac{a+3}{a-4}$ (with the restriction $a
eq -5, 4$)