Question

1. \\(\frac{5k^3 + 23k^2 - 42k}{4k^2 + 28k + 24}\\)

Explanation:

Step1: Factor Numerator and Denominator

First, factor the numerator \(5k^{3}+23k^{2}-42k\). We can factor out a common factor of \(k\) first:
\(5k^{3}+23k^{2}-42k = k(5k^{2}+23k - 42)\)
Now, factor the quadratic \(5k^{2}+23k - 42\). We need two numbers that multiply to \(5\times(-42)=-210\) and add to \(23\). The numbers are \(30\) and \(-7\).
So, \(5k^{2}+23k - 42=5k^{2}+30k - 7k - 42 = 5k(k + 6)-7(k + 6)=(5k - 7)(k + 6)\)
Thus, the numerator factors to \(k(5k - 7)(k + 6)\).

Next, factor the denominator \(4k^{2}+28k + 24\). Factor out a common factor of \(4\) first:
\(4k^{2}+28k + 24 = 4(k^{2}+7k + 6)\)
Now, factor the quadratic \(k^{2}+7k + 6\). We need two numbers that multiply to \(6\) and add to \(7\). The numbers are \(6\) and \(1\).
So, \(k^{2}+7k + 6=(k + 6)(k + 1)\)
Thus, the denominator factors to \(4(k + 6)(k + 1)\).

Step2: Cancel Common Factors

Now, the rational expression becomes:
\(\frac{k(5k - 7)(k + 6)}{4(k + 6)(k + 1)}\)
We can cancel the common factor of \((k + 6)\) (assuming \(k
eq - 6\)):
\(\frac{k(5k - 7)}{4(k + 1)}\)

Answer:

\(\frac{k(5k - 7)}{4(k + 1)}\) (for \(k
eq - 6,-1\))