Question

simplify. express your answer as a single term, without a denominator.
$qr \cdot q^{0}r$

Explanation:

Step1: Recall the zero - exponent rule

The zero - exponent rule states that for any non - zero number \(a\), \(a^{0}=1\). So, for the term \(q^{0}\), we have \(q^{0} = 1\) (assuming \(q
eq0\)).
The original expression is \(qr\cdot q^{0}r\). Substitute \(q^{0}=1\) into the expression: \(qr\cdot1\cdot r\).

Step2: Use the product rule for exponents (for the same base)

The product rule for exponents states that \(a^{m}\cdot a^{n}=a^{m + n}\). For the \(q\) terms: we have \(q^{1}\cdot q^{0}\), using the rule \(a^{m}\cdot a^{n}=a^{m + n}\), where \(a = q\), \(m = 1\) and \(n=0\), so \(q^{1}\cdot q^{0}=q^{1 + 0}=q^{1}=q\).
For the \(r\) terms: we have \(r^{1}\cdot r^{1}\), using the product rule \(a^{m}\cdot a^{n}=a^{m + n}\) with \(a = r\), \(m = 1\) and \(n = 1\), so \(r^{1}\cdot r^{1}=r^{1+1}=r^{2}\).
Combining the results for \(q\) and \(r\) terms, we get \(q\cdot r^{2}=qr^{2}\).

Answer:

\(qr^{2}\)