Question

($\sqrt{5} + 2) \cdot (\sqrt{5} - 5)$

Explanation:

Step1: Apply the distributive property (FOIL method)

We multiply each term in the first binomial by each term in the second binomial. So, \((\sqrt{5}+2)\cdot(\sqrt{5}-5)=\sqrt{5}\cdot\sqrt{5}+\sqrt{5}\cdot(-5)+2\cdot\sqrt{5}+2\cdot(-5)\)

Step2: Simplify each term

• For \(\sqrt{5}\cdot\sqrt{5}\), we know that \(\sqrt{a}\cdot\sqrt{a}=a\), so \(\sqrt{5}\cdot\sqrt{5} = 5\)
• For \(\sqrt{5}\cdot(-5)\), it is \(-5\sqrt{5}\)
• For \(2\cdot\sqrt{5}\), it is \(2\sqrt{5}\)
• For \(2\cdot(-5)\), it is \(-10\)

Now we combine these terms: \(5-5\sqrt{5}+2\sqrt{5}-10\)

Step3: Combine like terms

Combine the constant terms \(5 - 10=-5\) and the terms with \(\sqrt{5}\): \(-5\sqrt{5}+2\sqrt{5}=-3\sqrt{5}\)
So the expression becomes \(-5-3\sqrt{5}\)

Answer:

\(-5 - 3\sqrt{5}\)