Question

2)(2p² + 5p²)+(6p + 15)

Explanation:

Assuming the problem is to simplify \((2p^2 + 5p^2)+(6p + 15)\) (correcting possible typo in the exponent of the second \(p\) term, as having two \(p^2\) terms makes sense for combining like terms):

Step1: Combine like terms in the first parentheses

The first parentheses has \(2p^2\) and \(5p^2\). Combining them: \(2p^2 + 5p^2 = 7p^2\)

Step2: Write the simplified expression

Now we have \(7p^2 + 6p + 15\) (since the second parentheses is \(6p + 15\) and we add the combined like terms from the first parentheses)

If the original problem was supposed to be a multiplication, like \((2p^2 + 5p)(6p + 15)\) (correcting the typo for a more meaningful algebraic operation), here's the solution:

Step1: Use the distributive property (FOIL method)

Multiply each term in the first binomial by each term in the second binomial:
\(2p^2 \times 6p = 12p^3\)
\(2p^2 \times 15 = 30p^2\)
\(5p \times 6p = 30p^2\)
\(5p \times 15 = 75p\)

Step2: Combine like terms

Combine the \(p^2\) terms: \(30p^2 + 30p^2 = 60p^2\)
The expression becomes \(12p^3 + 60p^2 + 75p\)

Step3: Factor out the greatest common factor (GCF)

The GCF of \(12p^3\), \(60p^2\), and \(75p\) is \(3p\). Factoring out \(3p\):
\(3p(4p^2 + 20p + 25)\)

Step4: Factor the quadratic inside the parentheses

The quadratic \(4p^2 + 20p + 25\) is a perfect square trinomial: \((2p + 5)^2\)
So the fully factored form is \(3p(2p + 5)^2\)

Since the original image has a typo, but assuming the more likely multiplication of two binomials (correcting the exponents and operation), the final factored form is as above. If it was just combining like terms, the answer is \(7p^2 + 6p + 15\).

Answer:

If combining like terms: \(7p^2 + 6p + 15\)

If multiplying (corrected problem: \((2p^2 + 5p)(6p + 15)\)): \(3p(2p + 5)^2\)