Question

for questions 9-10, factor the trinomial. 9. $25a^2 + 120a + 144$ 10. $p^2 - 22p + 121$

Explanation:

Question 9: Factor \(25a^2 + 120a + 144\)

Step 1: Identify the form

The trinomial \(25a^2 + 120a + 144\) is a quadratic trinomial of the form \(Ax^2 + Bx + C\). We can check if it is a perfect square trinomial. A perfect square trinomial has the form \((mx + n)^2 = m^2x^2 + 2mnx + n^2\).

Step 2: Check coefficients

For \(25a^2\), the square root of \(25a^2\) is \(5a\) (since \((5a)^2 = 25a^2\)). For \(144\), the square root of \(144\) is \(12\) (since \(12^2 = 144\)). Now, check the middle term: \(2 \times 5a \times 12 = 120a\), which matches the middle term of the given trinomial.

Step 3: Write as a perfect square

Since \(25a^2 + 120a + 144\) fits the form \((mx + n)^2\) with \(m = 5\), \(x = a\), and \(n = 12\), we can factor it as \((5a + 12)^2\).

Step 1: Identify the form

The trinomial \(p^2 - 22p + 121\) is a quadratic trinomial of the form \(x^2 + Bx + C\). We check if it is a perfect square trinomial. A perfect square trinomial has the form \((x - n)^2 = x^2 - 2nx + n^2\) (since the middle term is negative).

Step 2: Check coefficients

For \(p^2\), the square root is \(p\) (since \(p^2 = p \times p\)). For \(121\), the square root is \(11\) (since \(11^2 = 121\)). Now, check the middle term: \(2 \times p \times 11 = 22p\), and since the middle term in the trinomial is \(-22p\), we have \( - 2 \times p \times 11=-22p\), which matches.

Step 3: Write as a perfect square

Since \(p^2 - 22p + 121\) fits the form \((x - n)^2\) with \(x = p\) and \(n = 11\), we can factor it as \((p - 11)^2\).

Answer:

\((5a + 12)^2\)

Question 10: Factor \(p^2 - 22p + 121\)