Question

1. if a prime number p is a factor of both (14n + 13) and (7n + 1), what is the value of p?

a 7
b 11
c 13
d it cannot be determined from the information given.

1. what is the third term in the binomial expansion of (x + 2y)^5?

a 10x^2y^3
b 10x^3y^2
c 40x^3y^2
d 80x^2y^3

1. 2√5(√2 + √5) =

Explanation:

Question 46

Step1: Use the property of factors

If \( p \) is a factor of both \( 14n + 13 \) and \( 7n + 1 \), then \( p \) is also a factor of any linear combination of them. Let's consider \( 14n + 13 - 2\times(7n + 1) \).
\[

$$\begin{align*} 14n + 13 - 2\times(7n + 1)&=14n + 13 - 14n - 2\\ &= 11 \end{align*}$$

\]

Step2: Determine the prime factor

Since \( p \) is a prime factor of \( 11 \) (and \( 11 \) is prime), the value of \( p \) must be \( 11 \).

Step1: Recall the binomial theorem

The binomial expansion of \( (a + b)^n \) is given by \( \sum_{k = 0}^{n} \binom{n}{k} a^{n - k}b^{k} \), where \( \binom{n}{k}=\frac{n!}{k!(n - k)!} \). For the third term, \( k = 2 \) (since we start from \( k = 0 \)).

Step2: Calculate the binomial coefficient and the term

For \( (x + 2y)^5 \), \( n = 5 \), \( a=x \), \( b = 2y \), and \( k = 2 \).
First, calculate \( \binom{5}{2}=\frac{5!}{2!(5 - 2)!}=\frac{5\times4}{2\times1}=10 \).
Then, the term is \( \binom{5}{2}x^{5 - 2}(2y)^{2}=10\times x^{3}\times4y^{2}=40x^{3}y^{2} \).

Step1: Distribute the multiplication

Using the distributive property \( a(b + c)=ab+ac \), we have \( 2\sqrt{5}(\sqrt{2}+\sqrt{5})=2\sqrt{5}\times\sqrt{2}+2\sqrt{5}\times\sqrt{5} \).

Step2: Simplify each term

For the first term: \( 2\sqrt{5}\times\sqrt{2}=2\sqrt{10} \).
For the second term: \( 2\sqrt{5}\times\sqrt{5}=2\times5 = 10 \).
So the result is \( 2\sqrt{10}+10 \) (assuming the option was likely \( 2\sqrt{10}+10 \), though the image cut off the options, but the calculation is as above).

Answer:

B. 11

Question 47