We need to find two numbers that multiply to 24 (last coefficient) and add to 10 (middle coefficient). Through trial and error, the two values are 6 and 4
6 + 4 = 10
6*4 = 24
So we can break up the 10ab into 6ab+4ab and then use factor by grouping
a^2 + 10ab + 24b^2
a^2 + 6ab + 4ab + 24b^2
(a^2+6ab) + (4ab+24b^2)
a(a+6b) + 4b(a+6b)
(a+4b)(a+6b)
Therefore, the original expression factors completely to (a+4b)(a+6b)
Answer:

Step-by-step explanation:
Given:
The equation is given as:

Therefore, the possible values of
are 2 and 
<u>ΔACB</u> <u>ΔCDA</u>
AC² + BC² = AB² AD² + CD² = AC²
BC² = AB² - AC² BC² + CD² = AC² (AD=BC is given)
BC² = AC² - CD²
AB² - AC² = AC² - CD² (both sides were = to BC²)
AB² + CD² = 2AC²
(3)² + (√2)² = 2AC² (AB=3 and CD=√2 were given)
9 + 2 = 2AC²
11 = 2AC²
= AC²
= AC
= AC