In this case, you have to find a common multiple of 3a^2 and 4ab. As only one of the numbers is divisible by two, this means that two cannot go outside of the bracket. The only other aspect that is in both sides of the equation is a, therefore, this goes outside the bracket. The best way to approach this is to divide both sides by a, and this will give you what is inside the bracket. 3a^2 divided by a is 3a, therefore, this is the first aspect in the bracket. -4ab divided by a, leaves -4b. Therefore, these go in the bracket.
3a^2- 4ab simplified is a(3a-4b)
Hope this helps
Answer:
(y + 14)(y - 14)
Step-by-step explanation:
A difference of squares has the general form
a³ - b² and factors as
(a + b)(a - b)
Given
y² - 196
y² = (y)² ⇒ a = y and 196 = 14² ⇒ b = 14
y² - 196
= y² - 14² = (y + 14)(y - 14) ← difference of squares
Let's check the dot product .
v.w = 5(4)-2(10) = 20 -20 =0
So the dot product is not 20 . And since the dot product is zero, so they are perpendicular.
And the y component of w is 10.
So correct options are first and third .
