So, you had done everything right so far (other than squaring the 2), but that was only half of the question.
to find the least common multiple, you need to first figure out what the prime factors have in common.

each have two twos. both have one 5, so we know our answer will look something like

now to figure out the other stuff... we have to represent the greatest amount of everything that is left, and we have 3s and 7s left over, so we need to figure out how many of each we need.
one has one 3 and one has two, so we need two threes. now our equation is

what's the only number we have to deal with? 7...
how many sevens does 60 have? 0, and 630 has 1, so we know we need one 7. our answer becomes
Answer:
(f - g) (x)
Step-by-step explanation:
Since (x) is common between them, you can bring it out. for example,
f(x) = x +1
g(x) = 2x + 3
f(x) - g(x) = x + 1 - 2x + 3
= -x + 4
(f - g) (x) = x + 1 - 2x + 3
= -x + 4
keeping in mind that in a rhombus all sides are equal, Check the picture below.
Answer:
4x²+13x+9
iwill give you good method to simplify this
firstable take the 4 from x² and multiply it by 9
x²+13x+36
we find this equation
so the values is 4 and 9 because 4+9=13
4×9=36
(x+ )×(x+ )
first we take 9 and devide it by 4
9/4 the four for the x and the nine is number
so its will be
(4x+9)×(x+ )
same thing with 4 devide by 4 so the value is 1
so its will be like this
(4x+9)×(x+1) and this is the answer hopefully you understand my explain with my English