is anyone willing to help with some exam questions?

Hi, i am really struggling with some maths questions. I dont wish to be ridiculed or mocked because i really am struggling and mocking me wont help matters.

As @braindeadpog has said, you can approach this question by multiplying out all the brackets, grouping the resulting terms into a single quadratic and then showing that the quadratic has two identical factors. But there is another way. You could note that 3x + 5 = 3x + 2 + 3, meaning that the x(3x + 5) term can be rewritten as x(3x + 2) + 3x. The complete expression should then consist of three terms, all of which have the factor (3x + 2). This give you another way forward.

Expand all the brackets and simplify (remember foil) and you will be left with a quadratic expression. You should be able to factorise this quadratic, (using ac method since the coefficient of x^2 is bigger than 1). The factorised form can be written as a square, hence showing it is a perfect square.

Alternatively, expand all brackets and simplify. Then factorise 9 from the x^2 and x terms, and complete the square. After simplifying the terms, you will note it to be in the form of 9(x+c)^2. Since 9 = 3^2. Think about what happens when two squares are multiplied together ( e.g. m^2 x n^2, can be writtern as m x m x n x n , or mn x mn, which simplifiers to mn ^2)

In terms on how to improve in such questions, revisit quadratic factorisation, expanding and simplifying brackets + algebraic proofs

any ideas??

work through the (2x+1)(3x+2) and then work through the x(3x+5) and then you'll be left with terms which you can collect and add together
then with the perfect square uhh lets say you have an equation like this:
ax^2 + bx + c
for it to be a perfect square half of b squared should be c
so one example would be:
x^2 + 6x + 9
(half of 6 is 3 and 3 squared is 9 which is in fact the third number and so its a perfect square)
i mean that's the easy way to explain it and i think you'll get a mark for proving it like that but if you want to get all technical then whatever values you have should satisfy b^2 = 4ac

What are yours? We cannot do questions for you as it's against the rules/

i am not asking for ANSWERS. i am asking for help. smhhhh

i got this way of doing it
6x^2+7x+2
ax^2+bx+c
a=6
b=7
c=2
to be a perf square should be b^2=4ac
7^2=4x6x2
which equals 48 so would it be equal or not?