Answer:
With the first four questions, all you need to do is find a value of x that gives you a zero value in any one of those brackets. Let's do it.
You said you finished 1, so let's skip it.
I can't give you a full answer for question 2, because you didn't capture the entire expression in the screenshot. I can see that the values 3 and 0.125 would give zero values, but I can't tell you the last.
For question three, we need to find x values where either 9x + 3 = 0, or x² - 9 = 0. There are three values that meet those conditions. You can solve it just by assigning a zero value and solving for x:
For the second term actually has two solutions, because a squaring a number also causes negatives to be converted to positives. Another way of seeing this is to notice that the second term is actually a difference of squares:
so in this case x could be -1/3, -3, or 3 to give a zero value.
For question four, you can only get a zero if you use imaginary numbers, as no real number can be squared to give you a negative. So the only answers to that are 5i, or -5i.
With questions five six seven and eight, we just need to make exceptions for zero denominators, so:
For question 5, you can x can be any real number except -4, 2 and 7
For 6, x can be anything but 3, 1/8th or 5
for 7, We again have that difference of squares, so the actual denominator is (9x + 3)(x + 9)(x - 9), meaning x can be anything but -1/3, -9, or 9
And finally for question 8, we again have a squaring, meaning that only an imaginary number will cause division by zero, meaning x can't be 5i or -5i.
