Question

I finished question 1 but I could use some help on the others​

Answer 1

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:

$9x + 3 = 0\\3x + 1 = 0\\3x = -1\\x = -1/3$

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:

$p(x) = (9x + 3)(x^2 - 9)\\= (9x + 3)(x + 3)(x - 3)$

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.