There's no if about it,

has a zero

so

is a factor. That's the special case of the Remainder Theorem; since

we'll get a remainder of zero when we divide

by

At this point we can just divide or we can try more little numbers in the function. It doesn't take too long to discover

too, so

is a factor too by the remainder theorem. I can find the third zero as well; but let's say that's out of range for most folks.
So far we have

where

is the zero we haven't guessed yet. Again we could divide

by

but just looking at the constant term we must have

so

We check

We usually talk about the zeros of a function and the roots of an equation; here we have a function

whose zeros are
Answer: not sure how helpful this is, but the second one is correct (if it's the one that's corresponding to the blue line). The pink line, however, is incorrect if it is corresponding to the first equation. The first equation must have a y-intercept of 1.
Step-by-step explanation:
The equation of a circle is (x - h)² + (y - k)² = radius² where (h, k) is the ordered pair for the center.
(x - (- 3))² + (y - 6)² = 10²
(x + 3)² + (y - 6) = 100
Answer:
look at the photo
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