Answer:
17: see below.
18: x = 2; x = -2; x = 3i; x = - 3i
19: x = sqrt(3); x= - sqrt(3); x = 1 ; x = - 1
Explanation:
17
Short answer No without doing anything. The 61 must be factorable using 6 as a factor. 61 is prime. But that's not really what you want to know although it is the answer. You look like you are doing synthetic division, so we'll answer it that way.
6 || 1 - 7 - 3 61
6 -6 -54
====================
1 -1 -9 7
The answer is x^2 - x - 9 with remainder of 7 / (x - 6)
18
x^4 + 5x^2 - 36 = 0
Substitute z for x^2
z^2 + 5z - 36 = 0 This factors.
(z + 9) (z - 4) = 0 Put the x^2 into the factors.
x^2 - 4 = 0 Add 4 to both sides
x^2 = 4 Take the square root of both sides.
x = +/- 2
===============
x^2 + 9 = 0
x^2 = - 9 Take the square root of both sides
sqrt(x^2) = sqrt(-9)
x = +/- 3i
19
19 is done exactly as 18 was done. You should get 4 real roots.
The graph is below. Very briefly it is going to factor as
(x^2 - 3)(x^2 - 1)
The results are x = sqrt(3); x= - sqrt(3); x = 1 ; x = - 1