Question 1206674
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Since the leading coefficient is 1, we list the factors of the last term to generate all possible rational roots.


List of possible rational roots:
1, -1, 5, -5, 7, -7, 35, -35
List the positive and negative versions of each.


Then plug each of them into the function to see which generates the result of 0.
Let's try x = 1
f(x) = x^3 - 5x^2 + 7x - 35
f(1) = (1)^3 - 5(1)^2 + 7(1) - 35
f(1) = -32
The nonzero result tells us that x = 1 is not a root.
Every other value but x = 5 will also generate nonzero results. 


f(x) = x^3 - 5x^2 + 7x - 35
f(5) = (5)^3 - 5(5)^2 + 7(5) - 35
f(5) = 0
Proving that x = 5 is a root and x-5 is a factor.


Let's apply synthetic division with this root.
<table border = "1" cellpadding = "5"><tr><td>5</td><td></td><td>1</td><td>-5</td><td>7</td><td>-35</td></tr><tr><td></td><td></td><td></td><td>5</td><td>0</td><td>35</td></tr><tr><td></td><td></td><td>1</td><td>0</td><td>7</td><td>0</td></tr></table>
The last item in the bottom row is 0 which is the remainder. The remainder 0 confirms that x = 5 is indeed a root.
The other items in the bottom row form the quotient.


Quotient = 1x^2 + 0x + 7 = x^2 + 7


We determine x^3-5x^2+7x-35 = (x-5)(x^2+7)


Solving x^2+7 = 0 leads to x = i*sqrt(7) and x = -i*sqrt(7) where i = sqrt(-1)


The three roots are
<font color=red>x = 5, x = i*sqrt(7) and x = -i*sqrt(7)</font>
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