The sum of all integral values of \(x\) for which \(\frac{18({x}^{2}-7x+10)}{{x}^{3}-6{x}^{2}+3x+10}\) has an integral value is...
Thanks in advance for the help everyone!
+532 Use synthetic division twice on the two polynomials to get the whole equation as 18/(x+1).
the roots of the denominator of the original cubic are -1, 2, and 5 so those will not work. the remaining ones are the the positive and negative factors of 18, subtracted by 1 so you get -19, -10, -5, -3, -2, -1(extraneous), 0, 1, 2(extraneous), 5(extraneous), 8, and 17. adding up the ones that work, you get
-19 - 10 - 5 - 3 - 2 + 0 + 1 + 8 + 17. with a little mental math, you get -13.
HOPE THIS HELPED!
Thanks for the help!
But I have some questions. What is synthetic division? And what do you mean by "the roots of the denominator of the original cubic are -1, 2, and 5". I think I know what you mean by that but the wording is off.
I'm not as good at math as you lol
+532 i mean x^3-6x^2+3x+10
and synthetic division is just a quicker way of doing long division, its almost the same thing
+129852 Here's my best attempt....
The top factors as 18 (x - 5) ( x - 2)
The bottom factors as (x + 1) ( x - 5) ( x - 2)
So we are left with 18 / (x + 1)
The divisors of 18 are ± (1, 2, 3, 6, 9, 18)
This means that we have integer values when x = -2 , -3 -4, -7, -10, -19, 0, 1, 5, 8, 17
Note that we cannot use x = -1 or x = 2 because they make an original denominator = 0
So the sum of these integer values are - 45 + 31 = - 14
