Cd = 1/2 * 1/3
= 1/6
...............
Answer:
There's no c in your uploaded question
Your function is
. The fundamental theorem of algebra says that there will be three roots, since the degree of the polynomial is 3. The problem provides two real roots, x = -2 and x = 3, so there must be one more.
The theorem also says that possible roots of the polynomial would be in this case, the factors of the constant (-6) over the factors of the coefficient of the term with the highest degree (1).
Factors of -6 are: 1, 2, 3, 6, -1, -2, -3, -6
Factors of 1 are: 1, -1
Possible rational roots are: 1, 2, 3, 6, -1, -2, -3, -6
I then use synthetic division to see which possible rational root is a real root by dividing
by the possible rational roots, and I get a root when the remainder is 0. Remember to put the placeholder of 0 for x^2 when dividing:
-1} 1 0 -7 -6
-1 1 6
-----------------
1 -1 -6 0
When I divide by the possible rational root of -1, I get a remainder of 0, which means -1 is a root.
To check:
(x + 2)(x - 3)(x + 1)
= (x^2 - x - 6)(x + 1)
= x^3 - x^2 - 6x + x^2 - x - 6
= x^3 - 7x - 6
2x-25=x+5 move your “x” on the right side to the left side but to move it you need to switch the sign to cancel it out on the right. That ends up giving you x-25=5. You want X by itself so you move over your “25” but also change the sign. Giving you x=30
To find your y you set the equation to 0. 9y+28=0. You want Y by itself, so you want to move the 28 over first. Switch the sign which gives you 9y=-28. To get y alone divide by 9. 28/9 = 3.1(repeating 1)