Two roots of the equation x^3+px+q are -1and 3.The third root is?
let f(x) = x^3 + px + q
-1 is a root, so
f(-1) = 0 = -1 -p+ q or
-p + q = 1
p - q = -1
also f(3) = 0
27 + 3p + q = 0
3p + q = -27
solve for p and q, I would add them to get p,
then you can find q
sub back into the original and then factor to get the last root
or
since -1 and 3 are roots, let the third root be a
(x+1)(x-3)(x-a) = x^3 + px + q
(x-a)(x^2 - 2x - 3) = x^3 + px + q
x^3 - 2x^2 - 3x - ax^2 + 2ax + 3a = x^3 + px + q
x^2(-2 - a) + x(-3 + 2a) + (3a) = px + q
since there was no x^2 term, -2-a = 0, a = -2
alos -3 + 2a = p
p = -3 - 4 = -7
q = 3a = -6
we would have x^3 - 7x - 6
check: What is (x+1)(x-3)(x+2)
= (x+2)(x^2 - 2x - 3)
= x^3 + 2x^2 - 3x - 2x^2 - 4x - 6
= x^3 - 7x - 6
as expected
To find the third root of the equation, we need to use the fact that the sum of the roots of a cubic equation is zero.
Given that -1 and 3 are two roots of the equation x^3 + px + q, we can write:
(-1) + 3 + r = 0,
where 'r' represents the third root that we are trying to find.
Simplifying the equation, we get:
2 + r = 0.
To solve for 'r', we can subtract 2 from both sides of the equation:
r = -2.
Therefore, the third root of the equation x^3 + px + q is -2.