If a polynomial function f(x) has roots 0, 4, and 3 + √11 , what must also be a root of f(x)?
Solution:
If you want rational coefficients then you should get the conjugate of any irrational zero.
The conjugate of 3 + √11 is 3 - √11
Explanation:
x = 3 + √11
Subtract 3 on both sides
x - 3 = 3 + √11 - 3
x - 3 = √11
By squaring on both sides
(x - 3)2 = 11
Now subtract 11 on both sides
(x - 3)2 - 11 = 0
To factor use the difference of squares
u2 - v2 = (u - v) (u + v)
[(x - 3) - 11][(x - 3) + 11] = 0
We get,
(x - 3) - 1 = 0 or (x - 3) + 11 = 0
Solve for x - 3 and x
Add √11 on both sides of first equation and subtract √11 on both sides of second equation
x - 3 = √11 or x - 3 = - √11
By adding 3 on both sides
x = 3 + √11 or x = 3 - √11
Therefore, 3 - √11 must also be a root of f (x).
If a polynomial function f(x) has roots 0, 4, and 3 + √11 , what must also be a root of f(x)?
Summary:
If a polynomial function f(x) has roots 0, 4, and 3 + √11 , 3 - √11 is also a root of f(x).
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