Solve the polynomial equation by finding all real roots
g(x) = x^4 + 15x^2 -16
SO:
(X^2+16) (X-1)....I'M STUCK
THANKS
you factored incorrectly, expand your answer, you won't get the original
x^4 + 15x^2 -16 = 0
(x-1)(x^3 + x^2 + 16x + 16) = 0
so x = 1
or
x^3 + x^2 + 16x + 16 = 0
by grouping
x^2(x+1) + 16(x+1) = 0
(x+1)(x^2 + 16) = 0
so we have a double root at x=1
and
x^2 = -16
x = ± √-16 = ± 4i , but those are not real, so
all you have in real roots is
x = 1
To solve the polynomial equation g(x) = x^4 + 15x^2 - 16, we can try factoring the equation and then solving for the roots.
You correctly factored x^4 + 15x^2 - 16 into (x^2 + 16)(x - 1). Now, we can set each factor equal to zero and solve for x to find the real roots.
Setting x^2 + 16 = 0:
x^2 = -16
Taking the square root of both sides:
x = ±√(-16)
Since we are looking for real roots, we know that the square root of a negative number is not a real number. Therefore, there are no real roots from this factor.
Next, setting x - 1 = 0:
x = 1
So, the only real root of the polynomial equation g(x) = x^4 + 15x^2 - 16 is x = 1.
Therefore, the solution to the equation g(x) = 0 is x = 1.