find all the other zeroes of the polynomial x^4+4x^3-4x^2-32x-32 if two of its zeroes are 2 root 2 and -2 root 2
If x = 2√2 and x = -2√2
then two of the factors are
(x - 2√2) and (x + 2√2)
or (x^2 - 8) must be a factor of your polynomial.
Do a long algebraic division to get
x^4+4x^3-4x^2-32x-32
= (x^2 - 8)(x^2 - 4x + 4)
we can get more roots from
x^2 - 4x + 4 = 0
(x-2)(x-2) = 0 ---> a perfect square
so x = 2
so the roots are 2 , ± 2√2
Thank you
To find the other zeros of the polynomial, we can make use of the fact that if a polynomial has complex or irrational zeros, they will always occur in conjugate pairs.
Given that two of the zeros are 2√2 and -2√2, we can deduce that the other two zeros must be their conjugates, which are -2√2 and 2√2.
So, the other two zeros of the polynomial x^4 + 4x^3 - 4x^2 - 32x - 32 are -2√2 and 2√2.
To find the other zeroes of the polynomial x^4 + 4x^3 - 4x^2 - 32x - 32, given that two of its zeroes are 2√2 and -2√2, we can use the fact that the sum and product of the zeroes of a polynomial are related to its coefficients.
Let's denote the remaining two zeroes as α and β.
We know that the sum of the zeroes is given by:
α + β = -(coefficient of x^3) / (coefficient of x^4)
In this case, the coefficient of x^3 is 4, and the coefficient of x^4 is 1. So:
α + β = -4 / 1 = -4
We also know that the product of the zeroes is given by:
α * β = constant term / (coefficient of x^4)
In this case, the constant term is -32, and the coefficient of x^4 is 1. So:
α * β = -32 / 1 = -32
Now we have a system of equations:
α + β = -4
α * β = -32
To solve this system, we can make use of the fact that the quadratic equation with roots α and β is:
x^2 - (sum of roots)x + product of roots = 0
Substituting the values into this equation, we get:
x^2 - (-4)x - 32 = 0
x^2 + 4x - 32 = 0
Now we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = 4, and c = -32. Substituting these values, we have:
x = (-4 ± √(4^2 - 4(1)(-32))) / (2(1))
x = (-4 ± √(16 + 128)) / 2
x = (-4 ± √144) / 2
x = (-4 ± 12) / 2
This gives us two possible values for x:
x1 = (-4 + 12) / 2 = 8 / 2 = 4
x2 = (-4 - 12) / 2 = -16 / 2 = -8
Therefore, the other two zeroes of the polynomial x^4 + 4x^3 - 4x^2 - 32x - 32 are 4 and -8.