Factor the polynomial as the product of factors that are irreducible over the real numbers. Then write the polynomial in completely factored form involving complex nonreal or imaginary numbers.
x^4 + 20x^2 -44=0
(x^2 + 22)(x^2 - 2)= 0
(x + √22i)(x - √22i)(x+√2)(x-√2) = 0
x = ± √22i or x = ± √2
To factor the polynomial x^4 + 20x^2 - 44 = 0, we can start by letting u = x^2. This allows us to rewrite the equation as u^2 + 20u - 44 = 0.
Next, we can solve this quadratic equation for u. We can either use the quadratic formula or factor the quadratic expression. Let's use factoring:
(u + 22)(u - 2) = 0
Now, we set each factor equal to zero and solve for u:
u + 22 = 0 or u - 2 = 0
Solving these equations, we get:
u = -22 or u = 2
Since u = x^2, we can substitute these values back into the equation to solve for x:
x^2 = -22 or x^2 = 2
For x^2 = -22, there are no real solutions because the square of a real number cannot be negative. However, we can introduce the imaginary unit i to represent the square root of -1. Therefore, we have:
x = ± √(-22)i
For x^2 = 2, we can take the square root of both sides to find the real solutions:
x = ± √2
So the completely factored form of the polynomial x^4 + 20x^2 - 44 = 0 involving complex nonreal or imaginary numbers is:
x = ± √2 or x = ± √(-22)i
To factor the given polynomial as the product of irreducible factors over the real numbers, we can start by substituting a variable to simplify the polynomial. Let's use the substitution \(u = x^2\). Rewriting the polynomial with this substitution, we have:
\(u^2 + 20u - 44 = 0\)
Now, we can solve this quadratic equation for \(u\) by applying the quadratic formula: \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 20\), and \(c = -44\). Simplifying the equation:
\(u = \frac{-20 \pm \sqrt{20^2 - 4(1)(-44)}}{2(1)}\)
\(u = \frac{-20 \pm \sqrt{400 + 176}}{2}\)
\(u = \frac{-20 \pm \sqrt{576}}{2}\)
\(u = \frac{-20 \pm 24}{2}\)
We have two possible solutions for \(u\):
1. When \(u = \frac{-20 + 24}{2} = 2\)
2. When \(u = \frac{-20 - 24}{2} = -22\)
Now, let's substitute back \(x^2\) for \(u\) in each solution to determine the values of \(x\):
1. When \(x^2 = 2\), taking the square root of both sides: \(x = \pm \sqrt{2}\)
2. When \(x^2 = -22\), taking the square root of both sides is not possible since the square root of a negative number is not defined in real numbers. Therefore, we have to consider complex nonreal or imaginary numbers.
To express the polynomial in completely factored form involving complex nonreal or imaginary numbers, we can use the solutions we found:
The factored form of the polynomial is:
\((x - \sqrt{2})(x + \sqrt{2})(x - \sqrt{-22})(x + \sqrt{-22}) = 0\)
Simplifying further, we can write it as:
\((x - \sqrt{2})(x + \sqrt{2})(x - \sqrt{-1} \sqrt{22})(x + \sqrt{-1} \sqrt{22}) = 0\)
Finally, using the imaginary unit \(i\), we can express the polynomial as:
\((x - \sqrt{2})(x + \sqrt{2})(x - i \sqrt{22})(x + i \sqrt{22}) = 0\)
This is the completely factored form of the given polynomial involving complex nonreal or imaginary numbers.