Question

f(x) = x⁴ + 2x² - 8

Explanation:

Step1: Let \( t = x^2 \)

Since \( x^4=(x^2)^2 \), we can substitute \( t = x^2 \) (where \( t\geq0 \)) into the function. Then the function \( f(x)=x^4 + 2x^2-8 \) becomes a quadratic function in terms of \( t \): \( f(t)=t^2 + 2t - 8 \).

Step2: Factor the quadratic function

To factor \( t^2 + 2t - 8 \), we need to find two numbers that multiply to \( - 8 \) and add up to \( 2 \). The numbers are \( 4 \) and \( - 2 \) because \( 4\times(-2)=-8 \) and \( 4+(-2)=2 \). So, \( t^2 + 2t - 8=(t + 4)(t - 2) \).

Step3: Substitute back \( t = x^2 \)

Substituting \( t = x^2 \) back into the factored form, we get \( f(x)=(x^2 + 4)(x^2 - 2) \).

Step4: Factor \( x^2 - 2 \) (using difference of squares)

The difference of squares formula is \( a^2 - b^2=(a + b)(a - b) \). For \( x^2 - 2 \), we can write it as \( x^2-(\sqrt{2})^2 \), so \( x^2 - 2=(x+\sqrt{2})(x - \sqrt{2}) \). The factor \( x^2 + 4 \) cannot be factored over the real numbers (since the discriminant of \( x^2 + 4 = 0 \) is \( 0^2-4\times1\times4=-16<0 \)). So the complete factorization of \( f(x) \) over the real numbers is \( f(x)=(x^2 + 4)(x+\sqrt{2})(x - \sqrt{2}) \). If we consider complex numbers, we can further factor \( x^2 + 4 \) as \( (x + 2i)(x - 2i) \) (using the difference of squares for complex numbers: \( a^2 + b^2=(a+bi)(a - bi) \) where \( i^2=-1 \)), so over the complex numbers, \( f(x)=(x + 2i)(x - 2i)(x+\sqrt{2})(x - \sqrt{2}) \).

Answer:

If factoring over the real numbers: \( \boldsymbol{(x^2 + 4)(x+\sqrt{2})(x - \sqrt{2})} \)

If factoring over the complex numbers: \( \boldsymbol{(x + 2i)(x - 2i)(x+\sqrt{2})(x - \sqrt{2})} \)