Question

question
factor to find all x-intercepts of the function.
$f(x) = x^4 + 8x$
answer attempt 1 out of 2
additional solution no solution
$x = $
submit answer

Explanation:

Step1: Factor out the common term

To find the x - intercepts, we first set \(f(x) = 0\), so we have the equation \(x^{4}+8x = 0\). The common factor of \(x^{4}\) and \(8x\) is \(x\). Factoring out \(x\), we get \(x(x^{3}+8)=0\).

Step2: Recognize the sum of cubes

We know that the sum of cubes formula is \(a^{3}+b^{3}=(a + b)(a^{2}-ab + b^{2})\). In the expression \(x^{3}+8\), we can rewrite \(8\) as \(2^{3}\), so \(x^{3}+8=x^{3}+2^{3}\). Using the sum of cubes formula with \(a = x\) and \(b = 2\), we have \(x^{3}+2^{3}=(x + 2)(x^{2}-2x + 4)\). So the factored form of the equation \(x^{4}+8x = 0\) is \(x(x + 2)(x^{2}-2x + 4)=0\).

Step3: Solve for x

We set each factor equal to zero:

• For the first factor \(x=0\), we get \(x = 0\).
• For the second factor \(x + 2=0\), we solve for \(x\) and get \(x=-2\).
• For the third factor \(x^{2}-2x + 4=0\), we use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) where \(a = 1\), \(b=-2\) and \(c = 4\). The discriminant \(\Delta=b^{2}-4ac=(-2)^{2}-4\times1\times4=4 - 16=-12<0\). Since the discriminant is negative, the quadratic equation \(x^{2}-2x + 4 = 0\) has no real solutions.

Answer:

The x - intercepts are \(x = 0\) and \(x=-2\). If we consider the answer box (maybe for one of them, but from the factoring we have two real x - intercepts. If we take the non - quadratic factor solutions, the real x - intercepts are \(x = 0\) and \(x=-2\). If we assume the answer box is for one, but based on the factoring process, the real x - intercepts are \(x = 0\) and \(x=-2\).