Question

solve algebraically for all values of x:
5x⁵ + 45x³ = 0
answer attempt 1 out of 2
○ additional solution ○ no solution
x =
√ ±

Explanation:

Step1: Factor out the GCF

The greatest common factor (GCF) of \(5x^5\) and \(45x^3\) is \(5x^3\). Factoring it out, we get:
$$5x^3(x^2 + 9)=0$$

Step2: Apply the zero - product property

The zero - product property states that if \(ab = 0\), then either \(a = 0\) or \(b = 0\). So we set each factor equal to zero:

• For \(5x^3=0\):

Divide both sides by 5: \(x^3 = 0\). Taking the cube root of both sides, we get \(x = 0\).

• For \(x^2+9 = 0\):

Subtract 9 from both sides: \(x^2=-9\). Taking the square root of both sides, we have \(x=\pm\sqrt{-9}=\pm 3i\) (where \(i\) is the imaginary unit with \(i^2=-1\)). But if we are looking for real - valued solutions, we consider only the real solution from the first factor. If we consider all complex solutions, the solutions are \(x = 0,x = 3i,x=-3i\). However, if we assume we are solving for real numbers (a common case in basic algebra problems unless stated otherwise), the real solution is \(x = 0\), and the complex solutions are \(x=\pm3i\). But since the problem says "solve algebraically for all values of \(x\)" (without specifying real or complex), we should present all solutions.

Answer:

The solutions are \(x = 0\), \(x = 3i\), and \(x=-3i\). If we consider only real solutions, the solution is \(x = 0\).