Question

solve:

1. ( x^4 - 8x^3 + 16x^2 = 0 )

Explanation:

Step1: Factor out \(x^2\)

We start by factoring out the greatest common factor from the left - hand side of the equation \(x^{4}-8x^{3}+16x^{2} = 0\). The greatest common factor of \(x^{4}\), \(-8x^{3}\) and \(16x^{2}\) is \(x^{2}\). So we have:
\(x^{2}(x^{2}-8x + 16)=0\)

Step2: Factor the quadratic

Now we factor the quadratic expression \(x^{2}-8x + 16\). We know that for a quadratic of the form \(ax^{2}+bx + c\) (here \(a = 1\), \(b=-8\), \(c = 16\)), we need to find two numbers that multiply to \(ac=1\times16 = 16\) and add up to \(b=-8\). The numbers are \(-4\) and \(-4\) since \((-4)\times(-4)=16\) and \(-4+( - 4)=-8\). So we can factor \(x^{2}-8x + 16\) as \((x - 4)^{2}\).
So the equation becomes \(x^{2}(x - 4)^{2}=0\)

Step3: Solve for \(x\)

Using the zero - product property, which states that if \(ab = 0\), then either \(a = 0\) or \(b = 0\) (or both).
If \(x^{2}=0\), then taking the square root of both sides, we get \(x=0\) (with multiplicity 2).
If \((x - 4)^{2}=0\), then taking the square root of both sides, we have \(x-4 = 0\), so \(x = 4\) (with multiplicity 2).

Answer:

The solutions of the equation \(x^{4}-8x^{3}+16x^{2}=0\) are \(x = 0\) (multiplicity 2) and \(x = 4\) (multiplicity 2), or in set notation \(\{0,4\}\) (where we can also note the multiplicities as needed).