Question

find the zeros and give the multiplicity of each for ( f(x)=x^{6}+x^{5}-6x^{4} ).
note: to be counted as correct, you must get all answers correct.

• one zero is ( x=quad ) and has a multiplicity of help (numbers)
• another zero is ( x=quad ) and has a multiplicity of: help (numbers)
• the last zero is ( x=quad ) and has a multiplicity of: help (numbers)

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Explanation:

Step1: Factor out the common term

Factor \(x^{4}\) from \(f(x)=x^{6}+x^{5}-6x^{4}\), we get \(f(x)=x^{4}(x^{2}+x - 6)\).

Step2: Factor the quadratic expression

Factor \(x^{2}+x - 6=(x + 3)(x - 2)\). So \(f(x)=x^{4}(x + 3)(x - 2)\).

Step3: Find the zeros and their multiplicities

Set \(f(x)=0\).

• For \(x^{4}=0\), \(x = 0\) and the multiplicity is \(4\) (since the exponent of \(x\) is \(4\)).
• For \(x+3=0\), \(x=-3\) and the multiplicity is \(1\) (since the exponent of \((x + 3)\) is \(1\)).
• For \(x - 2=0\), \(x = 2\) and the multiplicity is \(1\) (since the exponent of \((x - 2)\) is \(1\)).

Answer:

One zero is \(x = 0\) and has a multiplicity of \(4\).
Another zero is \(x=-3\) and has a multiplicity of \(1\).
The last zero is \(x = 2\) and has a multiplicity of \(1\).