Question

sec 5.3 graphs of polynomial functions: problem 7
(1 point)
find the zeros and give the multiplicity of each for ( f(x)=x^{6}+x^{5}-6 x^{4} ).
note: to be counted as correct, you must get all answers correct.

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

note: you can earn partial credit on this problem.
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Explanation:

Step1: Factor out the common term

Factor out \(x^{4}\) from \(f(x)=x^{6}+x^{5}-6x^{4}\).

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Step2: Factor the quadratic expression

Factor \(x^{2}+x - 6\) using the formula \(ax^{2}+bx + c=a(x - x_{1})(x - x_{2})\), where \(x_{1,2}=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c = 0\). For \(x^{2}+x - 6\) (\(a = 1\), \(b=1\), \(c=-6\)), \(x=\frac{-1\pm\sqrt{1^{2}-4\times1\times(-6)}}{2\times1}=\frac{-1\pm\sqrt{25}}{2}=\frac{-1\pm5}{2}\). The roots are \(x = 2\) and \(x=-3\). So \(x^{2}+x - 6=(x - 2)(x + 3)\). Then \(f(x)=x^{4}(x - 2)(x + 3)\).

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

Answer:

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