Question 170657


Finding the Degree:


{{{x(x+4)(x-3)}}} Start with the given expression



{{{x(x^2+x-12)}}} FOIL {{{(x+4)(x-3)}}} to get {{{x^2+x-12}}}



{{{x^3+x^2-12x}}} Distribute



Notice how the largest exponent is 3. So this means that the degree of {{{x^3+x^2-12x}}}(and {{{x(x+4)(x-3)}}}) is 3



So the degree is 3.


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Finding the zeros:



{{{x(x+4)(x-3)}}} Start with the given expression



{{{x(x+4)(x-3)=0}}} Set the given expression equal to zero




Now set each factor equal to zero:



{{{x=0}}}, {{{x+4=0}}} or {{{x-3=0}}}



Now solve for x for each factor:



{{{x=0}}}, {{{x=-4}}} or {{{x=3}}}



So the zeros of {{{x(x+4)(x-3)}}} are {{{x=0}}}, {{{x=-4}}} or {{{x=3}}}