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Consider the function h(x) = x^3 + 2x^2 - 15x.What are the zeros of h(x)?Smallest zero:Middle zero:Largest zero:

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We are given


h(x)=x^3+2x^2-15x

We want to find the zeros of h(x)

Solution

Finding the zeros of h(x) means that we should equate h(x) to zero and solve


\begin{gathered} i\mathrm{}e \\ h(x)=0 \end{gathered}

We now solve for h(x) = 0


\begin{gathered} h(x)=0 \\ h(x)=x^3+2x^2-15x=0 \\ x^3+2x^2-15x=0 \\ x(x^2+2x-15)=0 \end{gathered}

We also factorize the quadratic as well


\begin{gathered} x^2+2x-15 \\ x^2+5x-3x-15 \\ (x^2+5x)-(3x+15) \\ x(x+5)-3(x+5) \\ (x+5)(x-3) \end{gathered}

We come back to


\begin{gathered} x(x^2+2x-15)=0 \\ x(x+5)(x-3)=0 \\ \text{therefore,} \\ x=0 \\ or \\ x+5=0 \\ x=-5 \\ or \\ x-3=0 \\ x=3 \end{gathered}

Thus,

The zeros are x = 0, x = -5 and x = 3

Smallest zero = -5

Middle zero = 0

Largest zero = 3

User Khayam Khan
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