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Chapter 2: Problem 14

Determine the number of zeros of the polynomial function. $$h(t)=(t-1)^{2}-(t+1)^{2}$$

Short Answer

Expert verified
The number of zeros of the polynomial function \(h(t)=(t-1)^{2}-(t+1)^{2}\) is one, which is \(t=0\).

Step by step solution

01

Simplify the Polynomial Function

We shall start the problem by simplifying the given polynomial function \(h(t)=(t-1)^{2}-(t+1)^{2}\). We can do this by expanding it using the formula \(a^{2} - b^{2} = (a - b)(a + b)\). Substituting \(a = t-1\) and \(b = t+1\) into the formula, we then get \(h(t) = (2)(2t)\). Further simplification gives \(h(t)=4t\).
02

Setting the Simplified Polynomial Function to Zero

Next, to find the zeros of the polynomial, we set the simplified polynomial function, \(h(t)\), to zero and solve for \(t\). This gives us the equation \(4t=0\).
03

Solve for t

To isolate \(t\) on one side, we divide both sides of the equation by 4. This gives us the solution \(t=0\).

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