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Chapter 2: Problem 65

Write the polynomial as the product of linear factors and list all the zeros of the function. $$h(x)=x^{2}-2 x+17$$

Short Answer

Expert verified
The given polynomial factored is \(h(x) = (x - (1 + 4i))(x - (1 - 4i))\). The zeros of the polynomial are \(x = 1 + 4i\) and \(x = 1 - 4i\).

Step by step solution

01

Factoring the quadratic polynomial

Because the quadratic \(h(x)=x^{2}-2 x+17\) has a discriminant \(b^2 - 4ac = (-2)^2 - 4*(1)*17 = -64\) which is less than 0, this polynomial cannot be factored using real numbers. It can, however, be factored using imaginary numbers.
02

Determine the zeros using the quadratic formula

The quadratic formula is given by: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]. Plugging in the values \(a=1\), \(b=-2\), and \(c=17\), we get the following zeros: \(x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4*(1)*17}}{2*1}\) => \(x = \frac{2 \pm \sqrt{4 - 68}}{2}\) => \(x = \frac{2 \pm \sqrt{-64}}{2}\) => \(x = \frac{2 \pm 8i}{2}\) => \(x = 1 \pm 4i\)
03

Express the polynomial as a product of linear factors

Using the zeros, we can express the original polynomial as a product of linear factors, keeping in mind that each zero \(x=a+bi\) gives a factor of \((x-(a+bi))\). Substituting the zeros we get: \(h(x) = (x - (1 + 4i))(x - (1 - 4i))\)

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