Question

1.6: homework - polynomial and rational functions
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question 3
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use your graphing calculator to solve the equation graphically for all real solutions
$x^3 - 4x^2 - 3x + 14 = 0$
solutions: $x = $
make sure your answers are accurate to at least two decimals
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Explanation:

Step1: Define the function

Let \( f(x)=x^{3}-4x^{2}-3x + 14\). We need to find the real roots of the equation \(f(x) = 0\) by graphing the function \(y = f(x)\) and finding the \(x\)-intercepts.

Step2: Analyze the function (optional for graphing)

We can also try to find some rational roots using the Rational Root Theorem. The possible rational roots are factors of \(14\) divided by factors of \(1\), so \(\pm1,\pm2,\pm7,\pm14\).

• For \(x = 1\): \(f(1)=1 - 4-3 + 14=8

eq0\)

• For \(x = 2\): \(f(2)=8-16 - 6+14=0\). So \(x = 2\) is a root.

Step3: Factor the polynomial

Since \(x = 2\) is a root, we can factor \(x^{3}-4x^{2}-3x + 14\) as \((x - 2)(x^{2}-2x - 7)\) (using polynomial long division or synthetic division).

Step4: Solve the quadratic equation

For the quadratic equation \(x^{2}-2x - 7=0\), we use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\), where \(a = 1\), \(b=-2\), \(c=-7\).

\(x=\frac{2\pm\sqrt{(-2)^{2}-4\times1\times(-7)}}{2\times1}=\frac{2\pm\sqrt{4 + 28}}{2}=\frac{2\pm\sqrt{32}}{2}=\frac{2\pm4\sqrt{2}}{2}=1\pm2\sqrt{2}\)

\(2\sqrt{2}\approx2.828\), so \(1 + 2\sqrt{2}\approx3.83\) and \(1-2\sqrt{2}\approx - 1.83\)

We can also verify by graphing the function \(y=x^{3}-4x^{2}-3x + 14\). The \(x\)-intercepts (where \(y = 0\)) are at \(x\approx - 1.83\), \(x = 2\), and \(x\approx3.83\)

Answer:

The real solutions are \(x\approx - 1.83\), \(x = 2\), \(x\approx3.83\) (accurate to two decimal places)