27.2: Review of polynomials and rational functions
- Page ID
- 68483
Divide the polynomials: \(\dfrac{2x^3+x^2-9x-8}{2x+3}\)
- Answer
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\(x^{2}-x-3+\dfrac{1}{2 x+3}\)
Find the remainder when dividing \(x^3+3x^2-5x+7\) by \(x+2\).
- Answer
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\(21\)
Which of the following is a factor of \(x^{400}-2x^{99}+1\): \[x-1, \quad x+1, \quad x-0 \nonumber \]
- Answer
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\(x − 1\) is a factor, \(x + 1\) is not a factor, \(x − 0\) is not a factor
Identify the polynomial with its graph.
- \(f(x)=-x^{2}+2 x+1\) graph: _______________
- \(f(x)=-x^{3}+3 x^{2}-3 x+2\) graph: _______________
- \(f(x)=x^{3}-3 x^{2}+3 x+1\) graph: _______________
- \(f(x)=x^{4}-4 x^{3}+6 x^{2}-4 x+2\) graph: _______________
- Answer
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- \(\leftrightarrow \text { iii) }\)
- \(\leftrightarrow \text { iv) }\)
- \(\leftrightarrow \mathrm{i})\)
- \(\leftrightarrow \text { ii) }\)
Sketch the graph of the function: \[f(x)=x^4-10x^3-0.01x^2+0.1x \nonumber \]
- What is your viewing window?
- Find all roots, all maxima and all minima of the graph with the calculator.
- Answer
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Find all roots of \(f(x)=x^3+6x^2+5x-12\).
Use this information to factor \(f(x)\) completely.
- Answer
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\(f(x)=(x-1)(x+3)(x+4)\)
Find a polynomial of degree \(3\) whose roots are \(0\), \(1\), and \(3\), and so that \(f(2)=10\).
- Answer
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\(f(x)=(-5) \cdot x(x-1)(x-3)\)
Find a polynomial of degree \(4\) with real coefficients, whose roots include \(-2\), \(5\), and \(3-2i\).
- Answer
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\(f(x)=(x+2)(x-5)(x-(3-2 i))(x-(3+2 i))\) (other correct answers are possible, depending on the choice of the first coefficient)
Let \(f(x)=\dfrac{3x^2-12}{x^2-2x-3}\). Sketch the graph of \(f\). Include all vertical and horizontal asymptotes, all holes, and all \(x\)- and \(y\)-intercepts.
- Answer
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\(f(x)=\dfrac{3(x-2)(x+2)}{(x-3)(x+1)}\) has domain \(D=\mathbb{R}-\{-1,3\}\), horizontal asympt. \(y = 3\), vertical asympt. \(x = −1\) and \(x = 3\), no removable discont., \(x\)-intercepts at \(x = −2\) and \(x = 2\) and \(x = 3\), \(y\)-intercept at \(y = 4\), graph:
Solve for \(x\):
- \(x^4+2x< 2x^3+x^2\)
- \(x^2+3x\geq 7\)
- \(\dfrac{x+1}{x+4}\leq 2\)
- Answer
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- \((-1,0) \cup(1,2)\)
- \(\left(-\infty, \dfrac{-3-\sqrt{37}}{2}\right] \cup\left[\dfrac{-3+\sqrt{37}}{2}, \infty\right)\)
- \((-\infty,-7] \cup(-4, \infty)\)