1.3.1: Exercises 1.3
- Page ID
- 62180
Terms and Concepts
Exercise \(\PageIndex{1}\)
In your own words, explain the relationship between factors and roots.
- Answer
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Answers will vary.
Exercise \(\PageIndex{2}\)
If \(x=2\), \(x=5\),and \(x=-1\) are the only roots of the function \(f(x)\), what are the factors of \(f(x)\)?
- Answer
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The factors are \(x-2\), \(x-5\), and \(x+1\).
Exercise \(\PageIndex{3}\)
What does it mean for a quadratic to be irreducible?
- Answer
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It means it has no real number roots and that it cannot be factored.
Exercise \(\PageIndex{4}\)
What is the maximum number of linear factors that \(g(t) = t^6+t^4-2t^2 +1\) could have?
- Answer
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It can have a maximum of 6 linear factors since it is a sixth degree polynomial.
Exercise \(\PageIndex{5}\)
What is the maximum number of roots that \(g(t) = t^6+t^4-2t^2 +1\) could have?
- Answer
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It can have a maximum of 6 roots since it is a sixth degree polynomial.
Problems
In exercises \(\PageIndex{6}\) - \(\PageIndex{12}\), expand and simplify the given expressions.
Exercise \(\PageIndex{6}\)
\(3a(2b+5)(a-2b)\)
- Answer
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\(6a^2b-12ab^2+15a^2-30ab\)
Exercise \(\PageIndex{7}\)
\((2t+7)^2\)
- Answer
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\(4t^2 + 28t+49\)
Exercise \(\PageIndex{8}\)
\(2(x^2+3x+4)(2x+3)\)
- Answer
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\(4x^3+18x^2+34x+24\)
Exercise \(\PageIndex{9}\)
\((t^2-3t+1)(2t)-(t^2+2)(2t-3)\)
- Answer
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\(-3t^2-2t+6\)
Exercise \(\PageIndex{10}\)
\((x^3+x-2)(2)-(2x+1)(3x^2+1)\)
- Answer
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\(-4x^3-3x^2-5\)
Exercise \(\PageIndex{11}\)
\((x^2+3x-1)(4x)+(2x^2-5)(2x+3)\)
- Answer
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\(8x^3+18x^2-14x-15\)
Exercise \(\PageIndex{12}\)
\(3(\theta^2+4)^2(2\theta)\)
- Answer
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\(6\theta^5+48\theta^3+96\theta\)
In exercises \(\PageIndex{13}\) - \(\PageIndex{20}\), factor each function completely.
Exercise \(\PageIndex{13}\)
\(g(x)=4x^2+4x+1\)
- Answer
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\(g(x)=(2x+1)^2\)
Exercise \(\PageIndex{14}\)
\(y(z)=z^2-7z+10\)
- Answer
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\(y(z)=(z-5)(z-2)\)
Exercise \(\PageIndex{15}\)
\(f(k)=k^4-27k\)
- Answer
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\(f(k)=k(k-3)(k^2+3k+9)\)
Exercise \(\PageIndex{16}\)
\(\theta(\gamma) = 6\gamma^2-\gamma -2\)
- Answer
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\(\theta(\gamma) = (3\gamma-2)(2\gamma+1)\)
Exercise \(\PageIndex{17}\)
\(x(z) = 3z^3+6z^2-24z\)
- Answer
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\(x(z)=3z(z+4)(z-2)\)
Exercise \(\PageIndex{18}\)
\(y(x)=x^3+8\)
- Answer
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\(y(x)=(x+2)(x^2-2x+4)\)
Exercise \(\PageIndex{19}\)
\(f(x)=2x^3 -x^2 -5x-2\)
- Answer
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\(f(x)=(2x+1)(x-2)(x+1)\)
Exercise \(\PageIndex{20}\)
\(f(y)=y^3-5y^2-2y+24\)
- Answer
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\(f(y)=(y-3)(y-4)(y+2)\)
In exercises \(\PageIndex{21}\) - \(\PageIndex{24}\), determine the difference quotient of the given function.
Exercise \(\PageIndex{21}\)
\(g(t)=t^3+1\)
- Answer
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\(3t^2+3th+h^2\)
Exercise \(\PageIndex{22}\)
\(y(x)=2x^2-5\)
- Answer
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\(4x+2h\)
Exercise \(\PageIndex{23}\)
\(f(x)=x^3+x^2 -x\)
- Answer
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\(3x^2+3xh+h^2+2x+h-1\)
Exercise \(\PageIndex{24}\)
\(g(x)=4x^2+2x\)
- Answer
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\(8x+4h^2+2\)
In exercises \(\PageIndex{25}\) - \(\PageIndex{27}\), find all real roots of the given function.
Exercise \(\PageIndex{25}\)
\(g(x)=4x^2+2x\)
- Answer
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\(x=0\) and \(x=-\frac{1}{2}\)
Exercise \(\PageIndex{26}\)
\(f(x)=x^3+x^2 -x\)
- Answer
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\(x=0\), \(x=\frac{-1+\sqrt{5}}{2}\), and \(x=\frac{-1-\sqrt{5}}{2}\)
Exercise \(\PageIndex{27}\)
\(y(x)=4x^2-5\)
- Answer
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\(x=\frac{\sqrt{5}}{2}\) and \(x=-\frac{\sqrt{5}}{2}\)
In exercises \(\PageIndex{28}\) - \(\PageIndex{30}\), factor the given function, and relate the factors with the roots found in exercises \(\PageIndex{25}\) - \(\PageIndex{27}\).
Exercise \(\PageIndex{28}\)
\(g(x)=4x^2+2x\)
- Answer
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\(g(x) = 2x(2x+1)\); the factor \(x\) pairs with the root \(x=0\) and the factor \(2x+1\) pairs with the root \(x=-\frac{1}{2}\)
Exercise \(\PageIndex{29}\)
\(f(x)=x^3+x^2 -x\)
- Answer
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\(f(x) = x(x+ \frac{1-\sqrt{5}}{2})(x+\frac{1+\sqrt{5}}{2})\); the factor \(x\) pairs with the root \(x=0\), the factor \(x + \frac{1-\sqrt{5}}{2}\) pairs with the root \(x=\frac{-1+\sqrt{5}}{2}\), and the factor \(x+\frac{1-\sqrt{5}}{2}\) pairs with the root \(x=\frac{-1-\sqrt{5}}{2}\)
Exercise \(\PageIndex{30}\)
\(y(x)=4x^2-5\)
- Answer
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\(y(x) = 4(x-\frac{\sqrt{5}}{2})(x+\frac{\sqrt{5}}{2})\); the factor \(x-\frac{\sqrt{5}}{2}\) pairs with the root \(x=\frac{\sqrt{5}}{2}\) and the factor \(x+\frac{\sqrt{5}}{2}\) pairs with the root \(x=-\frac{\sqrt{5}}{2}\)