Lesson 15
The Remainder Theorem
Problem 1
For the polynomial function \(f(x)=x^3-2x^2-5x+6\), we have \(f(0)=6, f(2)=\text-4, f(\text-2)=0, f(3)=0,f(\text-1)=8, f(1)=0\). Rewrite \(f(x)\) as a product of linear factors.
Solution
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Problem 2
Select all the polynomials that have \((x-4)\) as a factor.
\(x^3-13x-12\)
\(x^3 + 8x^2 + 19x + 12\)
\(x^3+6x + 5x - 12\)
\(x^3-x^2-10x-8\)
\(x^2-4\)
Solution
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Problem 3
Write a polynomial function, \(p(x)\), with degree 3 that has \(p(7) = 0\).
Solution
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Problem 4
Long division was used here to divide the polynomial function \(p(x)=x^3+7x^2-20x-110\) by \((x-5)\) and to divide it by \((x+5)\).
\(\displaystyle \require{enclose} \begin{array}{r} x^2+12x+40 \\ x-5 \enclose{longdiv}{x^3+7x^2-20x-110} \\ \underline{-x^3+5x^2} \phantom{-20x-110} \\ 12x^2-20x \phantom{-110}\\ \underline{-12x^2+60x} \phantom{-110} \\ 40x-110 \\ \underline{-40x+200} \\ 90 \end{array}\)
\(\displaystyle \require{enclose} \begin{array}{r} x^2+2x-30 \\ x+5 \enclose{longdiv}{x^3+7x^2-20x-110} \\ \underline{-x^3-5x^2} \phantom{-20x-110} \\ 2x^2-20x \phantom{-110}\\ \underline{-2x^2-10x} \phantom{-110} \\ \text-30x-110 \\ \underline{30x+150} \\ 40 \end{array}\)
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What is \(p(\text-5)\)?
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What is \(p(5)\)?
Solution
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Problem 5
Which polynomial function has zeros when \(x=5,\frac23,\text-7\)?
\(f(x)=(x+5)(2x+3)(x-7)\)
\(f(x)=(x+5)(3x+2)(x-7)\)
\(f(x)=(x-5)(2x-3)(x+7)\)
\(f(x)=(x-5)(3x-2)(x+7)\)
Solution
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(From Unit 2, Lesson 5.)Problem 6
The polynomial function \(q(x)=3x^4+8x^3-13x^2-22x+24\) has known factors \((x+3)\) and \((x+2)\). Rewrite \(q(x)\) as the product of linear factors.
Solution
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(From Unit 2, Lesson 12.)Problem 7
We know these things about a polynomial function \(f(x)\): it has degree 3, the leading coefficient is negative, and it has zeros at \(x=\text-5,\text-1, 3\). Sketch a graph of \(f(x)\) given this information.
Solution
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(From Unit 2, Lesson 14.)