// Numbas version: exam_results_page_options {"name": "Polynomials: Factors, roots, division, remainders", "duration": 0, "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "allQuestions": true, "shuffleQuestions": false, "percentPass": 0, "timing": {"allowPause": true, "timeout": {"action": "none", "message": ""}, "timedwarning": {"action": "none", "message": ""}}, "pickQuestions": 0, "navigation": {"onleave": {"action": "none", "message": ""}, "reverse": true, "allowregen": true, "showresultspage": "oncompletion", "preventleave": true, "browse": true, "showfrontpage": true}, "feedback": {"showtotalmark": true, "advicethreshold": 0, "showanswerstate": true, "showactualmark": true, "allowrevealanswer": true}, "type": "exam", "questions": [], "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": [{"name": "The Fundamental Theorem of Algebra", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}], "functions": {}, "ungrouped_variables": ["degree", "c", "poly"], "tags": ["algebra", "Algebra", "Fundamental", "Polynomial", "polynomial", "roots"], "advice": "", "rulesets": {}, "parts": [{"stepsPenalty": "1", "prompt": "

The polynomial {poly} has [[0]] (complex) roots (counted with multiplicity).

\n

That is, it can be written as the product of a constant (complex) factor and [[0]] linear factors $x-r_i$ where $r_i\\in \\mathbb{C}$ is a root of the polynomial.

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The Fundamental Theorem of Algebra can be stated as follows: every non-zero, single-variable, degree $n$ polynomial with complex coefficients has, counted with multiplicity, exactly $n$ roots. 

\n

\n

Since real numbers are complex numbers (just with an imaginary part of zero) and the highest power in the given polynomial is $\\var{degree}$, the Fundamental Theorem of Algebra can be applied and therefore the polynomial has exactly $\\var{degree}$ roots (counted with multiplicity). That is, there isn't necessarily $\\var{degree}$ unique roots but the polynomial can be written as the product of $\\var{degree}$ linear factors $x-r_i$ where $r_i\\in \\mathbb{C}$ is a root of the polynomial.

\n

\n

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Suppose a polynomial $f(x)$ of degree $\\var{n}$ is divided by $\\simplify{x-{c}}$ and the remainder is $\\var{rem}$.

\n

This guarantees us that $f\\large($ [[0]] $\\large)=$ [[1]].

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The Remainder Theorem says that the remainder of dividing $f(x)$ by $x-a$ is equal to $f(a)$. So in this case we know $f(\\var{c})=\\var{rem}$.

\n

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Suppose a polynomial $f(x)$ evaluated at $\\var{c}$ is $0$, that is, $f(\\var{c})=0$. 

\n

Which of the following is guaranteed? 

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The Factor Theorem says that $x-a$ is a factor of a polynomial $f(x)$ if and only if $f(a)=0$. 

\n

Since we are told that $f(\\var{c})=0$ then we know $\\simplify{x-{c}}$ is a factor of $f(x)$.

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$\\simplify{x-{c}}$ is a factor of $f(x)$

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$\\simplify{x+{c}}$ is a factor of $f(x)$

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$x$ is a factor of $f(x)$

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$\\var{c}$ is a factor of $f(x)$

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Suppose $\\simplify{x-{d}}$ is a factor of a polynomial $g(x)$. That is, if we divided $g(x)$ by $\\simplify{x-{d}}$ the remainder would be $0$.

\n

This guarantees us that $g\\large($ [[0]] $\\large)=$ [[1]].

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The Factor Theorem says that $x-a$ is a factor of a polynomial $f(x)$ if and only if $f(a)=0$.  

\n

Since we are told $\\simplify{x-{d}}$ is a factor of $g(x)$, we know that $g(\\var{d})=0$.

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Suppose you divide a polynomial of degree $\\var{n}$ by $\\simplify{x-{c}}$, then the remainder must be a

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constant.

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linear function.

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quadratic function.

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polynomial of degree $\\var{n-1}$.

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When you divide the dividend by the divisor, the remainder must be a polynomial of a smaller degree than the divisor. 

\n

In the case that the divisor is $\\simplify{x-{c}}$ (a polynomial of degree 1) the remainder must be a polynomial of 0th degree, that is, a constant.

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Suppose you divide a polynomial of degree $\\var{n}$ by a polynomial of degree $\\var{m}$, then we can say the remainder is a polynomial of degree no more than

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$\\var{m}$

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$\\var{m-1}$

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$\\var{m-2}$

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$0$

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When you divide the dividend by the divisor, the remainder must be a polynomial of a smaller degree than the divisor. 

\n

In the case that the divisor is a polynomial of degree $\\var{m}$ the remainder must be a polynomial of degree $\\var{m-1}$ or less.

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Again, suppose you divide a polynomial of degree $\\var{n}$ by a polynomial of degree $\\var{m}$, then we can say the quotient is a polynomial of degree

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$\\var{n-m}$

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$\\var{n-m-2}$

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$\\var{n-m-1}$

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$\\var{n-m+1}$

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$\\var{n}$

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Due to the index laws, the degree of the dividend must equal the sum of the degrees of the divisor and quotient.

\n

This means that the degree of the quotient must be $\\var{n}-\\var{m}=\\var{n-m}$.

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There is a procedure called polynomial long division, if you are comfortable with long division of numbers then it isn't too different. You can see an example of the procedure here. However, if you don't recall long division we can rewrite the numerator so that it includes a multiple of the denominator. We do this term by term starting with the leading term (that is, $\\simplify{{r * m}x ^ 3}$).

\n

\\[\\begin{eqnarray*} \\simplify[std]{{r * m} * x ^ 3 + {n * r + m * s} * x ^ 2 + {n * s + t * r} * x + {t * n + be}}&=&\\simplify[std]{(x+{s})x^2+{n}x^2+{n*s+t}x+{t*n+be}}\\\\&=&\\simplify[std]{(x+{s})x^2+(x+{s})*{n}x+{t}x+{t*n+be}}\\\\ &=&\\simplify[std]{(x+{s})x^2+(x+{s})*{n}x+(x+{s})*{t}+{t*n+be-s*t}}\\\\ &=&\\simplify[std]{(x+{s})(x^2+{n}x+{t})+{t*n+be-s*t}} \\end{eqnarray*} \\]

\n

Hence
\\[\\frac{\\simplify[std]{{r * m} * x ^ 3 + {n * r + m * s} * x ^ 2 + {n * s + t * r} * x + {t * n + be}}}{\\simplify[std]{{r}x+{s}}}=\\simplify[std]{x^2+{n}x+{t}+{t*n+be-s*t}/({r}x+{s})}\\]

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Dividing a cubic polynomial by a linear polynomial. Find quotient and remainder.

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Input numbers as integers not decimals.

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$q(x)=\\;\\;$[[0]]

\n \n \n \n

Input all numbers as integers and not as decimals.

\n \n \n \n

$r=\\;\\;$[[1]]

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Divide $\\displaystyle{\\simplify[std]{{r * m} * x ^ 3 + {n * r + m * s} * x ^ 2 + {n * s + t * r} * x + {t * n + be}}}$ by $\\simplify[std]{{r}x+{s}}$ so that:
\\[\\frac{\\simplify[std]{{r * m} * x ^ 3 + {n * r + m * s} * x ^ 2 + {n * s + t * r} * x + {t * n + be}}}{\\simplify[std]{{r}x+{s}}}=q(x)+\\frac{r}{\\simplify[std]{{r}x+{s}}}\\]

\n

where $q(x)$ is the quotient polynomial and $r$ is the remainder ($r$ is a constant).

\n

The coefficients of $q(x)$ are integers, do not input as decimals.

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