// Numbas version: exam_results_page_options {"name": "JD's copy of Finding the missing value of a constant in a polynomial, using the Factor Theorem", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"showfrontpage": false, "allowregen": true, "preventleave": false}, "question_groups": [{"questions": [{"variables": {"d": {"group": "Ungrouped variables", "templateType": "anything", "definition": "random(-2..2 except 0 except a except b)", "name": "d", "description": "

Used in creation of the polynomial.

"}, "b": {"group": "Ungrouped variables", "templateType": "anything", "definition": "random(-2..3 except 0)", "name": "b", "description": "

Random number between -2 and 3 except 0 for creating polynomial.

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Random number between 2,3,4.

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Random number between -2 and 3, not including 0 for creating polynomial.

"}, "coef_x2": {"group": "Ungrouped variables", "templateType": "anything", "definition": "(w*d+a+w*b)*(-d)^2", "name": "coef_x2", "description": "

Number obtained by putting x=-d into the second term of the equation.

"}, "coef_x": {"group": "Ungrouped variables", "templateType": "anything", "definition": "(a*d+w*b*d+a*b)*(-d)", "name": "coef_x", "description": "

Number obtained by putting x=-d into the third term of the equation.

"}, "coef_x3": {"group": "Ungrouped variables", "templateType": "anything", "definition": "(w)*(-d)^3", "name": "coef_x3", "description": "

Number obtained by putting x=-d into the first term of the equation.

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The factor theorem states that if $f(x)$ is a polynomial and $f(p) = 0$, then $(x-p)$ is a factor of $f(x)$.

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Given that $(\\simplify{x+{d}})$ is a factor of $g(x) = \\simplify{{w}*x^3+({w}{d}+{a}+{w}{b})*x^2+({a}{d}+{w}{b}{d}+{a}{b})*x}+m$, find the value of $m$.

\n

$m =$ [[0]].

\n

", "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true}], "tags": ["Factor Theorem", "factor theorem", "polynomials", "Polynomials", "taxonomy"], "extensions": [], "metadata": {"description": "

Given a factor of a cubic polynomial, factorise it fully by first dividing by the given factor, then factorising the remaining quadratic.

", "licence": "Creative Commons Attribution 4.0 International"}, "type": "question", "ungrouped_variables": ["w", "a", "b", "d", "coef_x3", "coef_x2", "coef_x"], "rulesets": {}, "variablesTest": {"maxRuns": 100, "condition": ""}, "preamble": {"css": "", "js": ""}, "functions": {}, "advice": "

Using the factor theorem, we know that if $(x-a)$ is a factor of a polynomial $f(x)$, then $f(a)=0$.

\n

We are given that $(\\simplify{x+{d}})$ is a factor of $g(x) = \\simplify{{w}*x^3+({w}{d}+{a}+{w}{b})*x^2+({a}{d}+{w}{b}{d}+{a}{b})*x+m}$.

\n

By the factor theorem, this means that $g(\\simplify{-{d}}) = 0$.

\n

Substituting $x=\\simplify{-{d}}$ into $g(x)$ gives

\n

\\\begin{align} g(\\simplify{-{d}}) &= \\simplify[all,!collectNumbers]{{coef_x3}+{coef_x2}+{coef_x}+m}\\\\ &=\\simplify{{coef_x3}+{coef_x2}+{coef_x}+m}. \\end{align} \

\n

Therefore, as $g(\\simplify{-{d}}) = 0$, we have

\n

\\\begin{align} \\simplify{{coef_x3}+{coef_x2}+{coef_x}+m}&=0\\\\ m&=\\simplify{-({coef_x3}+{coef_x2}+{coef_x})}. \\end{align} \

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