// Numbas version: exam_results_page_options {"name": "Remainder Theorem ", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"advice": "
The remainder theorem states that if a polynomial $f(x)$ is divided by $(\\simplify{a*x-b})$ then the remainder is $f \\left( \\frac{b}{a} \\right)$.
\nThis means that if we substitute $x = \\frac{b}{a}$ into the equation for $f(x)$, the result will be equal to the remainder when $f(x)$ is divided by $(\\simplify{a*x-b})$.
\nTherefore, to calculate the remainder when $f(x) = \\simplify{{coef_x3}*x^3+{coef_x2}*x^2+{coef_x}*x+{const}}$ is divided by $(\\simplify{{a}*x+{k}})$, we use this same principle.
\nAs we are dividing $f(x)$ by $(\\simplify{{a}*x+{k}})$, using the remainder theorem tells us that substituting
\n\\[
\\begin{align}
x &= \\frac{b}{a}\\\\
&= \\simplify{-({k}/{a})}
\\end{align}
\\]
into our equation for $f(x)$ will give us the remainder when $f(x)$ is divided by $(\\simplify{{a}*x+{k}})$. Substituting this value of $x$ into $f(x)$ gives us
\n\\[
\\begin{align}
f(\\simplify{-({k}/{a})}) &= \\simplify[all,!collectNumbers, fractionnumbers]{{coef_x3*(-({k}/{a}))^3}+{coef_x2*(-({k}/{a}))^2}+{coef_x*(-({k}/{a}))}+{const}}\\\\
&= \\simplify[all,fractionnumbers]{{coef_x3*(-({k}/{a}))^3}+{coef_x2*(-({k}/{a}))^2}+{coef_x*(-({k}/{a}))}+{const}}.
\\end{align}
\\]
Therefore, the remainder when $f(x) = \\simplify{{coef_x3}*x^3+{coef_x2}*x^2+{coef_x}*x+{const}}$ is divided by $(\\simplify{{a}*x+{k}})$ is $\\displaystyle\\simplify[all,fractionnumbers]{{coef_x3*(-({k}/{a}))^3}+{coef_x2*(-({k}/{a}))^2}+{coef_x*(-({k}/{a}))}+{const}}$.
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"}}, "type": "question", "contributors": [{"name": "Mark Hodds", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/510/"}]}]}], "contributors": [{"name": "Mark Hodds", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/510/"}]}