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Numerator of s

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Dividing term 1.

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First remainder.

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Coefficient of x.

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Denominator of s.

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Dividing term 2.

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Coefficient of x^3.

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Simplifies first coefficient of s.

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Simplifies second coefficient of s.

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Second remainder.

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Consider the polynomial

\n

\$p(x) = \\simplify{{coef2_x3}x^3+s*x^2+{coef2_x}x+t}\\text{.}\$

\n

The polynomial:

\n
\n
• has remainder $\\var{rem1}$ when divided by $(\\simplify{x+{c}})$
• \n
• has remainder $\\var{rem2}$ when divided by $(\\simplify{x+{d}})$
• \n
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Using the remainder theorem for the remainder when $p(x)$ is divided by $(\\simplify{x+{c}})$, create an equation involving $s$ and $t$.

\n

[[0]]$s + t$ = [[1]].

\n

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The remainder theorem states that if a polynomial $f(x)$ is divided by $(\\simplify{a*x-b})$ then the remainder is $f(\\frac{b}{a})$.

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Using the remainder theorem for the remainder when $p(x)$ is divided by $(\\simplify{x+{d}})$, create another equation involving $s$ and $t$.

\n

[[0]]$s+t$ = [[1]].

\n

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Find the value of $s$. Reduce your answer to its simplest fractional form.

\n

$s =$ [[0]]

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Subtract the two simultaneous equations for $s$ and $t$, obtained in parts a) and b), from each other.

\n

Then rearrange this new equation to find the value of $s$.

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Find the value of $t$. Reduce your answer to its simplest fractional form.

\n

$t =$ [[0]]

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Substitute the value of $s$ from part c) into one of the simultaneous equations for $s$ and $t$.

\n

Then, rearrange this equation to find the value of $t$.

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This question tests the student's knowledge of the remainder theorem and the ways in which it can be applied.

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We are told that the polynomial:

\n
\n
• has a remainder $\\var{rem1}$ when divided by $(\\simplify{x+{c}})$
• \n
• has a remainder $\\var{rem2}$ when divided by $(\\simplify{x+{d}})$
• \n
\n

#### a)

\n

Firstly, substituting $x = \\simplify{-{c}}$ into $p(x)$ gives us

\n

\\begin{align}
p(\\simplify{-{c}}) &= \\simplify[all,!collectNumbers, fractionnumbers]{{coef2_x3*(-c)^3}+{(-c)^2}s+{coef2_x*(-c)}+t},\\\\
&= \\simplify[all,fractionnumbers]{{coef2_x3*(-c)^3}+{(-c)^2}s+{coef2_x*(-c)}+t}.
\\end{align}

\n

But, by the remainder theorem $p(\\simplify{-{c}}) = \\var{rem1}$ (using the first bullet point), so this becomes

\n

\\begin{align}
\\simplify[all,fractionnumbers]{{coef2_x3*(-{c})^3}+s*{(-{c})^2}+{coef2_x*(-{c})}+t} &= \\var{rem1},\\\\
\\simplify[all,fractionnumbers]{s*{x}+t} &= \\simplify[all,fractionnumbers]{{rem1}-{coef2_x3*(-{c})^3}-{coef2_x*(-{c})}}.
\\end{align}

\n

#### b)

\n

Similarly, substituting $x = \\simplify{-{d}}$ into $p(x)$, gives us

\n

\\begin{align}
p(\\simplify{-{d}}) &= \\simplify[all,!collectNumbers, fractionnumbers]{{coef2_x3*(-{d})^3}+{(-{d})^2}s+{coef2_x*(-{d})}+t},\\\\
&= \\simplify[all,fractionnumbers]{{coef2_x3*(-{d})^3}+{(-{d})^2}s+{coef2_x*(-{d})}+t}.
\\end{align}

\n

But, by the remainder theorem $p(\\simplify{-{d}}) = \\var{rem2}$ (using the second bullet point), so this becomes

\n

\\begin{align}
\\simplify[all,fractionnumbers]{{coef2_x3*(-{d})^3}+s*{(-{d})^2}+{coef2_x*(-{d})}+t} &= \\var{rem2},\\\\
\\simplify[all,fractionnumbers]{s*{y}+t} &= \\simplify[all,fractionnumbers]{{rem2}-{coef2_x3*(-{d})^3}-{coef2_x*(-{d})}}.
\\end{align}

\n

#### c)

\n

We now have two simultaneous equations for $s$ and $t$:

\n

\\begin{align}
\\simplify[all,fractionnumbers]{s*{x}+t} = \\simplify[all,fractionnumbers]{{rem1}-{coef2_x3*(-{c})^3}-{coef2_x*(-{c})}} \\\\
\\simplify[all,fractionnumbers]{s*{y}+t} = \\simplify[all,fractionnumbers]{{rem2}-{coef2_x3*(-{d})^3}-{coef2_x*(-{d})}}
\\end{align}

\n

Next, we subtract the second equation from the first equation.

\n

This allows us to cancel out the terms involving $t$ and gives us an equation only in terms of $s$, which we can then rearrange to find the value of $s$.

\n

Subtracting the two equations gives

\n

\$\\simplify{s*{(-{c})^2-(-{d})^2}} = \\simplify[all,fractionnumbers]{{rem1 - coef2_x3*(-c)^3-coef2_x*(-c)-rem2+coef2_x3*(-d)^3+coef2_x*(-d)}}.\$

\n

Then, we can rearrange this equation so that

\n

\$s = \\simplify[all,fractionnumbers]{{rem1 - coef2_x3*(-c)^3-coef2_x*(-c)-rem2+coef2_x3*(-d)^3+coef2_x*(-d)}/{{(-c)^2-(-d)^2}}}.\$

\n

#### d)

\n

We can calculate $t$ by substituting our value of $s$ into one of our original simultaneous equations. For example, let's use the equation

\n

\$\\simplify[all,fractionnumbers]{s*{(-{d})^2}+t} = \\simplify[all,fractionnumbers]{{rem2}-{coef2_x3*(-{d})^3}-{coef2_x*(-{d})}}.\$

\n

Substituting our value of $s$ into this equation gives us

\n

\\\begin{align} \\simplify[all,fractionnumbers,!noleadingMinus]{{numerator/denominator}+t} &= \\simplify[all,fractionnumbers]{{rem2-coef2_x3*(-d)^3-coef2_x*(-d)}},\\\\ t &= \\simplify[all,fractionnumbers]{{rem2-coef2_x3*(-d)^3-coef2_x*(-d) - numerator/denominator}}. \\end{align} \

\n

This same answer would've also been obtained if we had substituted our value of $s$ into the other equation instead.

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