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Remainder Theorem

5.3 Remainder theorem (EMCGV)

Remainder theorem

Given the following functions:

  • \(f(x) = x^{3} + 3x^{2} + 4x + 12\)
  • \(k(x) = x - 1\)
  • \(g(x) = 4x^{3} -2x^{2} + x -7\)
  • \(h(x) = x + 2\)
  1. Determine \(\dfrac{f(x)}{k(x)}\) and \(\dfrac{g(x)}{h(x)}\).
  2. Write your answers in the general form: \(a(x)=b(x).Q(x) + R(x)\).
  3. Determine \(f(1)\) and \(g(-2)\).
  4. What do you notice?
  5. Consider the degree of the quotient and the remainder - is there a rule?
  6. What conclusions can you draw?
  7. Write a mathematical equation to describe your conclusions.
  8. Complete the following sentence: a cubic function divided by a linear polynomial gives a quotient with a degree of \(\ldots \ldots\) and a remainder with a degree of \(\ldots \ldots\), which is called a constant.

The Remainder theorem

A polynomial \(p(x)\) divided by \(cx - d\) gives a remainder of \(p\left(\dfrac{d}{c}\right)\).

In words: the value of the remainder \(R\) is obtained by substituting \(x = \frac{d}{c}\) into the polynomial \(p(x)\).

\[R = p \left( \dfrac{d}{c} \right)\]

NOTE: PROOF NOT FOR EXAMS

Let the quotient be \(Q(x)\) and let the remainder be \(R\). Therefore we can write:

\begin{align*} p(x) &= (cx - d) \cdot Q(x) + R \\ \therefore p \left( \dfrac{d}{c} \right) &= \left[c\left( \dfrac{d}{c} \right) - d\right] \cdot Q \left( \dfrac{d}{c} \right) + R \\ &= (d - d)\cdot Q \left( \dfrac{d}{c} \right) + R \\ &= 0 \cdot Q \left( \dfrac{d}{c} \right) + R \\ &= R\\ \therefore p \left( \dfrac{d}{c} \right) &= R \end{align*}

Worked example 7: Finding the remainder

Use the remainder theorem to determine the remainder when \(p(x) = 3x^{3} + 5x^{2} - x + 1\) is divided by the following linear polynomials:

  1. \(x + 2\)
  2. \(2x -1\)
  3. \(x + m\)

Determine the remainder for each linear divisor

The remainder theorem states that any polynomial \(p(x)\) that is divided by \(cx - d\) gives a remainder of \(p\left(\dfrac{d}{c}\right)\):

  1. \begin{align*} p(x) &= 3x^{3} + 5x^{2} - x + 1 \\ p(-2) &= 3 \left( -2 \right)^{3} + 5\left( -2 \right)^{2} - \left( -2 \right) + 1 \\ &= 3 \left( -8 \right) + 5\left( 4 \right) +2 + 1 \\ &= -24 + 20 + 3 \\ \therefore R &= -1 \end{align*}
  2. \begin{align*} p(x) &= 3x^{3} + 5x^{2} - x + 1 \\ p \left( \frac{1}{2} \right) &= 3\left( \frac{1}{2} \right)^{3} + 5\left( \frac{1}{2} \right)^{2} - \left( \frac{1}{2} \right)+ 1 \\ &= 3\left( \frac{1}{8} \right) + 5\left( \frac{1}{4}\right) - \left( \frac{1}{2}\right) + 1 \\ &= \frac{3}{8} + \frac{5}{4} + \frac{1}{2} \\ &= \frac{3}{8} + \frac{10}{8} + \frac{4}{8} \\ \therefore R &= \frac{17}{8} \end{align*}
  3. \begin{align*} p(x) &= 3x^{3} + 5x^{2} - x + 1 \\ p(m) &= 3\left( -m \right)^{3} + 5\left( -m \right)^{2} - \left( -m \right) + 1 \\ \therefore R &= -3m^{3} + 5m^{2} + m + 1 \end{align*}

Worked example 8: Using the remainder to solve for an unknown variable

Given that \(f(x) = 2x^{3} + x^{2} + kx + 5\) divided by \(2x - 3\) gives a remainder of \(9\frac{1}{2}\), use the remainder theorem to determine the value of \(k\).

Use the remainder theorem to determine the unknown variable \(k\)

From the remainder theorem we know that \(f\left(\frac{3}{2}\right) = 9\frac{1}{2}\) and we can therefore solve for \(k\):

\begin{align*} f(x) &= 2x^{3} + x^{2} + kx + 5 \\ f\left( \frac{3}{2} \right) &= 2 \left( \frac{3}{2} \right)^{3} + \left( \frac{3}{2} \right)^{2} + k\left( \frac{3}{2} \right) + 5 \\ 9\frac{1}{2} &= 2 \left( \frac{27}{8} \right) + \left( \frac{9}{4} \right) + k\left( \frac{3}{2} \right) + 5 \\ 9\frac{1}{2} &= \frac{27}{4} + \frac{9}{4} + \frac{3k}{2} + 5 \\ 9\frac{1}{2} &= \frac{36}{4} + \frac{3k}{2} + 5\\ \therefore 9\frac{1}{2} - 9 - 5 &= \frac{3k}{2} \\ - 4\frac{1}{2} &= \frac{3k}{2} \\ - \frac{9}{2} \times \frac{2}{3} &= k \\ \therefore -3 &= k \end{align*}

Write the final answer

Therefore \(k = -3\) and \(f(x) = 2x^{3} + x^{2} - 3x + 5\).

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Remainder theorem

Exercise 5.4

Use the remainder theorem to determine the remainder \(R\) when \(g(x)\) is divided by \(h(x)\):

\begin{align*} g(x) &= x^{3} + 4x^{2} + 11x - 5 \\ h(x) &= x - 1 \end{align*}

\begin{align*} g(x) &= x^{3} + 4x^{2} + 11x - 5 \\ g(1) &= (1)^{3} + 4(1)^{2} + 11(1) - 5 \\ &= 1 + 4 + 11 - 5 \\ \therefore R &= 11 \end{align*}

\begin{align*} g(x) &= 2x^{3} - 5x^{2} + 8 \\ h(x) &= 2x - 1 \end{align*}

\begin{align*} g(x) &= 2x^{3} - 5x^{2} + 8 \\ g \left( \frac{1}{2} \right) &= 2\left( \frac{1}{2} \right)^{3} - 5\left( \frac{1}{2} \right)^{2} + 8 \\ &= 2\left( \frac{1}{8} \right) - 5\left( \frac{1}{4} \right) + 8 \\ &= \frac{1}{4} - \frac{5}{4} + 8 \\ \therefore R &= 7 \end{align*}

\begin{align*} g(x) &= 4x^{3} + 5x^{2} + 6x - 1 \\ h(x) &= x + 2 \end{align*}

\begin{align*} g(x) &= 4x^{3} + 5x^{2} + 6x - 1 \\ g (-2) &= 4(-2)^{3} + 5(-2)^{2} + 6(-2) - 1 \\ &= -32 +20 -12 - 1 \\ \therefore R &= -25 \end{align*}

\begin{align*} g(x) &= -5x^{3} - x^{2} -10x + 9 \\ h(x) &= 5x + 1 \end{align*}

\begin{align*} g(x) &= -5x^{3} - x^{2} -10x + 9 \\ g \left( -\frac{1}{5} \right) &= -5\left( -\frac{1}{5} \right)^{3} - \left( -\frac{1}{5} \right)^{2} -10\left( -\frac{1}{5} \right) + 9 \\ &= -5\left( -\frac{1}{125} \right) - \left( \frac{1}{25} \right) + 2 + 9 \\ &= \frac{1}{25} - \frac{1}{25} + 11 \\ \therefore R &= 11 \end{align*}

\begin{align*} g(x) &= x^{4} + 5x^{2} + 2x - 8 \\ h(x) &= x + 1 \end{align*}

\begin{align*} g(x) &= x^{4} + 5x^{2} + 2x - 8 \\ g (-1) &= (-1)^{4} + 5(-1)^{2} + 2(-1) - 8 \\ &= 1 + 5 - 2 - 8 \\ &= 6 -10 \\ \therefore R &= -4 \end{align*}

\begin{align*} g(x) &= 3x^{5} - 8x^{4} + x^{2} + 2 \\ h(x) &= 2 - x \end{align*}

\begin{align*} h(x) &= 2 - x \\ \text{If } x = 2: \quad h (2) &= 2 - 2 = 0 \\ g(x) &= h(x) \cdot Q(x) + R \\ g(2) &= h(2) \cdot Q(2) + R \\ \therefore g(2) &= 0 \cdot Q(2) + R \\ g(2) &= 0 + R \\ \therefore R &= 3(2)^{5} - 8(2)^{4} + (2)^{2} + 2 \\ &= 3(32) - 8(16) + 6 \\ &= 96 - 128 + 6 \\ \therefore R &= -26 \end{align*}

\begin{align*} g(x) &= 2x^{100} - x - 1 \\ h(x) &= x + 1 \end{align*}

\begin{align*} g(x) &= 2x^{100} - x - 1\\ g (-1) &= 2(-1)^{100} - (-1) - 1 \\ &= 2 + 1 - 1 \\ \therefore R &= 2 \end{align*}

Determine the value of \(t\) if \(x^{3} + tx^{2} + 8x + 21\) divided by \(x + 1\) gives a remainder of \(\text{16}\).

\begin{align*} a(x) &= x^{3} + tx^{2} + 8x + 21 \\ b(x) &= x + 1 \\ R &= 16 \\ a (-1) &= (-1)^{3} + t(-1)^{2} + 8(-1) + 21 \\ \therefore 16 &= -1 +t -8 + 21 \\ \therefore t &= 4 \end{align*}

Calculate the value of \(m\) if \(2x^{3} - 7x^{2} + mx - 26\) is divided by \(x - 2\) and gives a remainder of \(-\text{24}\).

\begin{align*} a(x) &= 2x^{3} - 7x^{2} + mx - 26 \\ b(x) &= x - 2 \\ R &= -24 \\ a (2) &= 2(2)^{3} - 7(2)^{2} + m(2) - 26 \\ \therefore -24 &= 16 -28 +2m -26 \\ 14 &= 2m \\ \therefore m &= 7 \end{align*}

If \(x^{5} - 2x^{3} - kx - 1\) is divided by \(x - 1\) and the remainder is \(-\frac{1}{2}\), find the value of \(k\) .

\begin{align*} a(x) &= x^{5} - 2x^{3} - kx - 1 \\ b(x) &= x - 1 \\ R &= -\frac{1}{2}\\ a (1) &= (1)^{5} - 2(1)^{3} - k(1) - 1 \\ \therefore -\frac{1}{2} &= 1 - 2 - k - 1 \\ \frac{3}{2} &= -k \\ \therefore k &= -\frac{3}{2} \end{align*}

Determine the value of \(p\) if \(18x^{3} + px^{2} - 8x + 9\) is divided by \(2x - 1\) and gives a remainder of \(\text{6}\).

\begin{align*} a(x) &= 18x^{3} + px^{2} - 8x + 9 \\ b(x) &= 2x - 1 \\ R &= 6 \\ a \left( \frac{1}{2} \right) &= 18\left( \frac{1}{2} \right)^{3} + p\left( \frac{1}{2} \right)^{2} - 8\left( \frac{1}{2} \right) + 9 \\ \therefore 6 &= 18 \left( \frac{1}{8} \right) + p \left( \frac{1}{4} \right) -4 + 9 \\ 6 &= \frac{18}{8} + \frac{p}{4} + 5 \\ 1 &= \frac{18}{8} + \frac{p}{4} \\ 4 &= 9 + p \\ \therefore p &= -5 \end{align*}

If \(x^{3} + x^{2} - x + b\) is divided by \(x - 2\) and the remainder is \(2\frac{1}{2}\), calculate the value of \(b\).

\begin{align*} \text{Let } f(x) &= x^{3} + x^{2} - x + b \\ R &= \frac{5}{2} \\ f (2) &= (2)^{3} + (2)^{2} - (2) + b \\ \therefore \frac{5}{2} &= 8 + 4 - 2 + b \\ \frac{5}{2} -10 &= b \\ \therefore b &= - \frac{15}{2} \end{align*}

Calculate the value of \(h\) if \(3x^{5} + hx^{4} + 10x^{2} - 21x + 12\) is divided by \(x - 2\) and gives a remainder of \(\text{10}\).

\begin{align*} a(x) &= 3x^{5} + hx^{4} + 10x^{2} - 21x + 12 \\ b(x) &= x -2 \\ R &= 10 \\ a (2) &= 3(2)^{5} + h(2)^{4} + 10(2)^{2} - 21(2) + 12 \\ \therefore 10 &= 3(32) + 16h + 40 -42 +12 \\ 10 &= 96 + 16h + 10 \\ -96 &= 16h \\ \therefore h &= -6 \end{align*}

If \(x^{3} + 8x^{2} + mx - 5\) is divided by \(x + 1\) and the remainder is \(n\), express \(m\) in terms of \(n\).

\begin{align*} a(x) &= x^{3} + 8x^{2} + mx - 5 \\ b(x) &= x + 1 \\ R &= n \\ a (-1) &= (-1)^{3} + 8(-1)^{2} + m(-1) - 5 \\ \therefore n &= -1 + 8 -m - 5 \\ n &= 2 - m \\ \therefore m &= 2 - n \end{align*}

When the polynomial \(2x^{3} + px^{2} + qx + 1\) is divided by \(x + 1\) or \(x - 4\), the remainder is \(\text{5}\). Determine the values of \(p\) and \(q\).

\begin{align*} a(x) &= 2x^{3} + px^{2} + qx + 1 \\ a(-1) &= 2(-1)^{3} + p(-1)^{2} + q(-1) + 1 \\ \therefore 5 &= -2 + p - q + 1 \\ 6 &= p - q \\ \therefore q &= p - 6 \ldots \ldots (1)\\ a(4) &= 2(4)^{3} + p(4)^{2} + q(4) + 1 \\ \therefore 5 &= 128 + 16p + 4q + 1 \\ -124 &= 16p + 4q \\ \therefore -31 &= 4p + q \ldots \ldots (2)\\ \text{Substitute eqn } (1) & \text{ into eqn } (2) \\ \therefore -31 &= 4p + (p - 6) \\ -25 &= 5p \\ \therefore -5 &= p \\ \text{And } q &= -5 -6 \\ & = -11 \end{align*}