Wemazi

Topic 6.12 · BC

Determine the coefficients

Builds on writing the partial-fraction form. 8 minutes.

Two numbers that make the pieces fit

The chalk curve is \(\dfrac{5x + 7}{(x - 1)(x + 2)}\). Its form is \(\dfrac{A}{x - 1} + \dfrac{B}{x + 2}\), but the constants \(A\) and \(B\) are not known yet, so its two faint pieces still lie flat on the x-axis.

Multiply both sides by \((x - 1)(x + 2)\). Each denominator cancels against its own factor, so the equation has no fractions: \(5x + 7 = A(x + 2) + B(x - 1)\). Both sides are the same polynomial, so you can substitute any number for \(x\).

Press Substitute x = 1. The term with \(B\) disappears, because \(x - 1 = 0\), and the equation becomes \(12 = 3A\), so \(A = 4\). The first piece grows to \(\dfrac{4}{x-1}\) and turns orange.

Press Substitute x = −2. The term with \(A\) disappears, because \(x + 2 = 0\), and the equation becomes \(-3 = -3B\), so \(B = 1\). Drag the orange handle on the x-axis. The two orange pieces add up to the chalk curve at every \(x\).

A polynomial identity

Take a proper fraction, which has a numerator of smaller degree than its denominator. Its denominator has two distinct linear factors, so its form has constants \(A\) and \(B\).

Multiply both sides by the denominator. The left side is the numerator, and the right side is a sum of two terms, each with one constant.

Move the sliders to change the numerator. The faint pieces change with it, but their orange bars still stack up to the chalk curve. The cleared equation is an identity: both sides are the same polynomial, so it holds at every x.

A root of a factor is the value of x that makes the factor 0. Substitute that root, and the other term is 0, so one constant remains. You can also match coefficients, and both routes give the same constants.

Notation

  1. For a denominator \((x - a)(x - b)\) and a numerator \(n(x) = n_1 x + n_0\), multiply both sides of the form by the denominator. One factor cancels in each term.

    \[n(x) = \cy{A}(x - b) + \cy{B}(x - a)\]

  2. The cleared equation is an identity, so it holds at \(x = a\) and \(x = b\) too, where the original fraction is undefined but a polynomial is not.

  3. To find \(A\) by substitution, put \(x = a\). The term with \(B\) contains the factor \(x - a\), so it is 0. Near \(x = a\) the chalk curve follows the orange piece, because the other piece stays finite there.

    \[\begin{aligned} n(a) &= \cy{A}(a - b) \\ \cy{A} &= \frac{n(a)}{a - b} \end{aligned}\]

  4. To find both constants by matching, expand the right side. Each coefficient must be equal on both sides, which gives two equations to solve.

    \[n_1 = A + B, \qquad n_0 = -Ab - Ba\]

  5. Test the answer. Substitute a value such as \(x = 0\) into the cleared equation, or add the pieces over the common denominator.

When it fails

Suppose the numerator is \(x^2 + 3\). Then the cleared equation is \(x^2 + 3 = A(x + 2) + B(x - 1)\), and the left side has an \(x^2\) term that the right side cannot produce.

Substituting the roots still gives \(A = \tfrac43\) and \(B = -\tfrac73\), but the red curve made of those two pieces lies one unit below the chalk curve everywhere. The fraction is not proper, so divide first.

Example 1: Substitute the roots

Find \(A\) and \(B\) in \(\dfrac{2x + 11}{(x - 2)(x + 3)} = \dfrac{A}{x - 2} + \dfrac{B}{x + 3}\).

  1. Multiply both sides by the denominator.

    \[2x + 11 = A(x + 3) + B(x - 2)\]

  2. Substitute \(x = 2\). The term with \(B\) is 0, because \(x - 2 = 0\).

    \[2\cdot 2 + 11 = 5A \quad\Longrightarrow\quad A = 3\]

  3. Substitute \(x = -3\). The term with \(A\) is 0.

    \[2(-3) + 11 = -5B \ \Longrightarrow\ B = -1\]

  4. To test the answer, add the pieces over the common denominator. The numerator must be 2x + 11, and the orange bars stack up to the chalk curve.

    \[3(x + 3) - (x - 2) = 2x + 11\]

Example 2: Use the root of the factor

Find \(A\) and \(B\) in \(\dfrac{7x - 18}{(x - 4)(x + 1)} = \dfrac{A}{x - 4} + \dfrac{B}{x + 1}\).

  1. Multiply both sides by the denominator, and substitute \(x = 4\) to find \(A\).

    \[\begin{gathered} 7x - 18 = A(x + 1) + B(x - 4) \\ 7\cdot 4 - 18 = 5A \ \Longrightarrow\ A = 2 \end{gathered}\]

  2. A tempting wrong move: To find \(B\), substitute \(x = 1\), because the factor is \(x + 1\). The term with \(A\) is then 0.

    \[B = \frac{7\cdot 1 - 18}{1 - 4} = \frac{11}{3}\]

    At \(x = 2\), the chalk curve has height \(\tfrac23\). The pieces \(\tfrac{2}{x-4}\) and \(\tfrac{11/3}{x+1}\) have heights \(-1\) and \(\tfrac{11}{9}\) there, so the red curve made of them has height \(\tfrac29\), below the chalk curve.

  3. The factor \(x + 1\) is zero at \(x = -1\), not at \(x = 1\). Substitute \(x = -1\).

    \[7(-1) - 18 = -5B \ \Longrightarrow\ B = 5\]

  4. At \(x = 2\), the pieces now have height \(-1 + \tfrac53 = \tfrac23\), which is the height of the chalk curve.

Practice

Find \(A\) and \(B\) in \(\dfrac{x + 29}{(x - 3)(x + 5)} = \dfrac{A}{x - 3} + \dfrac{B}{x + 5}\).

Multiply both sides by the denominator. At the root of each factor the other term is 0, so each substitution finds one constant and grows one faint piece.

Solution

  1. Multiply both sides by the denominator.

    \[x + 29 = A(x + 5) + B(x - 3)\]

  2. Substitute \(x = 3\) to find \(A\). Substitute \(x = -5\) to find \(B\).

    \[\begin{gathered} 32 = 8A \Rightarrow A = 4 \\ 24 = -8B \Rightarrow B = -3 \end{gathered}\]

  3. Test by matching coefficients. The coefficient of \(x\) is \(A + B\), and the constant term is \(5A - 3B\).

    \[\begin{gathered} 4 + (-3) = 1 \\ 5\cdot 4 - 3\cdot(-3) = 29 \end{gathered}\]

Questions

The figure shows \(\dfrac{x - 6}{x(x - 2)} = \dfrac{A}{x} + \dfrac{B}{x - 2}\) as the sum of two pieces. The orange point is at \(x = 4\), and the heights of the two pieces are stacked there.

  1. The two pieces at the orange point have heights \(0.75\) and \(-1\). What is the height of the chalk curve at the point?

    • \(-0.25\)
    • \(1.75\)
    • \(0.75\)
    Answer

    \(-0.25\)

  2. Near the dashed line at \(x = 2\), on its right side, the chalk curve falls steeply. What does this show about \(B\) in the piece \(\dfrac{B}{x - 2}\)?

    • \(B\) is negative.
    • \(B\) is positive.
    • \(B\) is zero.
    Answer

    \(B\) is negative.

  3. Find \(A\) and \(B\) in \(\dfrac{5x - 1}{(x + 1)(x - 2)} = \dfrac{A}{x + 1} + \dfrac{B}{x - 2}\).

    • \(A = 2,\ B = 3\)
    • \(A = 3,\ B = 2\)
    • \(A = -6,\ B = 11\)
    • \(A = 2.5,\ B = 2.5\)
    Answer

    \(A = 2,\ B = 3\)

Summary

You multiply both sides by the denominator, then find each constant by substituting a root or by matching coefficients.

This works because the cleared equation is an identity, so it holds at every x, including the roots.

Use it when you know the form and need the constants before you integrate.

This is one of four skills in topic 6.12 of Which & Why: AP Calculus.