Factor Theorem

Section 1 of 11

Question. $x - 2$ and $x - 3$ are both factors of $\;x^{3} + ax + b = 0$. Find $a$, $b$ and the 3rd factor.

Plan

(i) Sub in
(ii) Divide
(iii) $x + k$

Section 2 of 11

(i) Sub in

$f(x) = x^{3} + ax + b$

$x - 2 = 0 \;\;\Rightarrow\;\; f(2) = 2^{3} + 2a + b = 0$

$\phantom{x - 2 = 0 \;\;\Rightarrow\;\; f(2) = {}} 8 + 2a + b = 0$

$2a + b = -8$

$x - 3 = 0 \;\;\Rightarrow\;\; f(3) = 3^{3} + 3a + b = 0$

$3a + b = -27$

$\begin{aligned} 3a + b &= -27 \\ 2a + b &= -8 \end{aligned}$ subtract

$a = -19$

$a = -19, \;\; b = 30$

Section 3 of 11

(ii) Divide

$x^{3} + 0x^{2} - 19x + 30 \;\;\div\;\; (x - 2)$

x² + 2x − 15 ──────────────────── x − 2 ) x³ + 0x² − 19x + 30 x³ − 2x² ─────── 2x² − 19x 2x² − 4x ──────── −15x + 30 −15x + 30 ───────── 0

Quotient: $x^{2} + 2x - 15$

Section 4 of 11

(iii) The 3rd Factor $(x + k)$

$x^{2} + 2x - 15$

$(x + 5)(x - 3)$

3rd factor: $x + 5$

Section 5 of 11

The Theorem

Warm-up. Solve $\;x^{2} - 2x - 8 = 0$.

$(x - 4)(x + 2) = 0$

$x = 4 \qquad x = -2$

Now let $\;f(x) = x^{2} - 2x - 8$ and check $f$ at those roots:

$f(4) = 16 - 8 - 8 = 0$

$f(-2) = 4 + 4 - 8 = 0$

$f(1) = 1 - 2 - 8 = -9$

At the roots, $f = 0$. At a non-root like $x = 1$, $f \ne 0$.

Factor Theorem

$f(x) = ax^{3} + bx^{2} + cx + d$. If $k \in \mathbb{R}$ is a value such that $f(k) = 0$ then $\;x - k\;$ is a factor ($x = k$ is a root).

Section 6 of 11

Irrational Roots

Question. Solve $\;x^{3} - 3x^{2} - 2x + 4 = 0$.

Trial and error (guess). Need a value of $x$ to give an answer of $0$.

$f(1) = 1 - 3 - 2 + 4 = 0$

$\Rightarrow\;\; x = 1$ is a root

$\Rightarrow\;\; x - 1$ is a factor

x² − 2x − 4 ───────────────── x − 1 ) x³ − 3x² − 2x + 4 x³ − x² ─────── −2x² − 2x −2x² + 2x ───────── −4x + 4 −4x + 4 ─────── 0

$x^{2} - 2x - 4 = 0$

$a = 1, \;\; b = -2, \;\; c = -4$

$x = \dfrac{2 \pm \sqrt{4 + 16}}{2}$

$x = \dfrac{2 \pm 2\sqrt{5}}{2}$

$x = 1 \pm \sqrt{5} \qquad$ irrational

$x = 1, \;\; x = 1 + \sqrt{5}, \;\; x = 1 - \sqrt{5}$

Section 7 of 11

Complex Roots

Question. The cubic equation $\;x^{3} - 4x^{2} + 9x - 10 = 0\;$ has one integer root and two complex roots. Find the three roots.

First root is a factor of the constant. Start at $-3$ and end at $3$.

$f(2) = 2^{3} - 4(2)^{2} + 9(2) - 10 = 8 - 16 + 18 - 10 = 0$

$\Rightarrow\;\; x = 2$ is a root $\;\Rightarrow\;\; x - 2$ is a factor

x² − 2x + 5 ────────────────── x − 2 ) x³ − 4x² + 9x − 10 x³ − 2x² ─────── −2x² + 9x −2x² + 4x ───────── 5x − 10 5x − 10 ─────── 0

$x^{2} - 2x + 5 = 0$

$x = \dfrac{2 \pm \sqrt{4 - 20}}{2} = \dfrac{2 \pm \sqrt{-16}}{2}$

$\sqrt{-16} = \sqrt{16} \cdot \sqrt{-1} = 4i$

$x = \dfrac{2 \pm 4i}{2} = 1 \pm 2i$

$x = 2, \;\; x = 1 + 2i, \;\; x = 1 - 2i$

Section 8 of 11

Find a Parameter

Question. $y = x^{3} - 3x + k$ has $\;x - 2\;$ as a factor. Find $k$, hence solve $\;x^{3} - 3x + k = 0$.

$x - 2 = 0 \;\Rightarrow\; x = 2$

$f(x) = x^{3} - 3x + k$

$f(2) = 8 - 6 + k = 0$

$k = -2$

Solve $\;x^{3} - 3x - 2 = 0$:

x² + 2x + 1 ────────────────── x − 2 ) x³ + 0x² − 3x − 2 x³ − 2x² ─────── 2x² − 3x 2x² − 4x ───────── x − 2 x − 2 ────── 0

$x^{2} + 2x + 1 = 0$

$(x + 1)(x + 1) = 0$

$x = -1$

$x = 2, \;\; x = -1 \;\;$ (twice)

Section 9 of 11

Two Factors — Sum & Product

Question. $x - 1$ and $x - 2$ are both factors of $\;y = x^{3} + px + q$. Find $p$, $q$ and the 3rd factor.

Plan

Factor ⟹ Sub in ⟹ Let $= 0$ ⟹ Divide

$f(1) = 1 + p + q = 0 \;\Rightarrow\; p + q = -1$

$f(2) = 8 + 2p + q = 0 \;\Rightarrow\; 2p + q = -8$

$\begin{aligned} \phantom{-}2p + q &= -8 \\ -(p + q) &= -(-1) \end{aligned}$ subtract

$p = -7$

$q = 6$

The two factors give a quadratic. Roots $x = 1, \;\; x = 2$:

$(x - 1)(x - 2) = 0$

$x^{2} - 2x - 1x + 2 = 0$

$x^{2} - 3x + 2 = 0$

Every quadratic: $\;x^{2} - (\text{sum of roots})x + \text{product of roots} = 0$.

Divide $\;x^{3} + 0x^{2} - 7x + 6\;$ by $\;x^{2} - 3x + 2$:

x + 3 ───────────────────── x² − 3x + 2 ) x³ + 0x² − 7x + 6 x³ − 3x² + 2x ───────────── 3x² − 9x + 6 3x² − 9x + 6 ──────────── 0

$p = -7, \;\; q = 6$. 3rd factor: $\;x + 3$

Section 10 of 11

Factor Structure I

Question. $x^{2} - px + q$ is a factor of $\;x^{3} + 3px^{2} + 3qx + r$.

(i) Show $q = -2p^{2}$. (ii) Show $r = -8p^{3}$. (iii) Find the three roots in terms of $p$.

The other factor must be linear, say $\;x + k$. Multiply:

$(x + k)(x^{2} - px + q) = x^{3} + 3px^{2} + 3qx + r$

$x^{3} - px^{2} + qx + kx^{2} - pkx + kq = x^{3} + 3px^{2} + 3qx + r$

Match coefficients.

$x^{2}: \;\; -p + k = 3p \;\;\Rightarrow\;\; k = 4p$

$x: \;\; q - pk = 3q$

$\phantom{x:\;\;} q - 4p^{2} = 3q$

$\phantom{x:\;\;} -4p^{2} = 2q \;\;\Rightarrow\;\; q = -2p^{2}$

constant: $\;\; kq = r$

$\phantom{\text{constant: }\;\;} 4pq = r$

$\phantom{\text{constant: }\;\;} r = 4p(-2p^{2}) = -8p^{3}$

(iii) The three roots.

$(x + k)(x^{2} - px - 2p^{2}) = 0$

$x + k = 0 \;\Rightarrow\; x = -k = -4p$

$x^{2} - px - 2p^{2} = 0$

$x^{2} - 2px + px - 2p^{2} = 0$

$x(x - 2p) + p(x - 2p) = 0$

$(x - 2p)(x + p) = 0$

$x = 2p \qquad x = -p$

$x = -4p, \;\; x = 2p, \;\; x = -p$

Section 11 of 11

Factor Structure II

Question. $x^{2} + bx + c$ is a factor of $\;x^{3} - p$. Show that (i) $\;b^{3} = p$ (ii) $\;c^{3} = p^{2}$.

The other factor is linear, say $\;x + k$. Multiply:

$(x + k)(x^{2} + bx + c) = x^{3} + 0x^{2} + 0x - p$

$x^{3} + bx^{2} + cx + kx^{2} + kbx + kc = x^{3} + 0x^{2} + 0x - p$

Match coefficients.

$x^{2}: \;\; b + k = 0 \;\;\Rightarrow\;\; k = -b$

$x: \;\; c + kb = 0 \;\;\Rightarrow\;\; c = -kb = b^{2}$

constant: $\;\; kc = -p \;\;\Rightarrow\;\; -bc = -p \;\;\Rightarrow\;\; p = bc$

The two key facts

$c = b^{2} \qquad p = bc$

(i) Show $b^{3} = p$.

$p = bc = b \cdot b^{2} = b^{3}$ ✓

(ii) Show $c^{3} = p^{2}$. The letter $b$ is gone, so eliminate $b$ from the two facts.

From $\;p = bc\;$: $b = \dfrac{p}{c}$

Sub into $\;c = b^{2}$:

$c = \!\left(\dfrac{p}{c}\right)^{\!2} = \dfrac{p^{2}}{c^{2}}$

$c^{3} = p^{2}$ ✓

That's the Factor Theorem.

One root by trial, divide to drop a degree, then solve what's left. From this comes every solvable cubic — real, irrational, complex, or expressed in a parameter.