Quick Recall
1. From Lesson 2: when $f(x) \div (x - a)$ has remainder $0$, what does that tell you about $(x - a)$?
2. Evaluate $f(3)$ for $f(x) = x^3 - 6x^2 + 11x - 6$. What do you notice?
3. How many steps of long division did it take to check this in Lesson 2?
The Wall
Is $(x - 7)$ a factor of $x^3 - 6x^2 + 11x - 6$? You could do a full long division — but that's several minutes of work just to get a yes/no answer. And if you need to check ten possible factors, that's ten full divisions. There must be a shortcut that reads off the answer from a single substitution.
The Need
We need a theorem that links the remainder of $f(x) \div (x - a)$ directly to the value $f(a)$ — no long division required.
This means you can find the remainder by substituting $x = a$ — no long division needed.
Green: dividend $f(x)$. Indigo (dashed): divisor $d(x) = x - a$. Amber dot: the point $(a, r)$ where $f(a) = r$.
This is just the remainder theorem with remainder $= 0$: if $f(a) = 0$, the division is exact.
Green curve: $f(x) = x^3 - 6x^2 + 11x - 6$. Amber dot: $(a, f(a))$. Adjust $a$ to find the values where $f(a) = 0$ — those give the factors $(x - a)$.
The linear factor is always written with integer coefficients (as $(ax-b)$), not as $(x - \frac{b}{a})$.
For example, $(2x - 3)$ is a factor $\iff f\!\left(\dfrac{3}{2}\right) = 0$.
Green curve: $f(x) = 2x^3 - 3x^2 - 8x + 12$. Amber dot: $(b/a,\, f(b/a))$. Adjust $a$ and $b$ to test whether $(ax-b)$ is a factor.
- Solving equations: Find one factor via the factor theorem, then divide to get a quadratic — factorise or use the formula to find all roots.
- Finding unknowns: If you know a factor (or remainder), substitute to build an equation in the unknown coefficient, then solve.
A student says: "If $f(2) = 0$, then $(x + 2)$ is a factor of $f(x)$." Is this correct? Think first.
Individual work (5 min): A student was asked to show that $(x - 3)$ is a factor of $x^3 - 5x^2 - 2x + 24$ and then solve $x^3 - 5x^2 - 2x + 24 = 0$. Their work is shown in 4 steps below — one of the steps contains the first error. Click the step you think is wrong.
The first error is in Step 1: $27 - 45 - 6 + 24 = 0$, not $1$. Since $f(3) = 0$, $(x - 3)$ is indeed a factor. Step 2 also contains a logic error — if $f(3) \neq 0$, then $(x - 3)$ would NOT be a factor — but the arithmetic error in Step 1 comes first.
Let $f(x) = x^3 + 2x^2 - 9x - 18$.
(i) Show that $f(-3) = 0$, $f(3) = 0$ and $f(-2) = 0$.
(ii) Factorise $f(x)$ completely.
(iii) Solve the equation $f(x) = 0$.
Use the factor theorem: substitute each given value into $f(x)$. If $f(a) = 0$, then $(x - a)$ is a factor. Once you have three factors, write $f(x)$ as a product.
Let $f(x) = x^3 + 2x^2 - 9x - 18$.
$$f(-3) = (-3)^3 + 2(-3)^2 - 9(-3) - 18 = -27 + 18 + 27 - 18 = 0$$
$$f(3) = (3)^3 + 2(3)^2 - 9(3) - 18 = 27 + 18 - 27 - 18 = 0$$
$$f(-2) = (-2)^3 + 2(-2)^2 - 9(-2) - 18 = -8 + 8 + 18 - 18 = 0$$
Since $f(-3) = 0$, $f(3) = 0$ and $f(-2) = 0$, by the factor theorem $(x + 3)$, $(x - 3)$ and $(x + 2)$ are all factors. ✓
Since we have three linear factors, and $f(x)$ is a cubic:
$$f(x) = (x + 3)(x - 3)(x + 2)$$
We can verify by long division. Divide $f(x)$ by $(x + 3)$:
Then $x^2 - x - 6 = (x - 3)(x + 2)$, giving $f(x) = (x + 3)(x - 3)(x + 2)$.
$$(x + 3)(x - 3)(x + 2) = 0$$
$$x = -3, \quad x = 3, \quad x = -2$$
$(x + 3)(x - 3)(x + 2) = (x^2 - 9)(x + 2) = x^3 + 2x^2 - 9x - 18$. ✓
(ii) $f(x) = (x + 3)(x - 3)(x + 2)$. (iii) $x = -3,\; x = 3,\; x = -2$.
Pair work (8 min): Explore the polynomial $f(x) = 2x^3 - x^2 - 5x - 2$ and find all its linear factors. The constant term is $2$ and the leading coefficient is $2$, so candidates for $\frac{b}{a}$ could be $\pm 1, \pm 2$ (integer values) and $\pm \frac{1}{2}$ (fractional values). Click each candidate to test.
The equation $\mathrm{f}(x) = x^3 - 4x^2 + x + 6 = 0$ has three integer roots. Solve $\mathrm{f}(x) = 0$.
Since the roots are integers, they must be factors of the constant term $6$. Test $\pm 1, \pm 2, \pm 3, \pm 6$ using the factor theorem. Once you find one factor, divide to get a quadratic and solve.
Test integer factors of the constant term $6$: $\pm 1, \pm 2, \pm 3, \pm 6$.
$f(1) = 1 - 4 + 1 + 6 = 4 \neq 0$
$f(-1) = -1 - 4 - 1 + 6 = 0$ ✓
By the factor theorem, $(x + 1)$ is a factor.
Divide $x^3 - 4x^2 + x + 6$ by $(x + 1)$:
$$x^2 - 5x + 6 = (x - 2)(x - 3)$$
$$\therefore f(x) = (x + 1)(x - 2)(x - 3)$$
$$(x + 1)(x - 2)(x - 3) = 0$$
$$x = -1,\; x = 2,\; x = 3$$
When $2x^3 + 3x^2 + kx - 6$ is divided by $(x + 1)$ the remainder is $7$. Find the value of $k$.
Use the remainder theorem: the remainder when dividing by $(x+1)$ is $f(-1)$. Set $f(-1) = 7$ and solve for $k$.
Let $f(x) = 2x^3 + 3x^2 + kx - 6$. By the remainder theorem, dividing by $(x + 1)$ gives remainder $f(-1)$:
$$f(-1) = 2(-1)^3 + 3(-1)^2 + k(-1) - 6 = -2 + 3 - k - 6 = -5 - k$$
Set this equal to the given remainder:
$$-5 - k = 7$$
$$k = -12$$
With $k = -12$: $f(-1) = -2 + 3 + 12 - 6 = 7$. ✓
$$k = -12$$
Test your understanding of all four lesson objectives. Each question targets one objective — try each before moving on!
When $f(x) = x^3 - 2x^2 + 5x - 7$ is divided by $(x - 2)$, the remainder is:
By the remainder theorem, the remainder of $f(x) \div (x - a)$ is $f(a)$. Here $a = 2$: $f(2) = 2^3 - 2(2)^2 + 5(2) - 7 = 8 - 8 + 10 - 7 = 3$.
A student tests whether $(x - 5)$ is a factor of $f(x) = x^3 - 4x^2 + x + 6$. Which calculation correctly applies the factor theorem?
By the factor theorem, $(x - a)$ is a factor $\iff f(a) = 0$. Here $a = 5$: $f(5) = 125 - 100 + 5 + 6 = 36 \neq 0$, so $(x - 5)$ is NOT a factor. Option C invents a false zero; option B uses the wrong sign for $a$; option D substitutes $0$ instead of $5$.
To test whether $(2x - 3)$ is a factor of $f(x)$, which value should you substitute into $f$?
For $(ax - b)$ to be a factor, we need $f\!\left(\dfrac{b}{a}\right) = 0$. Here $(2x - 3) = 0 \Rightarrow x = \dfrac{3}{2}$, so substitute $x = \dfrac{3}{2}$ (equivalently $\dfrac{b}{a} = \dfrac{3}{2}$).
When $f(x) = x^3 + kx^2 - 5x + 3$ is divided by $(x - 2)$, the remainder is $9$. Find $k$.
By the remainder theorem, $f(2) = 9$. Substituting: $f(2) = 8 + 4k - 10 + 3 = 4k + 1 = 9 \Rightarrow 4k = 8 \Rightarrow k = 2$. (Check: with $k = 2$, $f(2) = 8 + 8 - 10 + 3 = 9$. ✓)
You are given that $\mathrm{f}(x) = x^3 - 19x + 30$.
(i) Calculate $\mathrm{f}(0)$ and $\mathrm{f}(3)$. Hence write down a factor of $\mathrm{f}(x)$.
(ii) Find $p$ and $q$ such that $\mathrm{f}(x) \equiv (x - 2)(x^2 + px + q)$.
(iii) Solve the equation $x^3 - 19x + 30 = 0$.
$f(0) = 0 - 0 + 30 = 30$.
$f(3) = 27 - 57 + 30 = 0$.
Since $f(3) = 0$, by the factor theorem $(x - 3)$ is a factor.
Divide $x^3 - 19x + 30$ by $(x - 2)$ (since the question uses $(x-2)$):
Check: $f(2) = 8 - 38 + 30 = 0$. ✓ So $(x - 2)$ is also a factor.
So $f(x) = (x - 2)(x^2 + 2x - 15)$, giving $p = 2$, $q = -15$.
$x^2 + 2x - 15 = (x + 5)(x - 3)$.
So $f(x) = (x - 2)(x - 3)(x + 5) = 0$.
$f(-5) = -125 + 95 + 30 = 0$. ✓
(i) $f(0) = 30$, $f(3) = 0$; factor is $(x - 3)$. (ii) $p = 2$, $q = -15$. (iii) $x = 2, \; x = 3, \; x = -5$.
(i) Show that $x = 2$ is a root of the equation $x^4 - 5x^2 + 2x = 0$ and write down another integer root.
(ii) Find the other two roots of the equation $x^4 - 5x^2 + 2x = 0$.
$f(2) = 16 - 20 + 4 = 0$. ✓ So $x = 2$ is a root.
Another integer root: $f(0) = 0$, so $x = 0$ is also a root.
Factor out $x$: $x^4 - 5x^2 + 2x = x(x^3 - 5x + 2)$.
Since $x = 2$ is a root of $x^3 - 5x + 2$, divide by $(x - 2)$:
So $x^3 - 5x + 2 = (x - 2)(x^2 + 2x - 1)$.
Solve $x^2 + 2x - 1 = 0$ using the quadratic formula:
$$x = \dfrac{-2 \pm \sqrt{4 + 4}}{2} = \dfrac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2}$$
(i) $x = 2$ is a root; another integer root is $x = 0$. (ii) $x = -1 + \sqrt{2}$ and $x = -1 - \sqrt{2}$.
(i) The polynomial $\mathrm{p}(x) = x^3 - 6x^2 + 9x + k$ has a factor $(x - 4)$. Find the value of $k$.
(ii) Find the other factors of the polynomial.
Use the factor theorem: since $(x-4)$ is a factor, $p(4) = 0$. This gives an equation in $k$.
Since $(x - 4)$ is a factor, $p(4) = 0$:
$$p(4) = 64 - 96 + 36 + k = 4 + k = 0$$
$$k = -4$$
With $k = -4$: $p(x) = x^3 - 6x^2 + 9x - 4$. Divide by $(x - 4)$:
So $p(x) = (x - 4)(x^2 - 2x + 1) = (x - 4)(x - 1)^2$.
$p(1) = 1 - 6 + 9 - 4 = 0$. ✓ (confirms $(x-1)$ is a factor)
(i) $k = -4$. (ii) $p(x) = (x - 4)(x - 1)^2$; the other factor is $(x - 1)^2$.
When $x^3 + px^2 + p^2x - 36$ is divided by $(x - 3)$ the remainder is $21$. Find a possible value of $p$.
Use the remainder theorem: $f(3) = 21$. This gives a quadratic equation in $p$.
Let $f(x) = x^3 + px^2 + p^2x - 36$. By the remainder theorem, $f(3) = 21$:
$$f(3) = 27 + 9p + 3p^2 - 36 = 3p^2 + 9p - 9 = 21$$
$$3p^2 + 9p - 30 = 0$$
$$p^2 + 3p - 10 = 0$$
$$(p + 5)(p - 2) = 0$$
$$p = 2 \quad \text{or} \quad p = -5$$
$p = 2$: $f(3) = 27 + 18 + 12 - 36 = 21$. ✓
$p = -5$: $f(3) = 27 - 45 + 75 - 36 = 21$. ✓
$$p = 2 \quad (\text{or } p = -5)$$
When $x^3 + ax^2 + bx + 8$ is divided by $(x - 3)$ the remainder is $2$ and when it is divided by $(x + 1)$ the remainder is $-2$. Find $a$ and $b$ and hence obtain the remainder on dividing by $(x - 2)$.
Apply the remainder theorem twice: $f(3) = 2$ and $f(-1) = -2$. This gives two simultaneous equations in $a$ and $b$.
Let $f(x) = x^3 + ax^2 + bx + 8$.
$f(3) = 2$: $27 + 9a + 3b + 8 = 2 \;\Rightarrow\; 9a + 3b = -33 \;\Rightarrow\; 3a + b = -11$ ... (1)
$f(-1) = -2$: $-1 + a - b + 8 = -2 \;\Rightarrow\; a - b = -9$ ... (2)
From (2): $b = a + 9$. Substitute into (1):
$$3a + (a + 9) = -11 \;\Rightarrow\; 4a = -20 \;\Rightarrow\; a = -5$$
$$b = -5 + 9 = 4$$
With $a = -5$, $b = 4$: $f(x) = x^3 - 5x^2 + 4x + 8$.
By the remainder theorem, remainder $= f(2)$:
$$f(2) = 8 - 20 + 8 + 8 = 4$$
$f(3) = 27 - 45 + 12 + 8 = 2$. ✓ $f(-1) = -1 - 5 - 4 + 8 = -2$. ✓
$a = -5$, $b = 4$. The remainder on dividing by $(x - 2)$ is $\mathbf{4}$.
Exit Ticket
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