Work through in order. Each part changes one thing and keeps the rest still, or keeps the maths still and changes how it looks. Then it asks what you noticed.
This sheet is about
A bracket $(x - a)$ is a factor of $f(x)$ exactly when substituting $x = a$ gives zero; the value decides, not the look of the bracket or the cubic.
Watch for
which value the bracket sends you to · zero exactly, or not a factor · a non-zero value is the remainder · the letter and the order of terms
①One problem, no method yetStart here
1
A box has volume $x^3 - 2x^2 - 5x + 6$ cubic centimetres. Two of its edges are $(x - 1)$ cm and $(x + 2)$ cm, and the third edge is also a bracket of the form $(x + c)$. Find the third edge, and show that the three edges multiply to give the volume.
②Which brackets are factors?Spot the difference
Changes: the sign inside the bracketStays the same: the cubic, and the number in the bracket
2
$f(x) = x^3 - 2x^2 - 5x + 6$. Tick every bracket that is a factor of $f(x)$.
For each bracket you ticked, write the value of $x$ you substituted.
$(x + 2)$ is a factor and $(x - 2)$ is not. What did you substitute for each one, and why is it not the same number?
③Which reasoning actually proves it?Spot the difference
Changes: what is offered as the proofStays the same: the cubic and the bracket $(x - 2)$
3
Four students decide whether $(x - 2)$ is a factor of $f(x) = x^3 - 2x^2 - 5x + 6$. Tick every line of reasoning that is correct.
$f(2) = -4$, not zero: not a factor$2$ divides the constant $6$: a factor$f(2) = -4$ is nearly zero: nearly a factorthere is a $-2x^2$ term: a factor
What would $f(2)$ have to equal for $(x - 2)$ to be a factor?
Three lines mention the number $2$ in a way that looks relevant. What is the only thing about $2$ that matters?
④Only the constant changesOne thing changes
Changes: the constant term $c$Stays the same: the terms $x^3 - 2x^2$ and the divisor $(x - 2)$
4
Find the remainder when $f(x) = x^3 - 2x^2 + c$ is divided by $(x - 2)$, and say whether $(x - 2)$ is a factor.
$c = 0$
remainder $=$
$c = 5$
remainder $=$
$c = -3$
remainder $=$
The remainder equals $c$ every time. Why does that happen here? Which $c$ makes $(x - 2)$ a factor?
⑤Same bracket, three different cubicsSame answer, new clothes
Changes: the cubicStays the same: the bracket $(x + 2)$, so the value substituted is $-2$
5
$(x + 2)$ is a factor of each cubic. Show it by substitution.
$x^3 + 2x^2 - x - 2$
$f(-2) =$
$2x^3 + 3x^2 - 2x$
$f(-2) =$
$x^3 + 8$
$f(-2) =$
The cubics look nothing alike. What did you do identically all three times?
⑥One fact in four languagesSame answer, new clothes
Changes: how the fact is said: value, bracket, solution, graphStays the same: the fact $f(3) = 0$
6
The sketch shows $y = f(x)$ for $f(x) = x^3 - 2x^2 - 5x + 6$. Tick every statement that says the same thing as $f(3) = 0$.
$(x - 3)$ is a factor of $f(x)$$x = 3$ is a solution of $f(x) = 0$the curve crosses the $x$-axis at $x = 3$$(x + 3)$ is a factor of $f(x)$
Read the other two crossings from the sketch and write $f(x)$ as a product of three brackets.
Which of the four could you check by a ten-second substitution, and why do the other three then follow?
⑦Three letters, three orders, one cubicSame answer, new clothes
Changes: the letter and the order the terms are written inStays the same: the cubic and the bracket being tested
7
Show that $(t - 1)$ is a factor of $p$, $(y - 1)$ is a factor of $g$ and $(w - 1)$ is a factor of $h$.
$p(t) = t^3 - 2t^2 - 5t + 6$
$p(1) =$
$g(y) = 6 - 5y - 2y^2 + y^3$
$g(1) =$
$h(w) = w^3 + 6 - 5w - 2w^2$
$h(1) =$
What changed between the three? Did any of it change the value you got?
⑧Factor or not, and the remainder, togetherAll at once
Changes: the bracket, its sign and whether the value is zero, all at onceStays the same: the cubic $x^3 - 2x^2 - 5x$ plus a constant
8
$f(x) = x^3 - 2x^2 - 5x + 6$. Tick every statement that is true.
divided by $(x - 1)$: remainder $0$divided by $(x + 1)$: remainder $0$divided by $(x - 2)$: remainder $4$$(x + 2)$ is a factor
Correct each false statement.
9
$g(x) = x^3 - 2x^2 - 5x + k$. Find the value of $k$ for which $(x + 1)$ is a factor of $g(x)$. For that value of $k$, find the remainder when $g(x)$ is divided by $(x - 2)$.
Each statement needs three things: the right value, is it zero, and its sign if not. Which did each false one get wrong?
Teacher page · Factor Theorem
Variation Theory · answers
Object of learning: A bracket $(x - a)$ is a factor of $f(x)$ exactly when substituting $x = a$ gives zero; the value decides, not the look of the bracket or the cubic.
which value the bracket sends you to
zero exactly, or not a factor
a non-zero value is the remainder
the letter and the order of terms(something to disregard)
① One problem, no method yetStart here
1. $(x - 3)$ cm$(x - 1)(x + 2) = x^2 + x - 2$; the constant of the volume is $6 = -2 \times c$, so $c = -3$. Check: $(x^2 + x - 2)(x - 3) = x^3 - 2x^2 - 5x + 6$. Taken before the theorem is named; the student reasons from expansion.
Draw out: The bracket $(x + 2)$ is zero when $x = -2$, so the test is $f(-2)$, never $f(2)$. The sign between the bracket and the value substituted always flips; the pairs $(x + 1)$, $(x - 1)$ and $(x + 3)$, $(x - 3)$ show it is the value, not the digit, that decides.
③ Which reasoning actually proves it?Spot the difference
3. only the first; $f(2)$ would have to be exactly $0$$f(2) = 8 - 8 - 10 + 6 = -4$. Dividing the constant is necessary for a whole-number root but not sufficient ($2$ divides $6$ and $f(2)$ is not $0$); "nearly a factor" is not a thing; a $-2x^2$ term says nothing about $(x - 2)$.
Draw out: Being a factor is a claim about the VALUE $f(2)$, not about $2$ appearing in the cubic or dividing its constant. The divisors of the constant are only the whole-number CANDIDATES ($1$ and $3$ pass, $2$ fails); zero exactly, or not a factor. The AQA mark scheme gives M0 for anything that is not the substitution.
Draw out: At $x = 2$ the terms $x^3 - 2x^2$ cancel to $0$, so the remainder is exactly $c$, sign included. A remainder of $0$ is the factor theorem; a remainder of $-3$ is $-3$, not $3$.
⑤ Same bracket, three different cubicsSame answer, new clothes
Draw out: The bracket fixes the value ($x = -2$) whatever the cubic looks like; the number of terms, a leading coefficient of $2$, a missing constant, none of it changes the test.
⑥ One fact in four languagesSame answer, new clothes
6. the first three; $f(x) = (x + 2)(x - 1)(x - 3)$$f(3) = 27 - 18 - 15 + 6 = 0$ and $f(-3) = -24$, so $(x + 3)$ is the odd one out. Crossings at $-2$, $1$, $3$; check $(x + 2)(x - 1)(x - 3) = (x^2 + x - 2)(x - 3) = x^3 - 2x^2 - 5x + 6$.
Draw out: $f(3) = 0$, "$(x - 3)$ is a factor", "$x = 3$ solves $f(x) = 0$" and "the curve crosses at $3$" are one fact in four representations. The crossing at $-2$ gives $(x + 2)$: the sign flip is visible on the graph too.
⑦ Three letters, three orders, one cubicSame answer, new clothes
Draw out: The letter names the input and the order of the terms is a choice of how to write it; neither is the maths. The same four numbers are added each time, so the value and the factor are unchanged.
⑧ Factor or not, and the remainder, togetherAll at once
8. first and fourth true; the $(x + 1)$ remainder is $8$, the $(x - 2)$ remainder is $-4$$f(1) = 0$; $f(-1) = -1 - 2 + 5 + 6 = 8$; $f(2) = 8 - 8 - 10 + 6 = -4$ (sign matters); $f(-2) = -8 - 8 + 10 + 6 = 0$.
9. k = -2; remainder -12$(x + 1)$ means $g(-1) = 0$: $-1 - 2 + 5 + k = 2 + k = 0$, so $k = -2$. Then $g(2) = 8 - 8 - 10 - 2 = -12$. Check $g(-1) = -1 - 2 + 5 - 2 = 0$.
Draw out: Three aspects at once: the bracket $(x + 1)$ sends you to $-1$ (sign), only an exact $0$ makes a factor (zero), and a non-zero value is the remainder with its sign kept (remainder). The second false statement is a pure sign slip on the remainder, the error misconceptions.json records most often for this topic.
Factor Theorem