Add Maths · 3 Factors of polynomials

3 Factors of polynomials

Before you start

Expanding brackets such as $(x - 2) (x^{2} + 4 x + 3)$ and collecting like terms.
Factorising a quadratic, including one whose $x^{2}$ coefficient is not 1, such as $2 x^{2} - 7 x + 3 = (2 x - 1) (x - 3)$ .
The quadratic formula (it is on the List of formulas), the discriminant $b^{2} - 4 a c$ , and simplifying surds such as $\sqrt{8} = 2 \sqrt{2}$ .
Substituting negatives and fractions by hand: ${(- 2)}^{3} = - 8$ , ${(- 2)}^{2} = 4$ , ${(\frac{1}{2})}^{3} = \frac{1}{8}$ .

Syllabus objectives covered (3)

Know and use the remainder and factor theorems.
Find factors of polynomials.
Solve cubic equations.

Part 1 of 7

Polynomials, and what p(a) means

A polynomial in $x$ is a sum of terms like $a x^{n}$ where every power $n$ is a whole number $0, 1, 2, \dots$ . Its degree is the highest power. In this topic it is almost always a cubic, degree 3, such as $p (x) = x^{3} + 2 x^{2} - 5 x + 1$ .
Naming it $p (x)$ lets you write 'substitute' in four characters. $p (2)$ means replace every $x$ by $2$ : $p (2) = 8 + 8 - 10 + 1 = 7$ .
Put negatives in brackets, because the powers decide the sign: $p (- 2) = {(- 2)}^{3} + 2 {(- 2)}^{2} - 5 (- 2) + 1 = - 8 + 8 + 10 + 1 = 11$ . An odd power keeps the minus; an even power removes it.
Two shortcuts: $p (0)$ is just the constant term, and $p (1)$ is the sum of the coefficients. Here $p (1) = 1 + 2 - 5 + 1 = - 1$ .

Part 2 of 7

Dividing a polynomial by (x − a)

Algebraic long division copies long division of numbers: divide the leading terms, multiply back, subtract, bring down the next term, repeat. You end with a quotient and a remainder.
Divide $x^{3} + 2 x^{2} - 5 x + 1$ by $(x - 2)$ . Leading terms first: $x^{3} \div x = x^{2}$ . That is the first term of the quotient.
Multiply the whole divisor by $x^{2}$ and subtract. Write the bracket, because subtracting $- 2 x^{2}$ is where signs go wrong.
$(x^{3} + 2 x^{2}) - (x^{3} - 2 x^{2}) = 4 x^{2}$
Bring down $- 5 x$ to get $4 x^{2} - 5 x$ . Then $4 x^{2} \div x = 4 x$ ; multiply back, $4 x (x - 2) = 4 x^{2} - 8 x$ ; subtract: $(4 x^{2} - 5 x) - (4 x^{2} - 8 x) = 3 x$ .
Bring down $+ 1$ to get $3 x + 1$ . Then $3 x \div x = 3$ ; multiply back, $3 (x - 2) = 3 x - 6$ ; subtract: $(3 x + 1) - (3 x - 6) = 7$ . Nothing is left to bring down, so $7$ is the remainder.
Write the result as an identity. Expanding the right-hand side gives the left, which is your check.
$x^{3} + 2 x^{2} - 5 x + 1 \equiv (x - 2) (x^{2} + 4 x + 3) + 7$
A missing power needs a placeholder. To divide $x^{3} - 7 x + 6$ , first write it as $x^{3} + 0 x^{2} - 7 x + 6$ . Without the $0 x^{2}$ the terms slide out of line and every later subtraction is wrong.

Part 3 of 7

The remainder theorem

Look again at $p (x) \equiv (x - 2) (x^{2} + 4 x + 3) + 7$ and put $x = 2$ . The bracket $(x - 2)$ becomes $0$ , the whole product vanishes, and only the remainder survives: $p (2) = 7$ . That is the value you got by plain substitution in the first section.
This works for every polynomial and every divisor $(x - a)$ . It is the remainder theorem. It is not on the List of formulas, so you must know it.
$When p (x) is divided by (x - a), the remainder is p (a) .$
So you never need to divide just to find a remainder. Find the value of $x$ that makes the divisor zero, and substitute it.
The sign trap: dividing by $(x + 3)$ means $x + 3 = 0$ , so $x = - 3$ and the remainder is $p (- 3)$ , not $p (3)$ .
The same idea works for a divisor like $(2 x + 1)$ . It is zero when $x = - \frac{1}{2}$ , so the remainder is $p (- \frac{1}{2})$ . For $p (x) = 4 x^{3} - 2 x + 5$ that is $4 (- \frac{1}{8}) + 1 + 5 = \frac{11}{2}$ .
Why is the remainder always a number, with no $x$ in it? You keep dividing until what is left has a lower degree than the divisor, and anything of lower degree than $(x - a)$ is a constant.

Part 4 of 7

The factor theorem

A factor divides exactly, leaving remainder zero. Combine that with the remainder theorem and you get the factor theorem, which you must also recall.
$(x - a) is a factor of p (x) ⟺ p (a) = 0$
It runs both ways. If $p (a) = 0$ , then $(x - a)$ is a factor. If you are told $(x - a)$ is a factor, you may write $p (a) = 0$ as an equation.
For a factor like $(a x - b)$ , test $p (\frac{b}{a})$ . So $(2 x - 1)$ is a factor of $p (x) = 2 x^{3} - 3 x^{2} - 11 x + 6$ because $p (\frac{1}{2}) = \frac{1}{4} - \frac{3}{4} - \frac{11}{2} + 6 = 0$ .
The 'told it is a factor' direction is how unknown coefficients are found. Each fact you are given ('is a factor', 'leaves remainder 5') becomes one equation. Two unknowns need two facts, and then you solve the two equations simultaneously.
When a question says Show that $(x - a)$ is a factor, write $p (a)$ out term by term, reach $0$ , and finish with the conclusion: 'so $(x - a)$ is a factor'. The arithmetic alone does not show the result.

Part 5 of 7

Finding a first factor by trial

To factorise a cubic from scratch you need one factor to start with. Find it by testing values of $x$ , but not at random.
If $(x - a)$ is a factor and the $x^{3}$ coefficient is $1$ , then $a$ must divide the constant term. Here is why: in $(x - a) (x^{2} + b x + c)$ the constant term is $- a c$ , which is a multiple of $a$ .
So list the factors of the constant, with both signs. For $x^{3} - 4 x^{2} + x + 6$ the candidates are $\pm 1, \pm 2, \pm 3, \pm 6$ . Try the small ones first: the arithmetic is quicker.
Try $x = 1$ : $1 - 4 + 1 + 6 = 4$ , not zero. Try $x = - 1$ : $- 1 - 4 - 1 + 6 = 0$ . So $(x + 1)$ is a factor. Stop there. One linear factor is enough, because the rest comes from the quadratic.
If the $x^{3}$ coefficient is not $1$ , a root can be a fraction: $\pm \frac{factor of the constant}{factor of the x^{3} coefficient}$ . For $2 x^{3} + \dots + 6$ that adds $\pm \frac{1}{2}$ and $\pm \frac{3}{2}$ to the list. Try the whole numbers first.
A non-zero result is not wasted. It is the remainder for that divisor, and questions often ask for exactly that.

Part 6 of 7

From one factor to the quadratic factor

The syllabus expects you first to write the cubic as a linear factor times a quadratic factor. There are two reliable ways to get the quadratic. Use the one you make fewer slips with.
Method 1: algebraic long division, exactly as above. Divide by the factor you found. The remainder must come out as $0$ , which checks your arithmetic for free.
Method 2: comparing coefficients. Take $x^{3} - 7 x + 6$ . Since $p (1) = 1 - 7 + 6 = 0$ , $(x - 1)$ is a factor. Write the quadratic factor with unknown coefficients.
$x^{3} + 0 x^{2} - 7 x + 6 \equiv (x - 1) (x^{2} + b x + c)$
The bracket starts with $x^{2}$ because $x \times x^{2}$ has to give the $x^{3}$ on the left. Now match the constant terms: $(- 1) \times c = 6$ , so $c = - 6$ .
Match the $x^{2}$ terms. On the right they come from $x \times b x$ and $(- 1) \times x^{2}$ , giving $b - 1$ . The left has $0 x^{2}$ , so $b - 1 = 0$ and $b = 1$ .
Use the $x$ terms as a check. On the right they are $x \times c$ and $(- 1) \times b x$ , giving $c - b = - 6 - 1 = - 7$ . That matches the $- 7 x$ on the left.
$x^{3} - 7 x + 6 = (x - 1) (x^{2} + x - 6)$
Factorise fully. In this syllabus that means you are not finished while any factor can still be split: $x^{2} + x - 6 = (x + 3) (x - 2)$ , so $x^{3} - 7 x + 6 = (x - 1) (x + 3) (x - 2)$ .

Part 7 of 7

Solving a cubic equation

To solve $p (x) = 0$ : make one side zero, factorise, then use the zero-product rule. A product is zero only if one of its factors is zero.
Three linear factors give three roots. From $(x - 1) (x + 3) (x - 2) = 0$ : $x = 1$ , $x = - 3$ or $x = 2$ .
The roots are where the curve $y = p (x)$ crosses the $x$ -axis, one crossing per linear factor. For $2 x^{3} - 3 x^{2} - 11 x + 6 = (x + 2) (2 x - 1) (x - 3)$ the crossings are at $x = - 2$ , $x = \frac{1}{2}$ and $x = 3$ , and the curve meets the $y$ -axis at the constant term, $6$ .
Each linear factor is one crossing of the $x$ -axis, and the $y$ -intercept is the constant term.
A cubic with a positive $x^{3}$ coefficient comes up from the bottom left and leaves at the top right, so it has to cross the $x$ -axis somewhere. That is why every cubic has at least one real root.
If the quadratic factor does not factorise, use the quadratic formula on it and leave the roots as surds. Paper 1 has no calculator: $1 + \sqrt{2}$ is the finished answer, not a decimal.
If the quadratic factor has $b^{2} - 4 a c < 0$ , it has no real roots, and the cubic has only the one root from the linear factor. Show the discriminant and say so. For $(x - 2) (x^{2} + 3) = 0$ , the discriminant of $x^{2} + 3$ is $0 - 12 = - 12 < 0$ , so $x = 2$ is the only real solution.
**Never divide both sides by $x$ ** (or by a bracket) to shorten a cubic. Dividing $x^{3} = 4 x$ by $x$ gives $x^{2} = 4$ and loses the root $x = 0$ . Instead write $x^{3} - 4 x = 0$ , so $x (x - 2) (x + 2) = 0$ and $x = 0$ , $2$ or $- 2$ .

Worked examples

Example 1. Find the remainder when $p (x) = 2 x^{3} + 5 x^{2} - 4 x + 3$ is divided by (a) $(x + 3)$ (b) $(2 x - 1)$ .

Show the working

(a) The divisor is zero when $x = - 3$ , so by the remainder theorem the remainder is $p (- 3)$ . Bracket the negative.
Each power of $- 3$ in turn: ${(- 3)}^{3} = - 27$ and ${(- 3)}^{2} = 9$ .
$p (- 3) = 2 (- 27) + 5 (9) - 4 (- 3) + 3 = - 54 + 45 + 12 + 3 = 6$
(b) $2 x - 1 = 0$ when $x = \frac{1}{2}$ , so the remainder is $p (\frac{1}{2})$ . Work the fractions by hand.
${(\frac{1}{2})}^{3} = \frac{1}{8}$ and ${(\frac{1}{2})}^{2} = \frac{1}{4}$ .
$p (\frac{1}{2}) = \frac{2}{8} + \frac{5}{4} - 2 + 3 = \frac{1}{4} + \frac{5}{4} + 1 = \frac{5}{2}$
Neither remainder is zero, so neither divisor is a factor. That comes free with the answer.

Answer: (a) $6$ (b) $\frac{5}{2}$

Example 2. The polynomial $p (x) = x^{3} + a x^{2} + b x - 6$ has $(x - 2)$ as a factor, and leaves a remainder of $- 12$ when divided by $(x + 1)$ . (a) Find $a$ and $b$ . (b) Solve $p (x) = 0$ .

Show the working

Factor theorem: $(x - 2)$ is a factor, so $p (2) = 0$ . Simplify straight away.
$8 + 4 a + 2 b - 6 = 0 \Rightarrow 2 a + b = - 1$
Remainder theorem: dividing by $(x + 1)$ means substituting $x = - 1$ , and the result is $- 12$ .
$- 1 + a - b - 6 = - 12 \Rightarrow a - b = - 5$
Solve the pair. The $b$ terms have opposite signs, so add the equations: $3 a = - 6$ , giving $a = - 2$ . Then $b = - 1 - 2 a = - 1 + 4 = 3$ .
So $p (x) = x^{3} - 2 x^{2} + 3 x - 6$ . Check both facts before moving on: $p (2) = 8 - 8 + 6 - 6 = 0$ and $p (- 1) = - 1 - 2 - 3 - 6 = - 12$ . Both hold.
(b) You already know $(x - 2)$ is a factor. Use new letters for the quadratic, so they cannot be confused with $a$ and $b$ .
$x^{3} - 2 x^{2} + 3 x - 6 \equiv (x - 2) (x^{2} + q x + r)$
Constants: $- 2 r = - 6$ , so $r = 3$ . $x^{2}$ terms: $q - 2 = - 2$ , so $q = 0$ . Check the $x$ terms: $r - 2 q = 3$ , which matches $+ 3 x$ .
$p (x) = (x - 2) (x^{2} + 3)$
The quadratic factor has discriminant $0^{2} - 4 (1) (3) = - 12 < 0$ , so $x^{2} + 3 = 0$ has no real roots. Only the linear factor gives a solution.

Answer: (a) $a = - 2$ , $b = 3$ . (b) $x = 2$ is the only real solution, since $x^{2} + 3 = 0$ has no real roots.

Example 3. Solve $2 x^{3} + x^{2} - 13 x + 6 = 0$ .

Show the working

Candidates: the factors of $6$ , which are $\pm 1, \pm 2, \pm 3, \pm 6$ . Halves are possible too, because the $x^{3}$ coefficient is $2$ . Start with the small whole numbers.
$p (1) = 2 + 1 - 13 + 6 = - 4$ and $p (- 1) = - 2 + 1 + 13 + 6 = 18$ . Neither is zero.
$p (2) = 16 + 4 - 26 + 6 = 0$ , so $(x - 2)$ is a factor.
Divide by $(x - 2)$ . First $2 x^{3} \div x = 2 x^{2}$ ; subtract $2 x^{2} (x - 2) = 2 x^{3} - 4 x^{2}$ from $2 x^{3} + x^{2}$ to leave $5 x^{2}$ , and bring down $- 13 x$ .
Next $5 x^{2} \div x = 5 x$ ; subtract $5 x (x - 2) = 5 x^{2} - 10 x$ to leave $- 3 x$ , and bring down $+ 6$ . Then $- 3 x \div x = - 3$ ; subtract $- 3 (x - 2) = - 3 x + 6$ to leave $0$ . The zero remainder confirms the factor.
$2 x^{3} + x^{2} - 13 x + 6 = (x - 2) (2 x^{2} + 5 x - 3)$
Factorise the quadratic. You need two numbers with product $2 \times (- 3) = - 6$ and sum $5$ : they are $6$ and $- 1$ . So $2 x^{2} + 5 x - 3 = (2 x - 1) (x + 3)$ .
$(x - 2) (2 x - 1) (x + 3) = 0$
Check a root you did not test, $x = - 3$ : $2 (- 27) + 9 + 39 + 6 = 0$ . Correct.

Answer: $x = 2$ , $x = \frac{1}{2}$ or $x = - 3$ .

Example 4. Solve $x^{3} + x^{2} - 7 x - 3 = 0$ , giving your answers in exact form.

Show the working

The candidates are $\pm 1$ and $\pm 3$ . $p (1) = - 8$ , $p (- 1) = 4$ , $p (3) = 12$ , and $p (- 3) = - 27 + 9 + 21 - 3 = 0$ . So $(x + 3)$ is a factor.
Compare coefficients in $x^{3} + x^{2} - 7 x - 3 \equiv (x + 3) (x^{2} + q x + r)$ . Constants: $3 r = - 3$ , so $r = - 1$ . $x^{2}$ terms: $q + 3 = 1$ , so $q = - 2$ .
Check with the $x$ terms: $r + 3 q = - 1 - 6 = - 7$ . It matches.
$x^{3} + x^{2} - 7 x - 3 = (x + 3) (x^{2} - 2 x - 1)$
No pair of whole numbers multiplies to $- 1$ and adds to $- 2$ , so $x^{2} - 2 x - 1$ does not factorise. Use the quadratic formula with $a = 1$ , $b = - 2$ , $c = - 1$ .
$x = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm 2 \sqrt{2}}{2} = 1 \pm \sqrt{2}$
Simplify the surd fully ( $\sqrt{8} = 2 \sqrt{2}$ ) and cancel the $2$ . Leave the roots exact. On a non-calculator paper that is the finished answer.

Answer: $x = - 3$ , $x = 1 + \sqrt{2}$ or $x = 1 - \sqrt{2}$ .

Where people go wrong

Mistake: Finding the remainder on division by $(x + 2)$ by working out $p (2)$ .

Instead: Set the divisor equal to zero: $x + 2 = 0$ gives $x = - 2$ , so the remainder is $p (- 2)$ .

The theorem says 'divided by $(x - a)$ , remainder $p (a)$ ', and $(x + 2)$ is $(x - (- 2))$ , so $a = - 2$ .

Mistake: Testing whether $(2 x - 1)$ is a factor by working out $p (1)$ or $p (2)$ .

Instead: $2 x - 1 = 0$ when $x = \frac{1}{2}$ , so test $p (\frac{1}{2})$ .

Only the value that makes the divisor zero makes the product term vanish and leaves the remainder.

Mistake: Leaving out a missing power in long division, e.g. dividing $x^{3} - 7 x + 6$ with no $x^{2}$ term.

Instead: Write $x^{3} + 0 x^{2} - 7 x + 6$ before you start.

Each subtraction lines up terms of the same power. Without the placeholder, $- 7 x$ ends up subtracted from an $x^{2}$ term.

Mistake: Stopping at $(x - 1) (x^{2} + x - 6)$ when asked to factorise fully or to solve.

Instead: Check whether the quadratic factor splits: $(x - 1) (x + 3) (x - 2)$ .

'Factorise' means factorise fully in this syllabus, and an unsplit quadratic hides up to two roots.

Mistake: Dividing both sides of $x^{3} = 4 x$ by $x$ .

Instead: Bring everything to one side and take out the common factor: $x (x^{2} - 4) = 0$ .

Dividing by $x$ quietly assumes $x \neq 0$ , which throws away the solution $x = 0$ .

Check yourself

Find the remainder when $x^{3} - 2 x + 7$ is divided by $(x - 3)$ .
Answer
Substitute $x = 3$ : $27 - 6 + 7 = 28$ .
Find the remainder when $4 x^{3} + x - 1$ is divided by $(2 x + 1)$ .
Answer
Substitute $x = - \frac{1}{2}$ : $4 (- \frac{1}{8}) - \frac{1}{2} - 1 = - \frac{1}{2} - \frac{1}{2} - 1 = - 2$ .
When $x^{3} + k x^{2} - 4$ is divided by $(x + 2)$ the remainder is $8$ . Find $k$ .
Answer
$p (- 2) = - 8 + 4 k - 4 = 8$ , so $4 k = 20$ and $k = 5$ .
Show that $(x + 3)$ is a factor of $x^{3} + 2 x^{2} - 5 x - 6$ , and then factorise it fully.
Answer
$p (- 3) = - 27 + 18 + 15 - 6 = 0$ , so $(x + 3)$ is a factor. Then $x^{3} + 2 x^{2} - 5 x - 6 = (x + 3) (x^{2} - x - 2) = (x + 3) (x - 2) (x + 1)$ .
List the values of $x$ worth testing to find a factor of $x^{3} - 3 x^{2} - 4 x + 12$ , then find one factor.
Answer
$\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$ . $p (1) = 6$ and $p (- 1) = 12$ , but $p (2) = 8 - 12 - 8 + 12 = 0$ , so $(x - 2)$ is a factor. (In full, it is $(x - 2) (x - 3) (x + 2)$ .)
Given that $(x - 1)$ is a factor of $3 x^{3} + 2 x^{2} - 7 x + 2$ , find the quadratic factor by comparing coefficients, then factorise fully.
Answer
Write $(x - 1) (3 x^{2} + q x + r)$ . Constants: $- r = 2$ , so $r = - 2$ . $x^{2}$ terms: $q - 3 = 2$ , so $q = 5$ . Check $x$ terms: $r - q = - 7$ . So it is $(x - 1) (3 x^{2} + 5 x - 2) = (x - 1) (3 x - 1) (x + 2)$ .
Explain why $x^{3} + x^{2} + x - 3 = 0$ has exactly one real solution.
Answer
$p (1) = 0$ , so it factorises as $(x - 1) (x^{2} + 2 x + 3)$ . The quadratic has discriminant $4 - 12 = - 8 < 0$ , so it has no real roots. The only real solution is $x = 1$ .
Solve $x^{3} = 9 x$ .
Answer
$x^{3} - 9 x = 0$ , so $x (x - 3) (x + 3) = 0$ and $x = 0$ , $3$ or $- 3$ . Dividing by $x$ would have lost $x = 0$ .

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. $p (x) = 2 x^{3} + a x^{2} + b x - 10$ , where $a$ and $b$ are constants. It is given that $(x - 2)$ is a factor of $p (x)$ , and that when $p (x)$ is divided by $(x + 1)$ the remainder is $12$ .

(a) Find the value of $a$ and the value of $b$ . 4 marks No calculator

Mark scheme

B1 $p (2) = 0$ used: $16 + 4 a + 2 b - 10 = 0$ oe
B1 $p (- 1) = 12$ used: $- 2 + a - b - 10 = 12$ oe
M1 solves their two linear equations simultaneously to find one unknown
A1 $a = 7$ and $b = - 17$

(b) Using your values of $a$ and $b$ , factorise $p (x)$ fully. 3 marks No calculator

Mark scheme

M1 divides $p (x)$ by $(x - 2)$ , or compares coefficients, to obtain a quadratic factor; FT their $a$ and $b$
A1 $2 x^{2} + 11 x + 5$
A1 $(x - 2) (2 x + 1) (x + 5)$

(c) Hence solve $p (x) = 0$ . 1 mark No calculator

Mark scheme

B1 $x = 2$ , $x = - \frac{1}{2}$ , $x = - 5$ ; FT their three linear factors

Question 2. $q (x) = x^{3} - 4 x^{2} - 2 x + 8$

(a) Verify that $(x - 4)$ is a factor of $q (x)$ . 1 mark No calculator

Mark scheme

B1 $q (4) = 64 - 64 - 8 + 8 = 0$ shown, with a conclusion that $(x - 4)$ is a factor

(b) Solve $q (x) = 0$ , giving your answers in exact form. 3 marks No calculator

Mark scheme

M1 obtains a quadratic factor by division, comparing coefficients or grouping
A1 $x^{2} - 2$
A1 $x = 4$ , $x = \sqrt{2}$ , $x = - \sqrt{2}$ cao

(c) Find the remainder when $q (x)$ is divided by $(2 x - 1)$ . 2 marks No calculator

Mark scheme

M1 substitutes $x = \frac{1}{2}$ into $q (x)$
A1 $\frac{49}{8}$ oe

Question 3. The polynomial $f (x) = 2 x^{3} + k x^{2} - 7 x + 3$ , where $k$ is a constant, has $(2 x - 1)$ as a factor.

(a) Show that $k = 1$ . 2 marks No calculator

Mark scheme

M1 substitutes $x = \frac{1}{2}$ and sets $f (\frac{1}{2}) = 0$
A1 $\frac{1}{4} + \frac{k}{4} - \frac{7}{2} + 3 = 0$ simplified correctly to $k = 1$ ; nfww

(b) Solve $f (x) = 0$ , giving your answers in exact form. 4 marks No calculator

Mark scheme

M1 divides by $(2 x - 1)$ , or compares coefficients, to obtain a quadratic factor
A1 $x^{2} + x - 3$
M1 applies the quadratic formula to their three-term quadratic
A1 $x = \frac{1}{2}$ , $x = \frac{- 1 + \sqrt{13}}{2}$ , $x = \frac{- 1 - \sqrt{13}}{2}$ cao

How this is examined

Turning 'is a factor' and 'leaves remainder R' into two equations for unknown coefficients: a four-mark Find at the start of a cubic question, usually followed by factorising or solving the same cubic Each condition written as an equation with the substitution visible, e.g. $p (2) = 0$ and $p (- 1) = 12$ expanded term by term. Then the pair solved for both unknowns. Check both conditions with your values before going on, because a wrong coefficient spoils every later part.
Getting from one linear factor to the quadratic factor, then factorising fully: the middle part of a cubic question: Factorise fully, or Show that $p (x) = (x - a) (\dots)$ The quadratic factor stated explicitly, found by long division (with the zero remainder shown) or by comparing coefficients, then the product of all the linear factors. Stopping at linear times quadratic, when the quadratic still splits, loses the final mark.
Solving the cubic and giving every root exactly: a Solve after the factorising, sometimes with a quadratic factor that needs the formula All roots listed from the factorised form, with surds simplified (e.g. $1 \pm \sqrt{2}$ ). Where the quadratic factor has $b^{2} - 4 a c < 0$ , the discriminant is shown and the single real root is stated.
The remainder for a divisor such as $(2 x + 1)$ : a one- or two-mark Find, with no division needed Substituting the value that makes the divisor zero, here $x = - \frac{1}{2}$ , with the fractions worked by hand to a single exact value.
Proving a given linear expression is a factor: a one- or two-mark Show that or Verify $p (a)$ written out term by term, evaluated to $0$ , and the conclusion stated: 'so $(x - a)$ is a factor'.

Words to know

Polynomial: An expression made of terms $a x^{n}$ with whole-number powers, such as $2 x^{3} - 5 x + 1$ .
Degree: The highest power of $x$ in a polynomial. A cubic has degree $3$ .
Quotient: The polynomial produced by a division, not counting the remainder.
Remainder theorem: When $p (x)$ is divided by $(x - a)$ , the remainder is $p (a)$ .
Factor theorem: $(x - a)$ is a factor of $p (x)$ exactly when $p (a) = 0$ .
Root: A value of $x$ that makes $p (x) = 0$ ; on a graph, where $y = p (x)$ meets the $x$ -axis.
Comparing coefficients: Finding unknown coefficients by matching the number in front of each power of $x$ on the two sides of an identity.

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