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1. Reset the foundations
Roots, factors, signs and missing coefficients.
Polynomials
Grade 12 CAPS: factorise cubics, find remainders, turn zero into a factor, and solve every root. Learn it as one connected route—not a list of rules.
Grade 12 CAPS: factorise cubics, find remainders, turn zero into a factor, and solve every root. Learn it as one connected route—not a list of rules.
From a divisor to a remainder; from zero to a factor; from a factor to every root of a cubic. Work through the slides in order, then move to the practice bank.
A focused one-week plan: learn one link at a time, then practise it without notes.
Roots, factors, signs and missing coefficients.
Read every divisor, then substitute accurately.
Confirm, divide, then factorise the quotient.
Choose candidates, solve non-monic and repeated-root cases.
Complete the Mastery Bank, then the short self-test.
A focused Grade 12 topic: cubics and theorem application up to degree 3.
Factorise third-degree polynomials. First use a common factor, grouping or an identity if one is visible.
Use the Remainder Theorem. Read the divisor, then evaluate \(p(x)\) at the correct input.
Use the Factor Theorem. A zero remainder proves a factor; then finish the cubic and state all roots.
The theorem work is easy only when these basics are automatic.
\(x^3-4x+7\) means \(x^3+0x^2-4x+7\). The zero coefficient protects you when dividing.
If \(r=-3\) is a root, the factor is \((x+3)\). The graph meets the x-axis at \(x=-3\).
\(6x^3-24x=6x(x^2-4)=6x(x-2)(x+2)\). Do not use a theorem when simpler factorisation works.
Factorise \(x^3-3x^2-4x+12\).
This is the link between the algebra on Paper 1 and the graph you meet elsewhere.
If \(p(-2)=0\), then \(\boxed{x+2}\) is a factor, \(\boxed{x=-2}\) is a root, and the graph has x-intercept \(\boxed{(-2,0)}\).
For the divisor \(x+4\), which value must you substitute into \(p(x)\)?
Which factor belongs to the root \(r=\frac23\)?
Understand the move once, then use it quickly in an exam. A formal proof is not required.
\(Q(x)\) is the quotient. \(R\) is a constant because a linear divisor cannot leave anything bigger than degree \(0\).
At the root of the divisor, the quotient part disappears. What remains is the remainder.
The remainder is one substitution—but only if the input value is correct.
| Divisor | Set equal to zero | Remainder |
|---|---|---|
| \(x-a\) | \(x=a\) | \(p(a)\) |
| \(x+a\) | \(x=-a\) | \(p(-a)\) |
| \(cx-d\) | \(x=\frac dc\) | \(p(\frac dc)\) |
| \(cx+d\) | \(x=-\frac dc\) | \(p(-\frac dc)\) |
Same theorem, three common divisor forms. The input changes; the method does not.
Remainder when \(x^3-4x+5\) is divided by \(x-2\):
Remainder when \(x^3-2x+5\) is divided by \(x+2\):
Find the remainder when \(2x^3-x^2+3x+1\) is divided by \(2x+1\).
\(2x+1=0\Rightarrow x=-\frac12\).
\(\displaystyle p(-\frac12)=2(-\frac18)-\frac14+3(-\frac12)+1=\boxed{-1}\).
This is the move that causes avoidable theorem errors in tests.
Find the remainder when \(p(x)=4x^3-4x^2-x+2\) is divided by \(2x-3\).
\(q(x)=2x^3+px^2-7x+4\) leaves remainder \(5\) when divided by \(2x+1\). Find \(p\).
What is the remainder when \(x^3-2x^2+3x-4\) is divided by \(x+2\)?
When dividing \(x^2+5x-1\) by \(2x-3\), which expression gives the remainder?
A zero remainder is not the final answer. It is your way into the cubic.
A root \(r=-4\) pairs with the factor \((x+4)\).
\(p(-2)=0\), therefore the remainder is zero; therefore \((x+2)\) is a factor of \(p(x)\).
If the answer is nonzero, state: not a factor.
Division is a short bridge. It turns one known factor into the quadratic you still need to finish.
\(g(x)=x^3-7x+6\). Since \(g(2)=0\), \((x-2)\) is a factor. Write the missing \(x^2\) coefficient as zero before dividing.
Divide \(x^3+0x^2-7x+6\) by \(x-2\)
| 2 | 1 | 0 | −7 | 6 |
| × / + | 2 | 4 | −6 | |
| result | 1 | 2 | −3 | 0 |
The bottom row gives \(x^2+2x-3\), with remainder \(0\). The final \(0\) agrees with the Factor Theorem.
\(\displaystyle g(x)=(x-2)(x^2+2x-3)=(x-2)(x+3)(x-1)\).
So \(g(x)=0\) has roots \(\boxed{x=2,\;-3,\;1}\).
One chain: factor check, quotient, complete factorisation.
Factorise \(F(x)=3x^3+x^2-8x+4\), given that \(3x-2\) is a factor.
If \(x-2\) is a factor of \(x^3+kx^2-5x-6\), find \(k\).
Is \(x+2\) a factor of \(x^3+3x^2-4x-12\)?
If \(p(3)=0\), which statement is certainly true?
Use a short decision path to find the first factor of a cubic efficiently.
Solve \(6x^3-5x^2-17x+6=0\). Try \(x=2\):
So \((x-2)\) is a factor. Division gives \(6x^2+7x-3=(3x-1)(2x+3)\).
Finding one factor is the midpoint. The solution is complete only after the quadratic is finished.
Open the cubic with a factor, then finish the quadratic.
Solve \(x^3-2x^2-5x+6=0\).
A cubic does not always end with three neat integer roots.
\(\displaystyle x^3-3x^2+4=(x-2)^2(x+1)\).
State the repeated root; its repeated factor matters.
\(\displaystyle x^3-2x^2-6x+4=(x+2)(x^2-4x+2)\).
All roots: \(\boxed{-2,\;2+\sqrt2,\;2-\sqrt2}\).
Solve \(2x^3+x^2-8x-4=0\).
Each piece of information creates one equation. Solve the pair rather than guessing coefficients.
\(P(x)=x^3+px^2-7x+q\). Given \(x+1\) is a factor and the remainder on division by \(x-2\) is \(-9\), find \(p\) and \(q\).
A learner has \((x-2)(x^2+x-6)=0\). What must happen next?
Which root is repeated in \((x-2)^2(x+1)=0\)?
These are small slips, but each one can break an otherwise correct polynomial question.
Set the entire divisor equal to zero. The visible plus sign is the trap.
Do not substitute \(3\). Solve the linear divisor before touching \(p(x)\).
Write \(x^3-4x+7\) as \(x^3+0x^2-4x+7\) before dividing.
Zero remainder proves a factor. Then obtain the quotient and finish the quadratic.
Factorise ends with brackets. Solve ends with every value of \(x\), after \(p(x)=0\).
Use this before submitting an exam answer.
Using \(+a\) for divisor \(x+a\).
Stopping after one factor.
Leaving out a zero coefficient.
Use a theorem line: \(p(\frac32)=5\).
Write the factor conclusion.
End a solve question with all values of \(x\).
Use the Mastery Bank to build fluency. Then take the test without notes and use the result to choose the exact slide to revisit.
Original questions from reading a divisor through to complete cubic solutions. Answers reveal only after your attempt.
Next step 2A short exam-style self-check. Mark it, then revisit the exact slide that matches a missed skill.
Next step 3Work independently across all levels, use the memo, then repair only the skill that cost marks.
Optional support after the Remainder Theorem slides. It plays here and can expand to full screen.
Use this only after you can identify the input from the divisor. Press play to stay on Equation Station; use Expand for a larger view.
The core teaching is above. These are the next steps, not a replacement for it.
18 original questions by skill, with concise reveal answers and methods.
Original mixed paper: direct remainders, factors, cubics, parameters and error analysis.
Use the short exam-style self-check when you want a fast confidence check.
Use its revision, cubic, remainder, factor and solving sequence for extra explanation and exercises.
Official state-owned learner books and teacher support for Grade 12 Mathematics.
Use official papers after the course and original practice are secure.
These skills strengthen the algebra underneath every polynomial question.
Short answers for the checks learners make while preparing for the Grade 12 CAPS exam.
Factorising cubics and applying the Remainder and Factor Theorems to polynomials of degree at most three. You apply the theorems; you do not prove them.
Use \(x=-a\). Set the divisor equal to zero before every substitution so the sign is visible in your working.
State that the divisor is a factor, divide by it, factorise the quadratic quotient, and give every root if the question asks you to solve.
No. Use the theorem to find or confirm a factor first. Division is the step after you know a factor.
Finish the interactive slides, open the Mastery Bank, then take the short self-test without notes.