Reducible Polynomials
A polynomial is only as complicated as its factors allow. By substituting $u = x^n$, a scary degree-6 expression collapses into a familiar quadratic — and suddenly the factorisation is obvious. In this lesson you'll master the substitution technique, understand the boundary between reducible and irreducible factors over the reals, and use the discriminant to confirm irreducibility.
How might you factorise $x^4 + x^2 - 2$? Without using a formula — what substitution could help? Think about the structure of the exponents before reading on.
This lesson rests on two core strategies. First: substitution — replace a repeated chunk of the polynomial with $u$ to reveal a simpler expression underneath. Second: the discriminant test — once you have a quadratic factor, compute $\Delta = b^2 - 4ac$ to decide whether it splits further over the reals.
Every reducibility question uses one of two checks: does a substitution $u = x^k$ reduce this to a quadratic? If yes, factorise the quadratic and substitute back. Then check each quadratic factor's discriminant — negative means irreducible over the reals.
Key facts
- A polynomial is reducible over $\mathbb{R}$ if it factors into lower-degree polynomials with real coefficients
- Over $\mathbb{R}$, the only irreducible factors have degree 1 or 2
- A quadratic is irreducible over $\mathbb{R}$ iff $\Delta < 0$
Concepts
- Why substitution $u = x^k$ reduces the effective degree
- The difference between reducible over $\mathbb{R}$ and reducible over $\mathbb{C}$
- How to use sum/difference of cubes to complete a factorisation
Skills
- Identify an appropriate substitution from a polynomial's structure
- Factorise quartic and degree-6 polynomials over the reals
- Use the discriminant to confirm which quadratic factors are irreducible
A polynomial $P(x)$ with real coefficients is reducible over $\mathbb{R}$ if it can be written as $P(x) = Q(x) \cdot R(x)$ where both $Q$ and $R$ have real coefficients and degree at least 1.
Fundamental theorem over $\mathbb{R}$: Every polynomial with real coefficients can be factored completely into a product of linear factors and irreducible quadratic factors over $\mathbb{R}$.
- Linear factors: $ax + b$ — always irreducible (degree 1)
- Quadratic factors: $ax^2 + bx + c$ — irreducible over $\mathbb{R}$ iff $\Delta = b^2 - 4ac < 0$
- No degree-3 or higher irreducible factors exist over $\mathbb{R}$
Every real polynomial = product of linear and irreducible quadratic factors over R; Quadratic ax^2+bx+c is irreducible over R iff b^2-4ac < 0
Pause — copy the real factorisation theorem into your book: every real polynomial = product of linear and irreducible quadratic factors; a quadratic $ax^2+bx+c$ is irreducible over $\mathbb{R}$ iff $b^2 - 4ac < 0$.
Quick check: Which of the following quadratics is irreducible over $\mathbb{R}$?
We just saw that every real polynomial factors into linear and irreducible quadratic factors; a quadratic is irreducible over $\mathbb{R}$ iff $b^2 - 4ac < 0$. That raises a question: when all powers in a polynomial are multiples of the same integer $k$ (e.g. all even), what substitution reduces the problem to a lower-degree polynomial? This card answers it → substitute $u = x^k$, factorise in $u$, then back-substitute $u = x^k$ to get the factorisation in $x$.
For polynomials where all powers are multiples of $k$, substitute $u = x^k$ to reduce to a quadratic (or lower-degree) polynomial in $u$.
Key patterns:
- $x^4 + bx^2 + c$ → let $u = x^2$ to get $u^2 + bu + c$
- $x^6 + bx^3 + c$ → let $u = x^3$ to get $u^2 + bu + c$
- $x^8 + bx^4 + c$ → let $u = x^4$ to get $u^2 + bu + c$
After factorising in $u$, substitute back to get factors in $x$. Then check each factor's discriminant to determine whether further splitting over $\mathbb{R}$ is possible.
Check: are all exponents multiples of the same k? If yes, substitute u = x^k.; After factorising in u, substitute x^k back for each u.
Pause — copy the substitution technique into your book: if all exponents are multiples of $k$, let $u = x^k$; factorise the resulting polynomial in $u$; back-substitute $x^k$ for each $u$ to get the full factorisation in $x$.
Did you get this? True or false: the substitution $u = x^2$ can be used to factorise $x^4 + x^2 - 2$.
Worked examples · 3 in a row, reveal as you go
Factorise $x^4 - 5x^2 + 6$ over the reals.
Factorise $x^6 - 7x^3 - 8$ over the reals.
Show that $x^4 + 4$ can be written as $(x^2 + 2x + 2)(x^2 - 2x + 2)$.
Fill the gap: For $x^6 + 9x^3 + 8$, the substitution $u = x^3$ gives $u^2 + 9u + 8 = (u+1)(u + $ $)$.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: $x^2 + 2x + 4$ is irreducible over the reals.
Activities · practice with the ideas
Factorise $x^4 - 13x^2 + 36$ over the reals. Show the substitution step and check each quadratic factor.
Factorise $x^4 + x^2 - 2$ over the reals fully.
Factorise $x^6 + 9x^3 + 8$ over the reals fully. Use sum of cubes where needed.
Which of the following are irreducible over $\mathbb{R}$? Justify each answer using $\Delta$. (a) $x^2+4x+5$ (b) $x^2-4x+3$ (c) $x^2+1$
Explain why every polynomial of odd degree with real coefficients must have at least one real root.
Earlier you were asked about factorising $x^4 + x^2 - 2$. With $u = x^2$: $u^2 + u - 2 = (u+2)(u-1)$, so $x^4 + x^2 - 2 = (x^2+2)(x^2-1)$. Now: $x^2+2$ has $\Delta = -8 < 0$ (irreducible), and $x^2 - 1 = (x-1)(x+1)$. Full factorisation: $(x^2+2)(x-1)(x+1)$.
Why can't $x^2 + 1$ be factored over the reals? Because it has no real roots — its discriminant is $\Delta = 0 - 4 = -4 < 0$. Over $\mathbb{C}$ it factors as $(x-i)(x+i)$, but those involve complex numbers outside the real number system.
Odd one out: Which of the following is the odd one out when factorising $x^6 - 7x^3 - 8$ over the reals?
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Factorise $x^4 - 13x^2 + 36$ over the reals. (2 marks)
Q2. Factorise $x^6 + 9x^3 + 8$ over the reals. (3 marks)
Q3. Show that $x^4 + 4$ can be written as $(x^2 + 2x + 2)(x^2 - 2x + 2)$. (2 marks)
Comprehensive answers (click to reveal)
Activity drills: 1. $(x-2)(x+2)(x-3)(x+3)$ via $u=x^2$, $u^2-13u+36=(u-4)(u-9)$ · 2. $(x^2+2)(x-1)(x+1)$ · 3. $(x+1)(x^2-x+1)(x+2)(x^2-2x+4)$ · 4. (a) $\Delta=-4<0$: irreducible (b) $\Delta=4>0$: reducible $(x-1)(x-3)$ (c) $\Delta=-4<0$: irreducible · 5. Over $\mathbb{R}$, factors are linear or irreducible quadratic (degree 2). So odd-degree polynomial must have at least one linear factor, giving a real root.
Q1 (2 marks): $u=x^2$: $u^2-13u+36=(u-4)(u-9)$ [1] $\Rightarrow (x^2-4)(x^2-9)=(x-2)(x+2)(x-3)(x+3)$ [1].
Q2 (3 marks): $u=x^3$: $u^2+9u+8=(u+1)(u+8)=(x^3+1)(x^3+8)$ [1]. $(x^3+1)=(x+1)(x^2-x+1)$; $(x^3+8)=(x+2)(x^2-2x+4)$ [1]. Check $\Delta$: both quadratics have $\Delta<0$, irreducible [1].
Q3 (2 marks): $x^4+4 = x^4+4x^2+4-4x^2 = (x^2+2)^2-(2x)^2$ [1] $= (x^2+2+2x)(x^2+2-2x) = (x^2+2x+2)(x^2-2x+2)$ [1].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
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