Factorising a polynomial is, for a high-degree equation, the equivalent of taking a piece of furniture apart into its pieces: $P(x)=2x(x+5)(x-4)^2$ tells you at a glance that $P(x)=0$ exactly when $x=0$, $x=-5$ or $x=4$. This section brings together all the techniques —common factor, special products, Ruffini, rational roots— into a single systematic procedure, and adds the cases that 4th ESO did not cover: multiplicities, fractional roots and quadratic factors with no real roots.
Before you start, make sure you know: Special products (2nd ESO)
Common factor and special products
Factorising always starts the same way: take out as much common factor as possible and, if what is left is a special product or a quadratic equation, solve it.
Practice★☆☆Not started
Factorise out the common factor and use algebraic identities to fully factorise the following polynomial: $P(x) = 3x^5 - 12x^3$
Factorise out the common factor and use algebraic identities to fully factorise the following polynomial:
Check, using the Remainder Theorem, whether $x = -1$ is a root of the polynomial $P(x) = x^4 + 3x^3 - 2x^2 + x + 7$. If so, factorise out the corresponding factor.
Check, using the Remainder Theorem, whether $x = -1$ is a root of the polynomial $P(x) = x^4 + 3x^3 - 2x^2 + x + 7$. If so, factorise out the corresponding factor.
Factorising "completely" means repeating common factor, special products and Ruffini until all the factors have degree $1$ (or degree $2$ with no real roots). If a root appears more than once, it is a repeated root, and its multiplicity is the number of times it appears.
Practice★☆☆Not started
State the roots of the polynomial $P(x) = 5(x+3)^2(x-1)(x^2+4)$ and their respective multiplicities. Justify why this polynomial has no other real roots.
State the roots of the polynomial $P(x) = 5(x+3)^2(x-1)(x^2+4)$ and their respective multiplicities. Justify why this polynomial has no other real roots.
Biquadratic polynomials and repeated differences of squares
A biquadratic polynomial only has terms of degree $4$, $2$ and $0$ ($ax^4+bx^2+c$): it is factorised as if it were quadratic, replacing $x^2$ with an auxiliary variable. And a difference of squares, if it turns into one again after the first step, can be applied twice.
Practice★☆☆Not started
Write the polynomial $P(x) = x^4 - 81$ in factorised form by applying the difference of two squares identity successively.
Write the polynomial $P(x) = x^4 - 81$ in factorised form by applying the difference of two squares identity successively.
Factorise the polynomial $P(x) = (x^2 - 4)^2 - (x + 2)^2$ using algebraic identities and factorising out common factors, without needing to expand the entire polynomial…
Factorise the polynomial $P(x) = (x^2 - 4)^2 - (x + 2)^2$ using algebraic identities and factorising out common factors, without needing to expand the entire polynomial first.
Rational roots: when the leading coefficient is not 1
If the leading coefficient of $P(x)$ is not $1$, then in addition to the divisors of the constant term you also have to try fractions: candidates $\frac{p}{q}$, with $p$ a divisor of the constant term and $q$ a divisor of the leading coefficient.
Practice★★☆Not started
Factorise the polynomial $P(x) = 3x^4 + 5x^3 - 17x^2 - 13x + 6$. Note that some of its roots may be fractional numbers.
Factorise the polynomial $P(x) = 3x^4 + 5x^3 - 17x^2 - 13x + 6$. Note that some of its roots may be fractional numbers.
In an economics study, the marginal profit curve of a technology company fits the polynomial $B(x) = 6x^4 - 11x^3 - 26x^2 + 21x + 10$, where $x$ represents the R&D…
In an economics study, the marginal profit curve of a technology company fits the polynomial $B(x) = 6x^4 - 11x^3 - 26x^2 + 21x + 10$, where $x$ represents the R&D investment in millions of euros. Factorise the polynomial to find the exact equilibrium points (including fractional roots).
Not every polynomial breaks down into factors of degree $1$: when a factor of degree $2$ is left with a negative discriminant, it is irreducible over the real numbers and the factorisation stops there. Sometimes, adding and subtracting an auxiliary term lets you see two quadratic factors where at first glance you cannot see any.
Practice★★★Not started
Prove that the polynomial $P(x) = x^4 + 4$ has no real roots. Then, factorise it as the product of two quadratic polynomials. Hint: Add and subtract $4x^2$ to the…
Prove that the polynomial $P(x) = x^4 + 4$ has no real roots. Then, factorise it as the product of two quadratic polynomials. Hint: Add and subtract $4x^2$ to the expression to force the appearance of a difference of two squares.
When a polynomial has a parameter and a divisibility condition, you first find the parameter (as in the section on Ruffini) and then factorise once all the coefficients are known.
Practice★★☆Not started
Determine the value of the parameter $m$ so that the polynomial $P(x) = x^3 - mx^2 + 4x - 4$ is divisible by the binomial $x - 2$. Once the value of $m$ is found, fully…
Determine the value of the parameter $m$ so that the polynomial $P(x) = x^3 - mx^2 + 4x - 4$ is divisible by the binomial $x - 2$. Once the value of $m$ is found, fully factorise the polynomial over the set of real numbers.
The volume of a rectangular box as a function of its length $x$ is given by the polynomial expression $V(x) = x^3 - 7x^2 + 10x$. Factorise $V(x)$ to find the simplest…
The volume of a rectangular box as a function of its length $x$ is given by the polynomial expression $V(x) = x^3 - 7x^2 + 10x$. Factorise $V(x)$ to find the simplest algebraic expressions representing the three dimensions of the box.
Calculate the values of the parameters $a$ and $b$ so that the polynomial $P(x) = x^4 + ax^3 + bx^2 - 12x + 8$ is divisible by the polynomial $Q(x) = x^2 - 4$. Then,…
Calculate the values of the parameters $a$ and $b$ so that the polynomial $P(x) = x^4 + ax^3 + bx^2 - 12x + 8$ is divisible by the polynomial $Q(x) = x^2 - 4$. Then, fully factorise $P(x)$.