Polynomials - Factorisation of Polynomials

The Factor Theorem states that for a polynomial $p(x)$ of degree $n \geq 1$, if $p(a) = 0$ for a real number $a$, then $(x - a)$ is a factor of $p(x)$. Conversely, if $(x - a)$ is a factor, then $p(a) = 0$. Visually, if you plot the polynomial on a graph, the factor $(x-a)$ corresponds to the point where the curve crosses the x-axis at $x = a$.

Factorisation by splitting the middle term is used for quadratic polynomials of the form $ax^2 + bx + c$. We look for two numbers $p$ and $q$ such that $p + q = b$ and $pq = ac$. This technique is like breaking a large rectangular area representing the polynomial into smaller, manageable sections that share common side lengths.

Factorisation by grouping involves rearranging terms of a polynomial into groups that have a common factor. For example, in $ax + ay + bx + by$, we group as $(ax + ay) + (bx + by)$, factor out common terms to get $a(x + y) + b(x + y)$, and finally $(a + b)(x + y)$. Imagine organizing a set of tiles into rows and columns where the 'common factor' represents a shared dimension.

The difference of two squares $x^2 - y^2$ is factorised into $(x - y)(x + y)$. This can be visualized as taking a square with side $x$, removing a smaller square with side $y$ from the corner, and then slicing and rearranging the remaining shape to form a rectangle with dimensions $(x + y)$ and $(x - y)$.

For cubic polynomials like $ax^3 + bx^2 + cx + d$, factorisation often begins with the Factor Theorem to find one linear factor $(x - k)$ by testing divisors of the constant term $d$. Once one factor is found, the polynomial is divided by it to obtain a quadratic polynomial, which is then factorised further.

Perfect square trinomials $x^2 + 2xy + y^2$ and $x^2 - 2xy + y^2$ are factorised into $(x + y)^2$ and $(x - y)^2$ respectively. Geometrically, $x^2 + 2xy + y^2$ represents the total area of a large square made of two smaller squares ($x^2$ and $y^2$) and two identical rectangles (each with area $xy$).

Algebraic identities involving three variables, such as $x^3 + y^3 + z^3 - 3xyz$, can be factorised into $(x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$. This identity is particularly useful when $x + y + z = 0$, leading to the result $x^3 + y^3 + z^3 = 3xyz$.