Polynomial Identities and the Remainder Theorem

The Remainder Theorem

When a polynomial $p(x)$ is divided by $x-a$, the remainder is just $p(a)$, no long division required.

The polynomials $ax^3+3x^2-3$ and $2x^3-5x+a$ leave the same remainder when divided by $x-4$. Find $a$.

Solution: Both remainders equal the polynomial evaluated at $x=4$:

Setting them equal:

Factoring with the cube identities

Recognizing a sum or difference of cubes pattern turns a hard-looking factoring problem into a one-line answer:

Factorize $64a^3 - 27b^3 - 144a^2b + 108ab^2$.

Solution: With $x=4a$ and $y=3b$, $x^3=64a^3$ and $y^3=27b^3$. The middle terms match $3xy(x-y) = 3(4a)(3b)(4a-3b) = 144a^2b - 108ab^2$, so the full expression is $(x-y)^3 = x^3-y^3-3xy(x-y)$:

Worked example: rectangle side lengths

A rectangle's area is $25a^2-35a+12$. Give possible expressions for its length and breadth.

Solution: Substituting $x=5a$ turns the expression into $x^2-7x+12$, which factors as $(x-4)(x-3)$ (since $-4\times-3=12$ and $-4+-3=-7$). Replacing $x$ with $5a$ again:

So the length and breadth could be $(5a-4)$ and $(5a-3)$.