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Remainder and Factor Theorems

If you divide a polynomial $f(x)$ by $(x-a)$ , the remainder is just $f(a)$ — no long division required:

f(x) = (x-a)\,q(x) + f(a)

A direct consequence is the Factor Theorem: $(x-a)$ is a factor of $f(x)$ exactly when $f(a) = 0$ . This is the fastest way to test a candidate root — plug it in, and if you get zero, you know $(x-a)$ divides out cleanly.

Vieta's formulas

Vieta's formulas relate a polynomial's coefficients directly to sums and products of its roots, without requiring you to find the roots themselves. For a quadratic $ax^2+bx+c$ with roots $r, s$ :

r + s = -\dfrac{b}{a}, \qquad rs = \dfrac{c}{a}

For a cubic $ax^3+bx^2+cx+d$ with roots $r, s, t$ :

r+s+t = -\dfrac{b}{a}, \quad rs+st+tr = \dfrac{c}{a}, \quad rst = -\dfrac{d}{a}

These are especially powerful when a problem asks for a symmetric combination of the roots (like $r^2+s^2$ or $\tfrac1r+\tfrac1s$ ) — you can often express the answer purely in terms of the sum and product, skipping the roots entirely.

Worked example

Problem: If $r$ and $s$ are the roots of $2x^2 - 7x + 3 = 0$ , find $r^2 + s^2$ .

Solution: By Vieta's, $r+s = \tfrac{7}{2}$ and $rs = \tfrac{3}{2}$ . Use the identity $r^2+s^2 = (r+s)^2 - 2rs$ :

r^2+s^2 = \left(\tfrac{7}{2}\right)^2 - 2\left(\tfrac{3}{2}\right) = \tfrac{49}{4} - 3 = \tfrac{37}{4}

No need to ever compute $r$ and $s$ individually (which aren't even rational here).

Practice problems

1.Is $(x-2)$ a factor of $f(x) = x^3 - 3x^2 - 4x + 12$ ?

2.If the roots of $x^2 - 9x + 14 = 0$ are $r$ and $s$ , find $r+s$ and $rs$ without solving for the roots.

3.What is the remainder when $x^4 + 2x^3 - x + 5$ is divided by $x+1$ ?

4.If $r, s$ are the roots of $x^2 - 5x + 2 = 0$ , find $\dfrac{1}{r} + \dfrac{1}{s}$ .

Polynomial Theory

Remainder and Factor Theorems

Vieta's formulas

Practice problems