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How can one prove that all linear polynomials have exactly one root? This is geometrically intuitive (just rotate a line around the x axis) however I'm not sure how to formally prove this.

George Newton
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  • Use Euclidean division with your polynomial $f$ and $(x-\alpha)$ where $\alpha$ is one of the roots. What is your remainder? – Prahlad Vaidyanathan Jan 13 '15 at 03:04
  • I presume you mean an affine polynomial? – copper.hat Jan 13 '15 at 03:05
  • @copper.hat: It is deeply entrenched terminology to call polynomials of degree $1$ "linear" even if they have a constant term. (Consider for example talking about a polynomial splitting into linear factors). That doesn't match the use of "linear" in linear algebra, but there's nothing to be done about that now. – hmakholm left over Monica Jan 13 '15 at 03:14
  • @HenningMakholm: Thanks! I know, I'm just being persnickety :-). – copper.hat Jan 13 '15 at 04:07

3 Answers3

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A linear polynomial is of the form $f(t) = at+b$ for some coefficients $a$, and $b$, $a \neq 0$ over a field $F$. A root of a polynomial $f$ is a value $t$ satisfying $f(t) = 0$. It is clear that

$$f(t) = 0 \Rightarrow t = -\frac{b}{a}. $$

Therefore, we have precisely one root, $t = -b/a$.

MathMajor
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Let $p(x) = a x + b$ be a linear polynomial where $a$ is non-zero.

Suppose that $x_1$ and $x_2$ are solutions to $p$. Then

$$a x_1 + b = 0 = a x_2 + b$$

Subtracting $b$ from both sides we have,

$$a x_1 = a x_2$$

And dividing both sides by non-zero $a$,

$$x_1 = x_2$$

This shows that roots are always equal. Note if $a$ is zero, there will be infinite or no roots, and this logic does not follow.

yberman
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I'd do it like this - suppose $ax_1+b=ax_2+b$ then $$a(x_1-x_2)=0$$ and this means $a=0$ or $x_1-x_2=0$

This last step is possible because for real numbers $xy=0$ implies $x=0$ or $y=0$. There are other mathematical contexts where this is true. But it is not always true, and sometimes we have non-trivial divisors of zero. For example $2\times 3 \equiv 0 \bmod 6$.

This way of writing may look the same as others, but avoids division and gives an equation which remains true in a lot of contexts. It identifies the precise property which allows the next step to be taken (no non-trivial divisors of zero).

Mark Bennet
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