Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm tutoring a high school precalculus student; our current topic is the roots of higher order polynomials.

The problem we're solving is:

Find a polynomial with the roots $\frac23$, -1, and $(3 + \sqrt{2}i)$

I started the solution by explaining that since one of the roots is a complex number, that root's complex conjugate must also be a root. So the polynomial (well, one of many possible answers) must be:

$(x-\frac 23)(x+1)(x-(3+\sqrt{2}i))(x-(3-\sqrt{i}))$

When I multiplied all these terms, I get:

$f(x) = x^4-5x^3+5x^2+11x- \frac 23x^3+ \frac{10}3x^2- \frac{10}3x- \frac {22}3$

Now, I put as the next step, "multiply all terms by 3 to get rid of the denominator."

This does in fact arrive at the correct answer:

$f(x)=3x^4-17x^3+25x^2+23x-22$

This solution matches the instructor's solution. What my college-educated brain can't figure out is: why does this work? I've got an equation and I'm multiplying everything on the right-hand side by 3 without doing the same to the left hand side. How does this maintain equality between the left and right hand sides of the equation?

I realize this is remedial math here; I've been doing this algebra for a while today and it's currently 1AM so hopefully I'm just overlooking something every high school kid knows because I'm cross eyed and tired. Still, if someone could point out the obvious, I'd be grateful.

share|improve this question
    
Think of it as this: $x^4-5x^3+5x^2+11x-\frac{2}3x^3+\frac{10}3x^2-\frac{10}3x-\frac{22}3=x^4-5x^3+5x‌​^2+11x-\frac{2}3x^3+\frac{10}3x^2-\frac{10}3x-\frac{22}3$ –  EpicGuy Oct 21 '13 at 7:33
    
$f(x_0)=0\iff3f(x_0)=0$. –  Michael Hoppe Oct 21 '13 at 10:54

2 Answers 2

In general, if you know that $p(x)$ is a polynomial with zeroes $\alpha_1, \dots, \alpha_n$, and you know that there are no other zeroes, then $p(x) = a(x-\alpha_1)\dots(x-\alpha_n)$ for some $a \neq 0$. This follows from the Factor Theorem. As the questions asks for a polynomial, both of the polynomials you have written are correct solutions, but it is incorrect to say that they are the same polynomial; they just have the same zeroes.

share|improve this answer

They aren't the same function, so it's not correct to call them the same letter. However, the new polynomial will still have the exact same roots as the old polynomial, so it's one of the many other possible answers for the question.

It would be better to write $f(x) = x^4 - 5x^3 + ...$, and then define

$$g(x) = 3f(x) = 3x^4 - 17x^3 + ...$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.