3
$\begingroup$

Find the value of $a^4-a^3+a^2+2$ when $a^2+2=2a$

My Attempt,

$$a^4-a^3+a^2+2=a^4-a^3+2a$$ $$=a(a^3-a^2+2)$$

What's next?

  • 1
    Is there any reason you wouldn't just solve $a^2 + 2 = 2a$ to get $a = 1 \pm i$ and plug those values in. Yes, there's probably a very clever manuplation to reduce the expression to something simple so you don't have to do calculations, but if you don't see it, you don't see it. So why don't you simply *solve* the darned thing? – 2017-01-28

4 Answers 4

0

Probably the most "uninspired" (and therefore easy to arrive at) solution is to solve the quadratic to give $a = 1 \pm i$

Then substitute one of these solutions into the quartic expression and show that it vanishes. There's no need to test the other (complex conjugate) root by substituting it into the quartic as it has real coeffients and will therefore be zero in the other case as well.

7

$a^4-a^3+a^2+2=(a^2+a+1)(a^2-2a+2)=0$.

We can check also $a=1+i$ and $a=1-i$.

  • 0
    It's not clear how you expect readers to easily *discover* that factorization. It if is meant to be derived from the Polynomial Division Algorithm, then the *quotient* is uneeded, Rather, we need only the *remainder* (= 0). See my comment on the question. This is the point of modular arithmetic. – 2017-01-28
  • 0
    @Bill Dubuque It's just $a^4-a^3+a^2+2=a^4-2a^3+2a^2+a^3-2a^2+2a+a^2-2a+2=(a^2-2a+2)(a^2+a+1)$. – 2017-01-28
  • 1
    That's a *verification*, not an explanation of **how to discover** such factorizations. As written, it is pulled out of a hat - like magic. But magic is not mathematics! How did you find that factorization? Will the same method work for any quotient? (as does the division algorithm). – 2017-01-28
  • 0
    @Bill Dubuque Yes, of course! It works for all polynomials. It's just dividing of polynomials. At least we need to work with the remainder. – 2017-01-28
  • 1
    What is "it"? You have not yet divulged what method you used to derive the first equation in your answer. It remains a mystery. Keep in mind that some unexplained answers like that are derived via computer algebra systems. Such answers generally are not helpful. – 2017-01-28
  • 0
    @Bill Dubuque I told it you $3$ hours ago. It's just dividing of polynomials. Why do you speak to me in that tone? – 2017-01-28
  • 2
    Because I believe that answers should *explain* what method they used, not simply pull the answer out of a hat - like magic. Your "answer" *still* lacks any such explanation. – 2017-01-28
  • 0
    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/52665/discussion-between-michael-rozenberg-and-bill-dubuque). – 2017-01-28
2

Obviously $a \not = 0$. Then we have $a^4 = 2a^3 - 2a^2$ by multiplying the condition by $a^2$, so the equation becomes: $a^3 - a^2 + 2$. Similarly $a^3 = 2a^2 - 2a$, so the equation becomes: $a^2 - 2a + 2 = 0$

  • 1
    This is equivalent to applylng the Polynomial Division Algorithm, i.e. you compute $\,(f\bmod g) = 0\,$ by dividing (and ignoring the unneeded quotient) - see my hint on the question. – 2017-01-28
2

Hint $\ $ We seek the value of $\,f(a)\,$ given that $\, g(a) = a^2-2a+2 = 0.\,$ By the Polynomial Division Algorithm we can write $\ f = q\,g + r\,$ so $\, g=0\,\Rightarrow\, f = r = f\bmod g.\,$ So it suffices to compute the remainder $\, r = (f\bmod g).\,$ Doing so easily yields $\,f = r = 0.$

Because $\, g = 0\,$ we can perform all arithmetic modulo $g.\,$ This is the point of modular arithmetic - when we are interested only in the remainder then we can ignore the quotients. This is exactly what is done (implicitly) in Stefan4024's answer.