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I'm a ninth grader so please try to explain the answer in simple terms . I cant fully understand the explanation in my book . It just assumes that the expression that should be added has a degree of 1.

I apologize if this question is too simple or just stupid but this is a genuine doubt.

MayankJain
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    This is not typical ninth-grade material. Have you learned about long polynomial division? – hmakholm left over Monica Sep 11 '13 at 17:14
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    @HenningMakholm I think in many European countries they learn long division by 9th grade. – N. S. Sep 11 '13 at 17:25
  • @N.S.: It's been a long time. But I don't remember being taught anything like that even in high school ... – hmakholm left over Monica Sep 11 '13 at 17:28
  • I remember having to beg to get material anything like this advanced in high school... – abiessu Sep 11 '13 at 17:29
  • There were a handful of problems like this in our textbooks but they were never assigned in my experience. I came up with similar ones when I sat boredin Algebra II a few years ago. – obataku Sep 11 '13 at 17:33
  • A linear difference will be enough because you're looking to get to "the next" multiple of a quadratic (with leading coefficient 1). Just like getting to the next multiple of 10 requires adding less than 10, getting the next multiple of a quadratic only requires adding "less than" a quadratic, i.e. linear. – John Sep 11 '13 at 19:54
  • I'm currently a high school student in the US, and where I'm from, we learn polynomial long division in middle school. To me, this would not be an unreasonable problem for a ninth grade algebra class (though it would be a particularly challenging one). – David Zhang Sep 12 '13 at 01:43
  • @HenningMakholm I'm a ninth grader from the same country as Mayank is, and we are taught polynomial division in the ninth grade. – P.K. Mar 01 '14 at 12:25

6 Answers6

11

Imagine that we add $ax+b$ to the given polynomial, obtaining a new polynomial $$P(x)= x^4 +2x^3-2x^2+x-1+ax+b.$$

Note that $x^2+2x-3=(x+3)(x-1)$. So if $x^2+2x-3$ divides our new polynomial $P(x)$, then $P(1)=0$ and $P(-3)=0$.

We have $$P(1)=a+b+1, \qquad\text{and}\qquad P(-3)=-3a+b+5.$$

Solve the system of linear equations $a+b+1=0$, $-3a+b+5=0$ for $a$ and $b$.

Remark: We have skipped a logical step that in principle should not be skipped. If $ax+b$ is to work, our argument shows that $a$ and $b$ must satisfy the two equations. We have not shown that if $a$ and $b$ satisfy the two equations, then $ax+b$ automatically works. This is in fact true, but requires some theory.

André Nicolas
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  • I can't upvote twice, but +1 for the extra remark :-) – abiessu Sep 11 '13 at 17:48
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    I was going to post something very similar. It amounts to extending the remainder theorem - $p(a)$ is the remainder when $p(x)$ is divided by $(x-a)$ to division by a higher degree polynomial which splits into linear factors. – Mark Bennet Sep 11 '13 at 17:49
  • About the remark, instead of proving theoretically that $ax+b$ automatically works, is typically much easier to simply check that the answer is right. – N. S. Sep 11 '13 at 17:55
  • My only concern is that you are assuming one only needs to add $ax+b$ just like his textbook does. You mention it is justified by theory -- why not give a brief explanation of how for him? I was never satisfied with "because the book says so" when I was in his place and it can't hurt. – obataku Sep 11 '13 at 17:56
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    There are already several answers, from various points of view, and it seems likely there will be more. An answer inevitably involves a compromise. The above answer probably erred in the direction of simplicity. But it was a choice made consciously. – André Nicolas Sep 11 '13 at 18:03
  • ... well of course, but I raise the concern here specifically because of this line:

    "It just assumes that the expression that should be added has a degree of 1."

    While your answer was helpful no doubt, I worry it leaves open one of his main questions in the original post...

     – obataku Sep 11 '13 at 19:39
  • @AndréNicolas Thanks for the answer but I'm afraid oldrinb is right – MayankJain Sep 12 '13 at 04:01
  • You are welcome. The issue is partly dealt with, without proof, in the answer by abiessu. – André Nicolas Sep 12 '13 at 04:03
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Polynomial divisibility follows many of the same rules as number divisibility. In particular, it is possible to perform "long division" with polynomials the same as with integers:

$$x^4+2x^3-2x^2+x-1 = x^2(x^2+2x-3)+x^2+x-1$$

This means there is remainder $x^2+x-1$ after dividing by $x^2+2x^3-3$. If we then subtract these two (i.e., divide and look for the remainder), we get $-x+2$ as the final remainder, which means that

$$x^4+2x^3-2x^2+x-1=(x^2+1)(x^2+2x-3)-x+2$$

So it should be fairly clear what you need to add to make the original expression divisible by $x^2+2x-3$.

In general, if your divisor is of degree $n$, then the remainder after division can be at most degree $n-1$, just like the remainder after integer division by $k$ will be at most $k-1$. This is why the book would tell you to look for a degree $1$ polynomial offset for a divisor of degree $2$.

abiessu
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  • This is one of the solutions I like most. You gave a more systematic treatment he may appreciate more. – obataku Sep 11 '13 at 17:43
  • Thank you! I just wish it were easier to represent the long division form in MathJAX/HTML... – abiessu Sep 11 '13 at 17:47
  • I agree that the remainder will be of degree 1 if we divide but we are not adding the remainder , are we ? Adding the remainder does not always give a number that is exactly divisible by the divisor. – MayankJain Sep 12 '13 at 12:25
  • No, we are adding the negative of the remainder. Adding the negative of the remainder does always give a divisible result. In this case, the negative of the remainder is $x-2$. – abiessu Sep 12 '13 at 13:33
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Hint: $x^4 + 2x^3 - 2x^2 + x - 1 = x^4 + 2x^3 - 3x^2 + x^2 + 2x - 3- x+2 = (x^2+1)(x^2 + 2x - 3) -(x-2).$

njguliyev
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3

Hint: break $-2x^2=-3x^2+x^2$ to get two sorta similar trinomials (one should look a lot like your $x^2+2x-3$).. which coefficients prevent you from successfully factoring $x^2+2x-3$ out of both parts? How can you 'fix' said coefficients?

obataku
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If you don't want to use polynomial long division, you can factor $x^2+2x-3$ by seeing it has roots at $1$ and $-3$, Then, you need to add a degree 1 polynomial to your polynomial such that the sum is divisible by $x-1$ and $x+3$. Those are relatively prime, so the sum will then be divisible by the product, which is $x^2+2x-3$. So,

$$ \begin{aligned} x^2 + 2x -3 &\mid x^4+2x^3−2x^2+x−1+ax+b\ \mbox{if}\\ (x-1)(x+3) &\mid x^4+2x^3−2x^2+x−1+ax+b\\ P(1) &= P(-3) = 0\ \mbox{where}\\ P(x) &= x^4+2x^3−2x^2+x−1+ax+b\\ P(1) &= a + b + 1 = 0,\\ P(-3) &= -3a + b + 5 = 0\\ a + b &= -1\\ -3a + b &= -5\\ a&=1\\ b&=-2 \end{aligned} $$

Added

The poster asked why I needed a degree $1$ polynomial. Remember that integers have this property: when I divide integer $a$ by integer $b$, I get an equation:

$$ a = bq + r,\quad 0\le r < |b|. $$

The absolute value is what we call a Euclidean norm, and division has the remainder have its norm smaller than the dividend. The integers $\mathbb{Z}$ are called a Euclidean ring.

Polynomials are also a Euclidean ring, where the norm is just the degree:

$$ a(x) = b(x)q(x) + r(x),\quad 0\le \deg r < \deg b. $$

So, if you had found an arbitrary solution $p(x)$ to your problem, then you could divide $p$ by $x^2+2x-3$ and get a remainder of degree less than $2$, and that would be unique. So, use $ax+b$ so you have only two integers to find.

Eric Jablow
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  • Thanks for the answer . This is the answer given in my textbook but I wanted to clear why we must add a degree 1 polynomial to the dividend ? – MayankJain Sep 12 '13 at 12:21
  • Actually, you can add that degree 1 polynomial plus any multiple of $x^2 + 2x -3$. I'll edit my post. – Eric Jablow Sep 12 '13 at 12:55
  • I am sorry but I am still not clear , I understand that the remainder will have a degree less than 2 but we aren't adding the remainder , are we ? Adding the remainder does not give a number that is divisible by the divisor right ? – MayankJain Sep 12 '13 at 13:33
  • Subtracting it will. And negating a polynomial keeps the degree the same. – Eric Jablow Sep 12 '13 at 14:44
  • Thanks a lot , I got that the polynomial that we are adding should have a degree which is less than 2 but the degree can also be zero i.e it can be a constant right ? Then how would ax + b work ? Sorry if I'm being stupid right now but I'm a little confused right now . You cleared half my doubt so I assume you'll be able to clear the rest too. – MayankJain Sep 12 '13 at 16:42
  • Well, if the equations for $a$ and $b$ end in $a=b=0$, then one needs to add $0x + 0=0$ to the original polynomial. Remember that $0\le \deg q(x) < 2$. Zero is an allowable value of $\deg q(x)$, so of $ax + b$. It all comes down to one algebraic fact: If $q(x)$ has only single roots, then $q(x)|p(x)$ if and only if every root of $q$ is a root of $p$, just as for distinct primes $p, q$, $pq | r$ if $p|r$ and $q|r$. – Eric Jablow Sep 12 '13 at 20:20
  • Ok , Thanks for clearing my doubt . $|$ does mean evenly divides right ? What if the two roots are not relatively prime ? – MayankJain Sep 13 '13 at 05:47
  • That does not matter. $4$ and $6$ are not relatively prime in $\mathbb Z$. However, $x-4$ and $x-6$ are relatively prime in $\mathbb{Q}[x]$. – Eric Jablow Sep 13 '13 at 19:13
  • How do we know that $x-4$ and $x-6$ are relatively prime ? – MayankJain Sep 14 '13 at 09:18
  • Because $\frac12 (x-4) - \frac 12 (x-6) = 1$. I think I'm getting too technical here. – Eric Jablow Sep 14 '13 at 12:45
  • I think you are as I didn't understand that . Is there any simple way you could explain it to me ? – MayankJain Sep 15 '13 at 14:49
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You can get to a quartic divisible by $x^2+2x-3$ by writing

$$\begin{align} x^4+2x^3-2x^2+x-1+\text{something}&=(x^2+2x-3)(x^2+ax+b)\cr &=x^4+(a+2)x^3+(2a+b-3)x^2+(2b-3a)x-3b\cr \end{align}$$

which leads to

$$\text{something} = ax^3+(2a+b-1)x^2+(2b-3a-1)x+(1-3b)$$

for any coefficients $a$ and $b$ (of the quotient) that your heart desires. What the book presumably has in mind is to make the "something" of degree as small as possible. You can obviously get rid of the $ax^3$ by setting $a=0$ and then the $(2a+b-1)x^2$ by setting $b=1$. This leaves

$$\text{something}=(2b-3a-1)x+(1-3b)= x-2$$

Barry Cipra
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