Factoring polynomial - Seems difficult

[Barchie]

Homework Statement

Consider the polynomial p(x) = x^5+45x^3+324x-3x^4 - 135x^2 - 972 . Given that p has some integer roots on the real and imaginary axes, factorise p into linear and quadratic factors with real coefficients.

Enter your answer a set of factors in the form { x-1, x+4, x^2+83,...} as appropriate to the given polynomial.

The Attempt at a Solution

Can anyone push me in the right direction here. I am not too sure where to start factoring this guy. I thought of a substitution, but i can't think of one that will work.

 

[Mentallic]

Homework Statement

Consider the polynomial p(x) = x^5+45x^3+324x-3x^4 - 135x^2 - 972 . Given that p has some integer roots on the real and imaginary axes, factorise p into linear and quadratic factors with real coefficients.

Enter your answer a set of factors in the form { x-1, x+4, x^2+83,...} as appropriate to the given polynomial.

The Attempt at a Solution

Can anyone push me in the right direction here. I am not too sure where to start factoring this guy. I thought of a substitution, but i can't think of one that will work.

It's given that the roots of the polynomial are on the real and imaginary axis, thus, they are of the form x-a for some integer a and x-ib for an integer. But since p has all real coefficients, if x-ib is a root, what other root can you conclude as well? Also, do you know the rational root theorem?

 

[epenguin]

It's no virtue you do it after we tell you how. It's the how that is the test.

Suppose the solution were like the example they give you. { x-1, x+4, x^2+83}.

What would be the polynomial (x⁴ + ...) that has those factors? Don't do any actual multiplications of numbers. Then maybe you'll see. (And maybe see that it was explained in class or book, maybe not.)

 

[eumyang]

$$x^5 + 45x^3 + 324x - 3x^4 - 135x^2 - 972$$

You don't need the rational roots test at all. You can factor this polynomial by grouping and find the (only) real root. Look at the first 3 terms and take out the GCF, and look at the last 3 terms and take out the GCF. If you don't know what I mean, here's an example with 2+2 terms instead of 3+3:
$$\begin{aligned} x^3 - x^2 + 4x - 4 &= x^2(x - 1) + 4(x - 1) \\ &= (x^2 + 4)(x - 1) \end{aligned}$$

After factoring out the (only) linear factor, you will see a quartic that is "quadratic like" -- one with only an x⁴ term, an x² term and a constant term. Factor it as if it's a quadratic by letting y = x². Here's another example showing what I mean:
$$x^4 + 13x^2 + 36$$
let y = x²:
$$y^2 + 13y + 36$$
$$(y + 4)(y + 9)$$
$$(x^2 + 4)(x^2 + 9)$$

(Hopefully that's not too many hints.)EDIT: Oops, the original problem said factor, not solve.

 

[Barchie]

It's given that the roots of the polynomial are on the real and imaginary axis, thus, they are of the form x-a for some integer a and x-ib for an integer. But since p has all real coefficients, if x-ib is a root, what other root can you conclude as well? Also, do you know the rational root theorem?

Perhaps i need to brush up on exactly what it is, but i have always been under the impression i was using the rational root theorem to solve these types of problems, however the method i would usually undertake would be to

1. Get a list of numbers for set p which would be factors of the constant
2. Get a list of numbers for set q which are taken by the factors of the coefficient of the highest power
3. Compile a list p/q, and then using synthetic division, find a factor, and continue until i have something in a form that i can use a quadratic formula on.

My issue is that 972 has 18 factors, and surely guessing until i find one isn't the best way. (maybe it is and I am just being lazy).

Also, these type of questions we are expected to answer calculatorless, even the task of finding all the factors can prove tiresome, and if they were to give a bigger constant id have no chance of doing em, what if the factors are the last number, i could be there forever and so on.

Is there a better way?

It's no virtue you do it after we tell you how. It's the how that is the test.

Suppose the solution were like the example they give you. { x-1, x+4, x^2+83}.

What would be the polynomial (x4 + ...) that has those factors? Don't do any actual multiplications of numbers. Then maybe you'll see. (And maybe see that it was explained in class or book, maybe not.)

Ok, so for this i would multiply each set out, i don't see how to do it without multiplication of numbers, and would be forever greatfull if you would impart on me this knowledge.

i would expect to get something along the lines of:

x^4+83x^2+3x^3+249x-4x^2-332
giving me:

x^4+3x^3+79x^2+249x-332 ...

Any pointers?

You don't need the rational roots test at all. You can factor this polynomial by grouping and find the (only) real root. Look at the first 3 terms and take out the GCF, and look at the last 3 terms and take out the GCF. If you don't know what I mean, here's an example with 2+2 terms instead of 3+3:


After factoring out the (only) linear factor, you will see a quartic that is "quadratic like" -- one with only an x4 term, an x2 term and a constant term. Factor it as if it's a quadratic by letting y = x2. Here's another example showing what I mean:

let y = x2:

(Hopefully that's not too many hints.)

EDIT: Oops, the original problem said factor, not solve.

Thanks anyway eumyang

 

[Barchie]

Update - I have got my answer the long winded way, it took me about 15/20 finding the factors and doing synthetic division for 1,-1,2,-2,3,-3 ... then i found (x-3) to be a factor.

after that i was left with x^4+45x^2+324 - to which luckily i found first try 9 and 36 to be factors that multiply to 324 and add to 45.

factoring then gave me my final answer.

Any tips on speeding up this process?

 

[eumyang]

Update - I have got my answer the long winded way, it took me about 15/20 finding the factors and doing synthetic division for 1,-1,2,-2,3,-3 ... then i found (x-3) to be a factor.

after that i was left with x^4+45x^2+324 - to which luckily i found first try 9 and 36 to be factors that multiply to 324 and add to 45.

factoring then gave me my final answer.

Any tips on speeding up this process?

As I said, factoring by grouping would have sped up the process. You notice that the polynomial is NOT written in standard form (decreasing exponents): it's
$$x^5 + 45x^3 + 324x - 3x^4 - 135x^2 - 972$$

and not
$$x^5 - 3x^4 + 45x^3 - 135x^2 + 324x - 972$$

So that gave me a clue that factoring by grouping may be possible. Since you already found your answer, I'll show what I had:
$$x^5 + 45x^3 + 324x - 3x^4 - 135x^2 - 972$$
$$= x(x^4 + 45x^2 + 324) - 3(x^4 + 45x^2 + 324)$$
$$= (x - 3)(x^4 + 45x^2 + 324)$$
$$= (x - 3)(x^2 + 9)(x^2 + 36)$$

If you're not familiar with factoring by grouping, kindly look it up.

 

[epenguin]

Any tips on speeding up this process?

I would ask: does x = 1 make this polynomial 0? Fairly obviously not because you can see the last term outweighs all the others. Does x = -1? No for same reason. So neither (x - 1) nor (x + 1) are factors.

Next I could try x = ±2. But wait a minute, trying every number is going to be very inefficient even if it will get me there in the end.

So look at the constant term in the equation, 972. If the factors are (x - α), (x - β),...(x - ε) then the constant term, is the product αβγδε. 2 is a factor of 972 so it is a factor of αβγδε so it is a factor of one of them, α say. It may be α or it may not. So try ±2 after all.

No, it's not that. So α is 2X something.

At this point or earlier, I would just factorise 972, Division by 2 gives 486. This still has 2 as a factor, divide by 2 again, you get 243 which by a trick I learned early at school I know is divisible by 3. Divide by 3 again ... eventually I find 972 = 2²X3⁵ (So that's 7 factors to make up 5 roots, so at least one root is not ±2 or ±3 but some product of them.

Anyway, that makes next candidate for root ±3. Lo and behold x = 3 works, x = -3 doesn't.

At this point I notice they have given me a hint. Why would they write the polynomial out that way? Normally you would write a polynomial starting with the x⁵ term + then the x⁴ term etc. They meant me to notice that the second three coefficients are -3X the first three.

I put previous paper aside and pretend I noticed this at first. Maybe I will have to do it the way I was doing another time. So I start a new page and I write my polynomial

x^5+45x^3+324x-3x^4 - 135x^2 - 972 = (x - 3)(x^4 + 45x^2 + 324)

I see that luckily in this case they have made the next step relatively simple for me too. I notice that that
(x^4 + 45x^2 + 324)

is a quadratic in x². Sum of roots is -45 and product is 324 (= 2²3⁴. I know how to do those 'by inspection'. Bit large numbers, but I could always use the quadratic root formula and again pretend afterwards.

One way or another I get
$$(x^2 + 9)(x^2 + 36)$$

That is $$(x^2 + 3^3)(x^2 + 2^23^2)$$ so all the numbers that should be there are.

Then I realize when I was thinking about the numerical substitutions before as well as ±1, ±2 etc. I had forgotten what the question also suggested, i, 2i (I don't need to worry about ± in this case because if + works so does - , as someone already explained.)

Maybe some way to make this more efficient can be achieved, especially with hindsight. Whether I would have it at my fingertips the next time I did a problem like this after months is another matter. I think I could polish and streamline this account to a point the student would say 'I could never have thought of that!'

 

[Mentallic]

As I said, factoring by grouping would have sped up the process. You notice that the polynomial is NOT written in standard form (decreasing exponents): it's
$$x^5 + 45x^3 + 324x - 3x^4 - 135x^2 - 972$$

and not
$$x^5 - 3x^4 + 45x^3 - 135x^2 + 324x - 972$$

So that gave me a clue that factoring by grouping may be possible. Since you already found your answer, I'll show what I had:
$$x^5 + 45x^3 + 324x - 3x^4 - 135x^2 - 972$$
$$= x(x^4 + 45x^2 + 324) - 3(x^4 + 45x^2 + 324)$$
$$= (x - 3)(x^4 + 45x^2 + 324)$$
$$= (x - 3)(x^2 + 9)(x^2 + 36)$$

If you're not familiar with factoring by grouping, kindly look it up.

Ahh this method is much better!

 

[epenguin]

The trouble is with my method and the others is it is not what the guys who set this problem meant us to do. That is, we have not used the information they gave us about the nature of the roots. I was doing that though in the first bit which I abandoned. As I said, we may have to use that approach in other problems.

As a half-way house instead of treating them totally with the contempt they deserve I could throw them a sop and use the info in the second part of the problem, of solving that quadratic in x²

(x⁴ + 45x² + 324)

(write y for x² if that makes it more familiar.)

Now 324 is positive so both constants in the factors are the same sign.

45 is positive so they must be both positive.

324 = 2²3⁴ we already noted; the 45 has to use these factors somehow.

We narrow that down by the fact that 45 is odd, so of the two things it is the sum of, one must be odd and one even. So our 2's have both to go in one bracket so we must have

(x⁴ + 45x² + 324)

= (x² + 3^something)(x² + 4.3^{something else})

(the something else may also equal the first something).

Out of 3's and 2's we've go to make 45 which is 3².5. The odd thing out here is 5 which can be made from 3's and 2's by 6 - 1 but we said all pluses so it's got to be 4 + 1. And the fact that there is a factor 3² in 45 is trying to tell you that 3² is in both brackets.

So (waves hands) you can see it should come from $$(x^2 + 3^3)(x^2 + 2^23^2)$$

(It has however been an advantage to have known the answer before finding it. )