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#1 2014-07-23 07:18:30

demha
Member
Registered: 2012-11-25
Posts: 186

Factoring Polynomials of Higher Degree

Hey guys, I need help with Factoring Polynomials of Higher Degree. I got around 20 questions and I'll post 5 at a time. For the first five I answered all but one:

1. 125x^3 + 64
A (5x - 4)(25x^2 + 20x - 16)
B(5x + 16)(25x^2 - 20x + 2)
C(5x + 8)(25x^2 - 40x + 16)
D(5x + 4)(25x^2 - 20x + 16)
E (5x + 2)(25x^2 - 40x + 16)
F cannot be factored

My answer is D.


2. 216a^3 + 1
A (3a + 1)(24a^2 - 8a + 1)
B(6a + 1)(36a^2 - 6a + 1)
C(18a + 1)(2a^2 - 6a + 1)
D(4a + 1)(20a^2 - 4a + 1)
E (6a - 1)(36a^2 + 6a + 1)
F cannot be factored

My answer is B.


3. 7x^5 - 64y
A 7xy(x^2 - 8)(x^2 + 8)
B 7x(x^2 - 8)(x^2 + 7)
C 7xy(x^4 - 8)
D 7x(x^2 - 9)(x^2 + 1)
E 7(x5 - 9x)
F cannot be factored

My answer is F.


4. 27x^3 - 8
A (3x - 2)(3x + 2)2
B(3x - 2)(9x^2 - 6x + 4)
C(3x + 2)(3x^2 - 3x - 4)
D(27x + 2)(x^2 - 2x + 4)
E (3x - 2)(9x^2 + 6x + 4)
F cannot be factored

My answer is E.


5. x^9 + 1
A (x + 1)(x^2 - x + 1)(x^6 - x^3 + 1)
B(x^3 + 1)(x^6 + 1)
C(x + 1)(x^2 - x + 1)(x^3-1)2
D(x + 1)3(x^2 - x + 1)3
E (x^3 + 1)(x^6 - x^3 + 1)
F cannot be factored

I am having a little trouble on this one. I'm not really sure how I would even begin with it.


"The thing about quotes on the Internet is you cannot confirm their validity"
~Abraham Lincoln

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#2 2014-07-23 07:33:05

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,776

Re: Factoring Polynomials of Higher Degree

1 D is correct.

2) B is correct.

3) F is correct.

4) E is correct.

5) That is not right.


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#3 2014-07-23 12:34:58

Agnishom
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From: The Complex Plane
Registered: 2011-01-29
Posts: 16,876
Website

Re: Factoring Polynomials of Higher Degree

5. Note that x^9 is a perfect cube and so is 1


'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
'Humanity is still kept intact. It remains within.' -Alokananda

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#4 2014-07-26 03:36:03

demha
Member
Registered: 2012-11-25
Posts: 186

Re: Factoring Polynomials of Higher Degree

I come up with (x^3)^3 + 1^3 but I'm not entirely sure where to go from here.


"The thing about quotes on the Internet is you cannot confirm their validity"
~Abraham Lincoln

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#5 2014-07-26 07:24:02

demha
Member
Registered: 2012-11-25
Posts: 186

Re: Factoring Polynomials of Higher Degree

Thanks ShivamS, but I want to understand how to find the answer. Like what steps am I supposed to take.


"The thing about quotes on the Internet is you cannot confirm their validity"
~Abraham Lincoln

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#6 2014-07-26 09:29:45

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,776

Re: Factoring Polynomials of Higher Degree

One way is by trying the answers and seeing which one is right.


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#7 2014-07-26 11:02:59

ShivamS
Member
Registered: 2011-02-07
Posts: 3,537

Re: Factoring Polynomials of Higher Degree

Hi.

Google sum of cubes. The first page from purplemath is what you need.

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#8 2014-07-27 05:20:01

demha
Member
Registered: 2012-11-25
Posts: 186

Re: Factoring Polynomials of Higher Degree

It's still not making too much sense for me. I tried putting in to this:
(a + b)(a2 – ab + b2)

But it just comes up into a huge mess:
(x^3)^3 + 1^3

((x^3)^3 + 1^3)(((x^3)^3)2 – ((x^3)^3)(1^3) + (1^3)^2)


"The thing about quotes on the Internet is you cannot confirm their validity"
~Abraham Lincoln

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#9 2014-07-27 05:38:03

ShivamS
Member
Registered: 2011-02-07
Posts: 3,537

Re: Factoring Polynomials of Higher Degree

In (x^3)^3 + 1^3,

a = x^3
b = 1

so it becomes (x^3 + 1)(x^6 - x^3 + 1)

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#10 2014-07-27 06:03:39

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,395

Re: Factoring Polynomials of Higher Degree

hi demha

The rules of algebra are just the rules of arithmetic.  As a result if expression A = expression B, you can substitute any value for x and you will get number = same number.

So choose a value for x and evaluate the expression in the question.  Then evaluate every multi choice answer with the same choice for x.  Providing you don't slip up with the calculations, if one multi choice has the same result as the question, that must be the answer.

But there is a small chance that your choice for x will lead to two multi choice answers giving the same, correct result.  If that happens you would have to make another choice for x to determine which answer is right.

eg.

5. x^9 + 1
A (x + 1)(x^2 - x + 1)(x^6 - x^3 + 1)
B(x^3 + 1)(x^6 + 1)
C(x + 1)(x^2 - x + 1)(x^3-1)2
D(x + 1)3(x^2 - x + 1)3
E (x^3 + 1)(x^6 - x^3 + 1)

Choose x = 1

Question = 1^9 + 1 = 2

A = 2 x 1 x 1 = 2
B = 2 x 2 = 4
C = 2 x 1 x 0 = 0
D = 2 x 3 x 1 = 6
E = 2 x 1 = 2

So we have two contenders: A and E

choose another x, say, x = 2

Question = 512 + 1 = 513

A = 3 x 3 x 57 = 513
E = 9 x 57 = 513

Oh dear.  Could this mean both answers work?

It would if x^3 + 1 = (x+1)(x^2-x+1).

I'll leave it to you to show that this is indeed true.

So which answer should you pick?

You'll have to 'read the small print'.  Does it say 'completely factorise' ?  If so then pick A.  If it doesn't, I'd still pick A as a superior factorisation.  But, if this is CompuHigh, it could just be that the question compiler made a mistake.  sad

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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