Divide and Check.

See below please

Question

Divide and Check.

See below please

e.mccormick · Answer

The saga continues....  hehe

anonymous · Answer

$$\frac{ x^7 + x^3 }{ x }$$

would it just be $$x^6 + x^2$$

e.mccormick · Answer

Yes. There is no real reason too do more.  IF and ONLY IF it said, "fully factored form" then you would need to do more.  $x^2(x^4+1)$ is the fully factored version of the same thing.

anonymous · Answer

Makes sense!

e.mccormick · Answer

Yah, that is the sort of thing they try and get you with.  They do not realy make "trick" questions.  The purpose is to get you to read the question fully and consider the answer carefully.  So if there is just simplify, all you do is make it simple.  If it says factor or simplified factor form, or anything like that, then they mean more.

anonymous · Answer

Find the GCF then factor:

$$2x^7 + 4x^6 - 6x^5 + 4x^4$$

$$x^4$$ = GCF

$$x^4(2x^3 + 4x^2 - 6x + 4)$$

Is that the farthest I can factor it? @e.mccormick

e.mccormick · Answer

You can also factor out a 2 easily.  then it is seeing if the $P_3$ factors.  FYI: $P_3$ means Polynomeal of the 3rd degree.

e.mccormick · Answer

$2x^4(x^3+2x^2-3x+2)$ is the easy part.  Have you done polynomeal long division yet?

anonymous · Answer

I did not even notice the numbers! I was only looking at the exponents! man I need to look more! and no i have not.

e.mccormick · Answer

OK.  If they have not gone into the division, it may not be something they mean you to factor.  There is actually a way to divide othose things and if there is no remainder it is a factor.

anonymous · Answer

well this is the last week of class so I do not think we are going into that.

e.mccormick · Answer

Other factors would be nasty fractions in this case, so I do not think they expect you to do it.

anonymous · Answer

For this one it says to completely factor the polynomial.

$$7x^3 + 63x^2 + 7x + 63)$$

$$7(x^3 + 9x^2 + x + 9)$$

Should I also factor inside the parenthesis or is this okay? @e.mccormick

e.mccormick · Answer

Yah, try a little grouping there.

e.mccormick · Answer

In fact, you probably could have factored by grouping at the start, but it makes little difference.

anonymous · Answer

so? $$(x^3 + 9x)(x + 9)$$

anonymous · Answer

whoops its $$(x^3 + 9x^2)(x + 9)$$

@e.mccormick

e.mccormick · Answer

Hehe. You caught something!  Always best if you see it before I do.

e.mccormick · Answer

Hmm.... that would be $x^4$... etc.

anonymous · Answer

wait so I would multiply them again??

e.mccormick · Answer

I am saying that could not be a factor if you try and check it.

anonymous · Answer

so we would just leave as is?

anonymous · Answer

I dont like grouping lol

e.mccormick · Answer

Welcome to the club.

anonymous · Answer

Some of these are fun but than again, some equations look the same but you have to solve differently which confuses me.

e.mccormick · Answer

$$7x^3+63x^2+7x+63\Rightarrow 7(x^3+9x^2)+7(x+9)$$Now, do you see another common factor there?

e.mccormick · Answer

Yah, they have all these different ways to skin a polynomial in math.

anonymous · Answer

I don't see any other common factors. I think maybe we can factor the first parenthesis but im not sure

e.mccormick · Answer

$$7(x^3+9x^2)+7(x+9)\Rightarrow 7[x^2(x+9)+(x+9)]$$

anonymous · Answer

thats what I was seeing but wasnt sure where the 7 would go!!

e.mccormick · Answer

I could have left the 7 in or taken it out.  Eventually it would come out in fully factored form.
$$7[x^2(x+9)+(x+9)]\Rightarrow 7(x^2+1)(x+9)$$

e.mccormick · Answer

Now you can foil the last part and see what happens if you want to check.  That was what I started to do to check yours and I got $x^4$ as the first term, so I knew something had gone wrong.

anonymous · Answer

I checked it and it works!!

e.mccormick · Answer

I agree that grouping is a pain in the anatomy and can be hard to see.  However, when there are lots of numbers that divide up easily, that is a big clue that you may need to do it.

anonymous · Answer

true! when it comes to factoring trinomials completely, i know that there are two ways to do it. for this one can you help? (p.s, I really appreciate you sticking with me for this long)

$$x^2 + 28x + 5$$

anonymous · Answer

or is the trinomial a prime?

e.mccormick · Answer

Well, 5 is a prime number.  So the factors of 5 are only 1 and 5.  Does 5+1 or 5-1 or 1+5 or 1-5 = 28?

anonymous · Answer

nope

e.mccormick · Answer

Well, because that is what makes the center term is combinations like that....  it can't be done.  There is another test if you know the quadratic formula.

anonymous · Answer

I know that sometimes we multiply those two numbers or something like that

e.mccormick · Answer

Yah, the first and third.  In $Ax^2+Bx+C$ the B is mathematically related to A and C if there are factors.

anonymous · Answer

regardless though this polynomial is a prime?

e.mccormick · Answer

Yah.  That one looks prime.
FYI:  this was the other thing I was talking about.
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