#1**+10 **

Hi Quinn,

Do you want these factorized?

Are you absolutely certain that you have written the questions correctly.

$$(1)\quad(x+1)(x+3)(x+5)(x+7)+15\\\\

(2) \quad 20xy^2+y^2-x+20y^3+xy-y$$

Melody
Sep 28, 2014

#2**+15 **

$$\\(2) \quad 20xy^2+y^2-x+20y^3+xy-y\\\\

=20y^3+20xy^2\;\;+y^2+xy\;\;-y-x\\\\

=20y^2(y+x)\;\;+y(y+x)\;\;-1(y+x)\\\\

=(20y^2+y-1)(y+x)\\\\

$Using the quadratic formula to help me I determine that the roots are -0.25 and 0.2 $\\\\

=20(y-0.2)(y+0.25)(y+x)\\\\

=5(y-0.2)*4(y+0.25)(y+x)\\\\

=(5y-1)(4y+1)(y+x)\\\\$$

Melody
Sep 28, 2014

#4**+15 **

$$(1)\qquad (x+1)(x+3)(x+5)(x+7)+15$$

http://www.wolframalpha.com/input/?i=%28x%2B1%29%28x%2B3%29%28x%2B5%29%28x%2B7%29%2B15

I am just thinking about how to do this without the help of an online calc.

You will need to use the remainder theorum.

You have to look for a value of x that will make that expression = 0

hence (x+1)(x+3)(x+5)(x+7) must equal -15

for this to be true an odd number of them must be negative, which means x must be negative but bigger than -7.

I know x will not be -1,-3,-5, or -7 because they will make the expression = 0 (NOT -15)

try -2

-1*1+3*5=-15

so (x+2) is one factor.

try -4

-3*-1*1*3 no no good

try -6

-5*-3*-1*1=-15 good

so (x+2)(x+6) is a factor

Now I do not know if there is an easier way but I would expand and simplify the original question

then I would divide by x^2+8x+12

And then i would find the next factor.

does that make sense.

Melody
Sep 28, 2014

#7**0 **

Pardon?

The question is done.

Do you need me to show you how to do the division?

I assume that you can do the expansion yourself.

Why don't you do the expansion, put the answer to that up and if you need me to I can help you with the algebraic division.

Otherwise tell me what your problem really is. :)

Melody
Sep 28, 2014

#8**+10 **

Nice job on that second one Melody !!!

There may be some factoring trick for the first, but I would have probably taken Melody's approach and used the Factor Theorem, realizing that there can be no positive real roots because all the linear terms in the original question would be > 0. And multiplying these together and adding a "+15" to them could never give us "0."

The expansion of the original question is x^4 + 16X^3 + 86x^2 +176x + 120, and the factorization of this is, indeed, (x+2) (x+6) (x^2 + 8x + 10)

Using "p/q," the first root ... (-2)... could be found pretty easily...and after some synthetic division, the resulting polynomial would be ... x^3 + 14x^2 + 58x + 60......and the next rational roots that would come to mind to "test" would be -3, -4, -5 and -6. And, as Melody found, -6 is the correct one.

Then, we would use synthetic division again to reduce the polynomial to x^2 + 8x + 10.... and this polynomial has no rational zeroes. (But it does have two "real" roots!!)

Again...good job, Melody !!! Gin and tonics really DO help one's math skills.....(it's an "inside" joke !!!)

I propose a "toast".........

CPhill
Sep 28, 2014