# Thread: Help with Polynomial, Factorizing and Equations (semi-urgent)

1. ## Help with Polynomial, Factorizing and Equations (semi-urgent)

I need help with the following problems.

1.) Factor: y + 1000y^4

2.) Use the general factoring strategy to help factor this polynomial:

x^3 + 2x^2 + 2x^3 - 6x -15 + 3x^2 + 5

3.) Use the general factoring strategy (and the zero product property) to help solve this equation:

x^2 - 20x = -x^2 - 9x - 5

4.) Some of these trinomials are factorable, some are prime. Factor those that can be factored:

A.] 6 + 5t - 6t^2

B.] 6 + 5t + 6t^2

C.] 6 - 5t + 6t^2

D.] 6 - 5t - 6t^2

5.)

Use factoring to identify the zero's of this polynomial: 3x^5 - 30x^3 + 27x

Any help in explaining how you do these problems is greatly appreciated! I missed the first couple days of class due to illness and came back swamped, and lost as far as what to do.

2. Originally Posted by zagsfan20
I need help with the following problems.

1.) Factor: y + 1000y^4
y(1 + 1000y^3)

You'll need to recognize it's of the form x^3 + y^3; see formula

Thus,

y*(10y + 1)*(100y^2 - 10y + 1)

3. Originally Posted by zagsfan20

2.) Use the general factoring strategy to help factor this polynomial:

x^3 + 2x^2 + 2x^3 - 6x -15 + 3x^2 + 5
Combine like-terms:

3x^3 + 5x^2 - 6x - 10

Factor:

(x^2 - 2)*(3x + 5)

4. Originally Posted by zagsfan20
3.) Use the general factoring strategy (and the zero product property) to help solve this equation:

x^2 - 20x = -x^2 - 9x - 5
x^2 - 20x = -x^2 - 9x - 5; bring all terms to one side

2x^2 - 11x + 5 = 0

Several ways to solve this. Let's change it up and use the quadratic formula; I do not know the "general factoring strategy" that your teacher showed you.

x = [11 +/- sqrt(121 - 4*2*5)]/(2*2)

x = [11 +/- sqrt(81)]/4

x = [11 +/- 9]/4

x = 20/4 AND x = 2/4

Thus, x = 5 and x = 1/4.

We could have also done this by factoring, which I personally find to be quicker.

5. Originally Posted by zagsfan20
4.) Some of these trinomials are factorable, some are prime. Factor those that can be factored:

A.] 6 + 5t - 6t^2

B.] 6 + 5t + 6t^2

C.] 6 - 5t + 6t^2

D.] 6 - 5t - 6t^2
A.] Can be factored;

(-2t + 3)*(3t + 2)

B.] CANNOT be factored.

C.] CANNOT be factored.

D.] Can be factored;

Very similar to part A, although there is a sign change.

(-2t + 3)(3t - 2)

6. Originally Posted by zagsfan20
I need help with the following problems.

1.) Factor: y + 1000y^4

2.) Use the general factoring strategy to help factor this polynomial:

x^3 + 2x^2 + 2x^3 - 6x -15 + 3x^2 + 5

3.) Use the general factoring strategy (and the zero product property) to help solve this equation:

x^2 - 20x = -x^2 - 9x - 5

4.) Some of these trinomials are factorable, some are prime. Factor those that can be factored:

A.] 6 + 5t - 6t^2

B.] 6 + 5t + 6t^2

C.] 6 - 5t + 6t^2

D.] 6 - 5t - 6t^2

5.)

Use factoring to identify the zero's of this polynomial: 3x^5 - 30x^3 + 27x

Any help in explaining how you do these problems is greatly appreciated! I missed the first couple days of class due to illness and came back swamped, and lost as far as what to do.

You posted this in another thread!! Please refer to rule #1, here. I'll leave it to you to decide if it was Aftershock or me that wasted our time.

-Dan

7. Originally Posted by zagsfan20
5.)[/B]

Use factoring to identify the zero's of this polynomial: 3x^5 - 30x^3 + 27x
3x^5 - 30x^3 + 27x = 0

3x*(x^4 - 10x^2 + 9) = 0

3x*[(x^2)^2 - 10(x^2) + 9)]

3x*[(x^2 - 4x + 3)*(x^2 + 4x + 3)]

Separate above two terms and factor each;

(x - 3)*(x - 1) and (x + 1)*(x + 3)

Thus:

3x*(x - 3)*(x - 1)*(x + 1)*(x + 3) = 0

Thus, the zeros are:

x = -3, x = -1, x = 0, x = 1, x = 3

8. Originally Posted by topsquark
You posted this in another thread!! Please refer to rule #1, here. I'll leave it to you to decide if it was Aftershock or me that wasted our time.

-Dan
Meh. What a shame. Well, Dan, our results agree .

9. Thank you to both of you guys!

Much appreciated.

10. Originally Posted by AfterShock
Meh. What a shame. Well, Dan, our results agree .
Always a nice thing to have happen.

-Dan