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

• Jan 11th 2007, 04:03 PM
zagsfan20
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.

• Jan 11th 2007, 04:44 PM
AfterShock
Quote:

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)
• Jan 11th 2007, 04:46 PM
AfterShock
Quote:

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)
• Jan 11th 2007, 04:54 PM
AfterShock
Quote:

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.
• Jan 11th 2007, 04:58 PM
AfterShock
Quote:

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)
• Jan 11th 2007, 05:10 PM
topsquark
Quote:

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!! :mad: 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
• Jan 11th 2007, 05:12 PM
AfterShock
Quote:

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
• Jan 11th 2007, 05:16 PM
AfterShock
Quote:

Originally Posted by topsquark
You posted this in another thread!! :mad: 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 :p .
• Jan 11th 2007, 05:19 PM
zagsfan20
Thank you to both of you guys!

Much appreciated.
• Jan 11th 2007, 05:29 PM
topsquark
Quote:

Originally Posted by AfterShock
Meh. What a shame. Well, Dan, our results agree :p .

Always a nice thing to have happen. :D

-Dan