1. ## Need Help! factoring

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
1.) Factor: y + 1000y^4
$\displaystyle y + 1000y^4$

The only common factor to both terms is a y, so:
$\displaystyle y + 1000y^4 = y(1 + 1000y^3)$

Now recall that $\displaystyle a^3 + b^3 = (a + b)(a^2 + ab + b^2)$. Thus:
$\displaystyle y + 1000y^4 = y(1 + 1000y^3) = y(1 + 10y)(1 + 10y + 100y^2)$

-Dan

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
I've been trying to figure out why this is split up the way it is, but I can't figure it. Anyway, here's my approach:

$\displaystyle x^3 + 2x^2 + 2x^3 - 6x -15 + 3x^2 + 5 = 3x^3 + 5x^2 - 6x -10$

I'm going to rearrange the order of the terms here:
$\displaystyle 3x^3 + 5x^2 - 6x - 10 = (3x^3 - 6x) + (5x^2 -10)$

$\displaystyle = 3x(x^2 - 2) + 5(x^2 - 2) = (3x + 5)(x^2 - 2)$

-Dan

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
I'm not sure what is meant by the "general factoring strategy." You'll have to ask your instructor about that.

$\displaystyle x^2 - 20x = -x^2 - 9x - 5$

$\displaystyle 2x^2 - 11x + 5 = 0$

Start listing the factors of 2 x 5 = 10 and see which add to make -11.
Factors.....Sum
1, 10 ........11
2, 5............7
etc.

I get that -1 x -10 = 10 and -1 + -10 = -11.
Thus:
$\displaystyle 2x^2 - 11x + 5 = 0$

$\displaystyle 2x^2 - x - 10x + 5 = 0$

$\displaystyle (2x^2 - x) - (10x - 5) = 0$ <-- Watch the negative signs here.

$\displaystyle x(2x - 1) - 5(2x - 1) = 0$

$\displaystyle (x - 5)(2x - 1) = 0$

Now, if $\displaystyle ab = 0$ then one or both of a and b must be 0. Thus:
$\displaystyle x - 5 = 0$ or $\displaystyle 2x - 1 = 0$

$\displaystyle x = 5$ or $\displaystyle x = 1/2$

-Dan

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
For all of these, just to make sure we are working with the same format, write the polynomial in standard order: with the higher degree terms to the left.

So A becomes: $\displaystyle -6t^2 + 5t + 6$

Now use the same method I showed you in 3:
I want two factors of -6 x 6 = -36 that add to 5.
Factors......Sum
1,-36.........-35
2,-18.........-16
etc.

I get that 9 x -4 = -36 and 9 + -4 = 5. Thus:
$\displaystyle -6t^2 + 5t + 6 = -6t^2 + 9t - 4t + 6 = ... = (-3t - 2)(2t + 3) = -(3t + 2)(2t + 3)$

Similarly:
B) $\displaystyle 6t^2 + 5t + 6$ cannot be factored since there is no pair of factors of 6 x 6 = 36 that add to 5.

C) $\displaystyle 6t^2 - 5t + 6$ cannot be factored since there is no pair of factors of 6 x 6 = 36 that add to -5.

D) $\displaystyle -6t^2 - 5t + 6 = (-3t + 2)(2t + 3)$
(-6 x 6 = -36. -9 x 4 = -36 and -9 + 4 = -5)

-Dan

6. Originally Posted by zagsfan20
5.)

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

$\displaystyle = 3x(x^4 - 10x^2 + 9)$

The 4th degree factor is called a "biquadratic" since it has only 4th, 2nd, and 0 degree terms in it. We can think of these as being very similar to quadratic polynomials. Let's just consider this term for a moment.

$\displaystyle x^4 - 10x^2 + 9$

Make the substitution $\displaystyle y = x^2$. Thus:
$\displaystyle x^4 - 10x^2 + 9 = y^2 - 10y + 9$

And now we can factor this like we do any other quadratic.
$\displaystyle y^2 - 10y + 9 = y^2 - y - 9y + 9 = ... = (y - 9)(y - 1)$
(using the same method as in the last two posts.)

Now recall that $\displaystyle y = x^2$. So:
$\displaystyle x^4 - 10x^2 + 9 = y^2 - 10y + 9 = (y - 9)(y - 1) = (x^2 - 9)(x^2 - 1)$

So:
$\displaystyle 3x^5 - 30x^3 + 27x = 3x(x^4 - 10x^2 + 9) = 3x(x^2 - 9)(x^2 - 1)$

We aren't quite done yet. Recall that $\displaystyle a^2 - b^2 = (a + b)(a - b)$. The last two terms are of this form. So:
$\displaystyle 3x^5 - 30x^3 + 27x = 3x(x^4 - 10x^2 + 9) = 3x(x^2 - 9)(x^2 - 1)$$\displaystyle = 3x(x + 3)(x - 3)(x + 1)(x - 1)$

Now we wish to explore the zeros of this. That's simply the solution to
$\displaystyle 3x^5 - 30x^3 + 27x = 0$. Using the factored form, that's just simply all the possible factors set equal to zero:
$\displaystyle 3x = 0$ or $\displaystyle x + 3 = 0$ or $\displaystyle x - 3 = 0$ or $\displaystyle x + 1 = 0$ or $\displaystyle x - 1 = 0$.

Thus the zeros of the polynomial are $\displaystyle 0, \pm 3, \pm 1$.

-Dan

7. Thanks a ton for the help!

I think I'm starting to get a better understanding, but its still a work in progress.

This is a great website, very helpful!