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.
Thanks in advance!
I'm not sure what is meant by the "general factoring strategy." You'll have to ask your instructor about that.
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:
<-- Watch the negative signs here.
Now, if then one or both of a and b must be 0. Thus:
or
or
-Dan
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:
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:
Similarly:
B) cannot be factored since there is no pair of factors of 6 x 6 = 36 that add to 5.
C) cannot be factored since there is no pair of factors of 6 x 6 = 36 that add to -5.
D)
(-6 x 6 = -36. -9 x 4 = -36 and -9 + 4 = -5)
-Dan
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.
Make the substitution . Thus:
And now we can factor this like we do any other quadratic.
(using the same method as in the last two posts.)
Now recall that . So:
So:
We aren't quite done yet. Recall that . The last two terms are of this form. So:
Now we wish to explore the zeros of this. That's simply the solution to
. Using the factored form, that's just simply all the possible factors set equal to zero:
or or or or .
Thus the zeros of the polynomial are .
-Dan