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
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!
Start listing the factors of 2 x 5 = 10 and see which add to make -11.
1, 10 ........11
I get that -1 x -10 = 10 and -1 + -10 = -11.
<-- Watch the negative signs here.
Now, if then one or both of a and b must be 0. Thus:
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.
I get that 9 x -4 = -36 and 9 + -4 = 5. Thus:
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.
(-6 x 6 = -36. -9 x 4 = -36 and -9 + 4 = -5)
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:
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 .