I hate to be a pain but I need help on this ASAP... thanks
I need the solutions to the following problems. I have worked hard on them and come up with answers but do not know how to proceed.
1. Find the product of the roots of x^5-7x^2+6x-12=0
My answer: 12.
2. When x^27 - 6x^13 -x^4 + 4 is divided by x -1, the remainder is . What value will fill the empty box?
My answer: -2
3. In the equation 2x^4 - 6x^2 + 4x - 3 = 0, the sum of the roots is ___.
My answer: 2 + sqrt(3)
4. If (2 - i) and sqrt(3) are two roots of x^5 - 3x^4 - 2x^3 + 14x^2 - 3x - 15 = 0, find the equation's one rational root.
My answer: -1.
Final question to move on to next page: Evaluate
I cannot determine an answer for this one.
Thanks in advance
Hello, khfan93!
. . . . Right!
The leading coefficient must be 1 (one).
If it is not 1, divide through by the leading coefficient.
Write all the terms of the polynomial (insert 0 coefficients where needed).
. .
Insert alternating signs on the coefficients (start with +).
. .
. .
. . . . . . . . . . . . . . . . . . . . . .
. . . . sum of the roots. . . . . product of the roots
Therefore, the product of the roots is .
. . .
. . . . Correct!
When a polynomial is divided by , the remainder is: .
Substitute into the polynomial:
. .
. . .
. . . . no
. . . . . . . .
We have: .
. . . . . . . . . . .
Therefore, the sum of the roots is 0.
. . . .
. . . . Yes!
Complex roots always appear in conjugate pairs.
If is a root, then is also a root.
Irrational roots always appear in conjugate pairs.
If is a root, then is also a root.;
So we have four of the roots; let the fifth root be
In the equation, we see that the product of the roots is 15.
So we have: .
. .