# PreCalcCaching Box 4 - Polynomials and Sigma help

• Aug 30th 2010, 12:43 PM
khfan93
PreCalcCaching Box 4 - Polynomials and Sigma help
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
2. When x^27 - 6x^13 -x^4 + 4 is divided by x -1, the remainder is http://mathbits.com/caching/PCmath44b2.gif. What value will fill the empty box?
3. In the equation 2x^4 - 6x^2 + 4x - 3 = 0, the sum of the roots is ___.
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.

Final question to move on to next page: Evaluate http://mathbits.com/caching/pcmathsum9.gif
I cannot determine an answer for this one.

• Aug 30th 2010, 02:23 PM
khfan93
I hate to be a pain but I need help on this ASAP... thanks
• Aug 30th 2010, 03:05 PM
OnMyWayToBeAMathProffesor
I do not understand what your question is. Are you having an issue with the last one? Could you be more clear?
• Aug 30th 2010, 03:11 PM
pickslides
I agree with your answer to number 2. How did you get the rest? Are you willing to show your workings?
• Aug 30th 2010, 04:07 PM
Soroban
Hello, khfan93!

Quote:

$\text{1. Find the product of the roots of: }\;x^5-7x^2+6x-12\:=\:0$

$\text{My answer: 12}$ . . . . 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).

. . $x^5 + 0x^4 + 0x^3 - 7x^2 + 6x - 12 \;=\;0$

. . $+ \quad\; - \quad\;\;\; + \quad\; - \quad\;\;\;+ \quad\;\;-
$

. . $x^5 + 0x^4 + 0x^3 - 7x^2 + 6x - 12 \:=\:0$
. . . . . $\Uparrow$. . . . . . . . . . . . . . . . . $\Uparrow$
. . . . sum of the roots. . . . . product of the roots

Therefore, the product of the roots is $12$.

Quote:

$\text{2. When }x^{27} - 6x^{13} -x^4 + 4\text{ is divided by }(x -1),$
. . . $\text{the remainder is: }\frac{\boxed{\,}}{x-1}$

$\text{What value will fill the empty box?}$

$\text{My answer: -2}$ . . . . Correct!

When a polynomial $P(x)$ is divided by $(x-a)$, the remainder is: . $P(a)$

Substitute $x = 1$ into the polynomial:

. . $P(1) \:=\:1^{27} - 6\cdot1^{13} - 1^4 + 4 \:=\:1 - 6 - 1 + 4 \;=\;-2$

Quote:

$\text{3. In the equation }2x^4 - 6x^2 + 4x - 3 \:=\: 0$
. . . $\text{the sum of the roots is }\_\_.$

$\text{My answer: }\:2 + \sqrt{3}$ . . . . no

. . . . . . . . $+ \quad\;\; -$
We have: . $x^4 + 0x^3 - 3x^2 + 2x - \frac{3}{2} \:=\:0$
. . . . . . . . . . . $\Uparrow$

Therefore, the sum of the roots is 0.

Quote:

$\text{4. If }(2 - i) \text{ and }\sqrt{3}\,\text{ are two roots of:}$
. . . . $x^5 - 3x^4 - 2x^3 + 14x^2 - 3x - 15 \:=\: 0$

$\text{find the equation's one rational root.}$

$\text{My answer: -1}$ . . . . Yes!

Complex roots always appear in conjugate pairs.
If $(2-i)$ is a root, then $(2+i)$ is also a root.

Irrational roots always appear in conjugate pairs.
If $\sqrt{3}$ is a root, then $\text{-}\sqrt{3}$ is also a root.;

So we have four of the roots; let the fifth root be $r.$

In the equation, we see that the product of the roots is 15.

So we have: . $(2-i)(2+i)(\sqrt{3})(\text{-}\sqrt{3})r \;=\;15$

. . $(5)(\text{-}3)r \:=\:15 \quad\Rightarrow\quad \text{-}15r \:=\:15 \quad\Rightarrow\quad r \:=\:\text{-}1$