Polynomial Factoring

rage

New member
Joined
Nov 18, 2006
Messages
6
I have two questions. The first problem I want to make sure I am correct. It looks right to me; however I could be wrong.
1. Perform the indicated operation. If possible, simplify your answer.
5 - 2x + x
x-7 3x-21 2x-14

5 - 2x + x
x-7 3(x-7) 2(x-7)

= 5*6-2x*5+1*2
6(x-7)

= 32 – 10x
6(x-7)

= 2(16-5x)
6(x-7)

= 16-5x
3(x-7)

Problem 2:
4. Solve for x.
x^2 + 8 - 1 = 2(x+4)
x = x
 
in future, type your problems involving fractions as follows ...

5/(x-7) - 2x/(3x-21) + x/(2x-14)

or use latex ...

\(\displaystyle \L \frac{5}{x-7} - \frac{2x}{3x-21} + \frac{x}{2x-14}\)

\(\displaystyle \L \frac{5}{x-7} - \frac{2x}{3(x-7)} + \frac{x}{2(x-7)}\)

\(\displaystyle \L \frac{30}{6(x-7)} - \frac{4x}{6(x-7)} + \frac{3x}{6(x-7)}\)

\(\displaystyle \L \frac{30 - 4x + 3x}{6(x-7)} = \frac{30-7x}{6(x-7)}\)


(x^2 + 8)/x - 1 = 2(x + 4)/x

\(\displaystyle \L \frac{x^2+8}{x} - 1 = \frac{2(x+4)}{x}\)

\(\displaystyle \L \frac{x^2+8}{x} - \frac{x}{x} = \frac{2(x+4)}{x}\)

multiply every term by x to clear the fraction ...

\(\displaystyle \L x^2 + 8 - x = 2(x+4)\)

solve the equation ... remember that x cannot equal 0.
 
Hello, rage!

Silly errors in the first one . . .


1. Perform the indicated operation. If possible, simplify your answer.
. . \(\displaystyle \L\frac{5}{x\,-\,7} \,-\,\frac{2x}{3x\,-\,21} \,+\,\frac{x}{2x\,-\,14}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

. . \(\displaystyle \L\frac{5}{x\,-\,7}\,-\,\frac{2x}{3(x\,-\,7)} \,+\,\frac{x}{2(x\,-\,7)}\)

\(\displaystyle \L= \;\frac{5\cdot6 \,-\,2x\cdot\not{5} \,+\,1\cdot\not{2}}{6(x\,-\,7)}\;\nwarrow\) here!

You should have: \(\displaystyle \L\:\frac{6}{6}\,\cdot\,\frac{5}{x\,-\,7}\;-\;\frac{2}{2}\,\cdot\,\frac{2x}{3(x\,-\,7)} \;+\;\frac{3}{3}\,\cdot\,\frac{x}{2(x\,-\,7)}\)

. . \(\displaystyle \L=\;\frac{30\,-\,4x\,+\,3x}{6(x\,-\,7)} \;= \;\frac{30\,-\,x}{6(x\,-\,7)}\)





\(\displaystyle \text{4. Solve for }x:\;\;\L\frac{x^2\,+\,8}{x}\,-\,1\;=\;\frac{2(x\,+\,4)}{x}\)

Note that \(\displaystyle x\,\neq\,0\)

Multiply through by the LCD, \(\displaystyle x:\;\;x^2\,+\,8\,-\,x\;=\;2x\,+\,8\)

We have a quadratic: \(\displaystyle \,x^2\,-\,3x\;=\;0\)

. . which factors: \(\displaystyle \,x(x\,-\,3)\;=\;0\)

. . and has roots: \(\displaystyle \,x\:=\:0,\,3\)


Since \(\displaystyle x\,\neq\,0\), the only solution is: \(\displaystyle \,\fbox{x\,=\,3}\)

 
Top