a few questions involving factoring

G

Guest

Guest
For pm^3+m^2+pm+1, how do you factor these type of questions? quad formula or is there some method?

Do (1,-1), (2,-1),(3,1),(4,5) lie on the graph of a quadratic funation? [huh??] How do you know what point lay on a quad function?At first I thought no since there's two -1s for the y value but the answer is yes, so how do you know??

solve for x,x can equal complex numbers:

a) 2x^3-54=0
at first I thought that you can just do 2x^3=54
x^3=54/2
x= 3 but the text book's answer is 3, (-3+or-3i, 3squarerooted)/2 what did they do?

that's all my questions for now,
thanks a lot for the help
 
bittersweet said:
Do (1,-1), (2,-1),(3,1),(4,5) lie on the graph of a quadratic funation?
This one is kind of fun. Second differences are constant for points on a quadratic.

(1,-1), (2,-1),(3,1),(4,5)

Since the x-values are sequential and equally spaced, just deal with the y-values two at a time.

-1 - (-1) = 0
1 - (-1) = 2
5 - 1 = 4

Then

2 - 0 = 2
4 - 2 = 2

That's it. There IS a quadratic containing those four points.
 
Hello, bittersweet!

For pm3+m2+pm+1\displaystyle pm^3\,+\,m^2\,+\,pm\,+\,1, how do you factor this type of question?
Do you remember "factoring by grouping"?

Factor 'in pairs': m2(pm+1)+(pm+1)\displaystyle \:m^2(pm\,+\,1)\,+\,(pm\,+\,1)


Both groups have a common factor, (pm+1)\displaystyle (pm\,+\,1).

    \displaystyle \;\;Factor it out:   (pm+1)(m2+1)\displaystyle \;(pm\,+\,1)(m^2\,+\,1)


Solve for x.    x\displaystyle x.\;\;x can equal complex numbers.

a)    2x354  =  0\displaystyle a)\;\;2x^3\,-\,54\;=\;0
We have: x327  =  0\displaystyle \:x^3\,-\,27\;=\;0 . . . a difference of cubes

Factor: \(\displaystyle \:(x\,-\,3)(x^2\,+\,3x\,+\,9)\;=\;0\)


And we have two equations to solve:

[1]  x3=0        x=3\displaystyle [1]\;x\,-\,3\:=\:0\;\;\Rightarrow\;\;x\,=\,3

[2]  x2+3x+9=0\displaystyle [2]\;x^2\,+\,3x\,+\,9\:=\:0
        \displaystyle \;\;\;\;Quadratic Formula: x  =  3±324(1)(9)2(1)  =  3±272\displaystyle \:x\;=\;\frac{-3\,\pm\,\sqrt{3^2\,-\,4(1)(9)}}{2(1)} \;= \;\frac{-3\,\pm\,\sqrt{-27}}{2}
        \displaystyle \;\;\;\;Therefore: x  =  3±3i32\displaystyle \:x\;=\;\frac{-3\,\pm\,3i\sqrt{3}}{2}
 
Top