a few questions involving factoring

[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

 

[tkhunny]

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.

 

[soroban]

Hello, bittersweet!

For \(\displaystyle pm^3\,+\,m^2\,+\,pm\,+\,1\), how do you factor this type of question?

Do you remember "factoring by grouping"?

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


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

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

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

\(\displaystyle a)\;\;2x^3\,-\,54\;=\;0\)

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

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


And we have two equations to solve:

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

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

 

[Guest]

oooo thanks I forgot those rules