I have one more question for now

[kiley0118]

I also need help on this one:

2x^3 - x^2 - 18x + 9 = 0

Any help would be great.

Thanks

 

[Denis]

I also need help on this one:
2x^3 - x^2 - 18x + 9 = 0
Any help would be great.
Thanks

Look up "cubic equation".

I can eyeball x = 1/2 from (2x - 1), which works;
so chances are good for (x - 1), (x + 9) or (x + 1),(x - 9).

 

[soroban]

Hello, kiley0118!

2x³ - x² - 18x + 9 .= .0
.

Factor "by grouping": .x²(2x - 1) - 9(2x - 1) .= .0

Factor: .(2x - 1)(x² - 9) .= .0

Factor: .(2x - 1)(x - 3)(x + 3) .= .0

Solutions: .x .= .1/2, 3, -3

 

[Denis]

Youtchly elegant, soroban!
Is there a special "name" for such cubics?

Any idea why, at sites that deal with solving cubics, this is not mentionned?

 

[soroban]

Hello, Denis!

Is there a special "name" for such cubics?
Any idea why, at sites that deal with solving cubics, this is not mentioned?

Hmmm, never heard of a name for a factorable cubic . . .

Nor do I understand why they omit this technique.
Maybe it is assumed that everyone tries factoring first?

I was told to try "grouping" with any polynomial with four or more terms.

Example: .x⁴ - 2x³ + x² - 3x + 3

. . . . . . = .x²(x² - 2x + 1) - 3(x - 1)

. . . . . . = .x²(x - 1)² - 3(x - 1)

. . . . . . = .(x - 1) [x²(x - 1) - 3]

. . . . . . = .(x - 1)(x³ - x² - 3)