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Question from Ryan, a student:

what are all the zeros of f(x)= x^3-1/2x^2+1/3x-1/6

Hi Ryan,

So much of factoring and in fact other parts of mathematics is pattern recognition. It's like hearing the first line of a song and knowing what group you are hearing even though you may have never heard the song before.

When I see f(x)= x3 1/2x21/3x - 1/6 what immediately strikes me is that 1/3x - 1/61/3(x - 1/2).

Does this help?

If you need further assistance write back,
Penny

Ryan wrote back

I am still stuck I need to see he steps and the answer for it ti make sense
thank you so much

Ryan, let me look at a similar problem.

What are all the zeros of g(x) = 3/2 x3 + 3 x2 - 1/8 x - 1/4?

Again what I see in the last two terms is that

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

Now when I look at the first two terms I see a similar factoring

3/2 x3 + 3 x2 = 3x2 (1/2 x + 1)

so here is my solution.

3/2 x3 + 3 x2 - 1/8 x - 1/4 = 0
(3/2 x3 + 3 x2) - (1/8 x + 1/4) = 0
3x2 (1/2 x + 1) - 1/4 (1/2 x + 1) = 0
(3x2 - 1/4)(1/2 x + 1) = 0

Thus either

3x2 - 1/4 = 0 or 1/2 x + 1 = 0. The first gives x = ±√3/2 and the second gives x = -1/2.

Penny

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