Question

Can some one help me understand this. I need to help my little brother

2 and -1/3 are zeros of the polynomial \(\sf 3x^3-2x^2-7x-2\). Find the third zero.

Answer 1

Ninth mein hai ya tenth mein?

Answer 2

9th me

Answer 3

Plugging @Jhannybean for "teh lulz".

The sum of roots is \(\frac{2}{3}\) so the third root is \(\frac{2}{3}-\left(2 -\frac{1}{3}\right)\)

Answer 4

Thus the answer is -1.

Answer 5

We could also have done it using the product of roots. The product of roots is \(2/3\) so the third root is\[\frac{2/3}{2\cdot (-1/3)} = \boxed{-1}\]

Answer 6

Oh, ninth mein... hmm. I don't think they teach that in ninth - it's taught only in tenth. Here's another method!

Answer 7

Yes..

Answer 8

As I recall, they teach the so-called Factor Theorem in 9th grade. It says that if \(k\) is a root of a polynomial, then \(x-k\) must divide it. We know that \(2, -1/3\) are roots. Let the third root be \(k\). Then the polynomial can be written as\[3(x-2)(x+1/3)(x-k) = 3x^3 - 2x^2 - 7x - 2\]

Answer 9

Now comes the interesting part. Is he trying to understand the solution for himself or for his school-exams (as he has to write in his 9th exam-sheet)?

Answer 10

Exams..

Answer 11

Is that it? o_O

Answer 12

No. That's not it.\[3(x-2)(x+1/3)(x-k) = 3x^2 - 2x^2 - 7x - 2\]\[\Rightarrow x-k = \frac{3x^2 - 2x^2 - 7x - 2}{3(x-2)(x+1/3)}\]

Answer 13

\[x -k = \frac{3x^3 - 2x^2 - 7x - 2}{(x-2)(3x+1)} = \frac{3x^3 - 2x^2 - 7x - 2}{3x^2 - 5x - 2}\]