# Thread: [SOLVED] How to show that the n factor is always positive?

1. ## [SOLVED] How to show that the n factor is always positive?

Considering the polynomial, $p(x) = 2x^4 + x^3 - 3x^2 - 2x -2$,

Show that the third factor of P(x) is always positive. The other two factors are $x-\sqrt{2}$ and $x+\sqrt{2}$. Hence find the set values of x such that P(x)>0

2. Originally Posted by mark1950
Considering the polynomial, $p(x) = 2x^4 + x^3 - 3x^2 - 2x -2$,

Show that the third factor of P(x) is always positive. The other two factors are $x-\sqrt{2}$ and $x+\sqrt{2}$. Hence find the set values of x such that P(x)>0
Show that the third factor of P(x) is always positive.

divide $p(x)$ by $(x-\sqrt{2})(x+\sqrt{2}) = x^2-2$

... find the set values of x such that P(x)>0

solve the inequality $x^2 - 2 > 0$

3. Originally Posted by mark1950
Considering the polynomial, $p(x) = 2x^4 + x^3 - 3x^2 - 2x -2$,

Show that the third factor of P(x) is always positive. The other two factors are $x-\sqrt{2}$ and $x+\sqrt{2}$. Hence find the set values of x such that P(x)>0
The product of the two given factors is $x^2 - 2$. Therefore $p(x) = 2x^4 + x^3 - 3x^2 - 2x -2 = (x^2 - 2)(2x^2 ........ + 1)$ and it shouldn't be too hard to fill in the gap. Then note that the third factor is an irreducible quadratic ....

4. Oh thanks guys.

5. Originally Posted by skeeter
Show that the third factor of P(x) is always positive.

divide $p(x)$ by $(x-\sqrt{2})(x+\sqrt{2}) = x^2-2$

... find the set values of x such that P(x)>0

solve the inequality $x^2 - 2 > 0$
I found out the third factor and did the discriminant thingy, b2 - 4ac and found that it's less than 0. This shows that it is always above the x-axis and therefore it is always positive. Is it correct or did I miss out something?

6. Originally Posted by skeeter
... find the set values of x such that P(x)>0

solve the inequality $x^2 - 2 > 0$
Um, if u don't mind me asking, why did you choose $x^2 - 2 > 0$? Can we choose other factors instead of that?

Since they used the word, Hence, shouldn't we be using something from 2x^2 + 2x + 1?

7. Originally Posted by mark1950
I found out the third factor and did the discriminant thingy, b2 - 4ac and found that it's less than 0. This shows that it is always above the x-axis and therefore it is always positive. Is it correct or did I miss out something?
If the leading term of an irreducible quadratic function q(x) is positive then q(x) > 0 for all x.

If the leading term of an irreducible quadratic function q(x) is negative then q(x) < 0 for all x.

Originally Posted by mark1950
Um, if u don't mind me asking, why did you choose $x^2 - 2 > 0$? Can we choose other factors instead of that?

Since they used the word, Hence, shouldn't we be using something from 2x^2 + 2x + 1?
The only thing that can make p(x) negative is the $x^2 - 2$ factor ....

8. Originally Posted by mark1950
Um, if u don't mind me asking, why did you choose $x^2 - 2 > 0$? Can we choose other factors instead of that?
Since they used the word, Hence, shouldn't we be using something from 2x^2 + 2x + 1?
By this point, you had already proved that $2x^2+ 2x+ 1$, the "third factor" was always positive. That meant that whether the entire polynomial was positive or not depended entirely on the other two factors.