Polynomials Help! (2 Viewers)

Octagon · Jan 29, 2012

A polynomial P(x) is given by P(x)= x⁵-5px + q, where p and q are real.
Prove that P(x) = 0 has 3 distinct real zeros if q⁴is less than 256p⁵.
Thnx guys!

kingkong123 · Jan 29, 2012

having a go at it. where's the question from?

SpiralFlex · Jan 29, 2012

$\fbox{Using this property: The y coordinates of the stationary points multiplied is less than zero.}$

$\therefore y_{1}*y_{2}<0\; $(Where$\; y_{1}, y_{2}\;$ are our y coordinates of our stationary points.)$$

$Why is this?$

$Let us investigate$

$x^5-5px+q=0$

$5x^4-5p=0$

$$For stationary points,$\; y'=0$

$$So$\; x=\pm p^\frac{1}{4}$

$If I substitute in the coordinates, I will get some y coordinates namely,$

$$When x =p^\frac{1}{4}$

$p^\frac{1}{4}(-4p)+q$

$$When x =-p^\frac{1}{4}$

$p^\frac{1}{4}(4p)+q$

$Now these are our y coordinates of the stationary points.$

$The product of them should be less than zero. I will explain.$

$(p^\frac{1}{4}(4p)+q)(p^\frac{1}{4}(-4p)+q)<0$

$q^2-16p^2\sqrt{p}<0$

$q^4<256p^5$

$Let's see what this actually means, this is just ONE possibility of your graph. There are many more, but generally something we notice stays the same.$

$The graphs have two stationary points, we have just found them. And they are OPPOSITE about the line y=0 not necessarily symmetrical.$

$So if we have a positive y value times a negative y values, we would expect the resultant value to be negative.$

nightweaver066 · Jan 29, 2012

Excel MX2 Patel.

Polynomials Help! (2 Viewers)

Octagon

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kingkong123

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SpiralFlex

Well-Known Member

nightweaver066

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