Polynomial questions (1 Viewer)

[bos1234]

Given that P(x) = 2x^3 + x^2 -4x - 2 has a rational zero, find this zero and factorise P(x) over the set of real numbers.

But there are soooooooooooo many rational numbers. What do i do to go about finding the PROBABLE factors and why is this the case????

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Show the roots of the following equations are always real

(x-2p)(x-2q)=4r^2

i grouped them etc..

then said delta >= 0

and then?

similar one..
P(1-y-y^2) = q(1+y)

thanks bye

 

[bos1234]

so does that mean the numerator must be a factor of the constant and denominator a factor of the co-efficient???

thanks for the detailed proof.. i didnt know there was a theorem like this. THANKS!!!!!!

 

[Riviet]

so does that mean the numerator must be a factor of the constant and denominator a factor of the co-efficient???

Yes, the constant term and leading coefficient respectively.

 

[jyu]

Given that P(x) = 2x^3 + x^2 - 4x - 2 has a rational zero, find this zero and factorise P(x) over the set of real numbers.

thanks bye

Given that P(x) = 2x^3 + x^2 - 4x - 2 has a rational zero, instead of using the factor theorem, factorise by grouping the terms appropriately (not always possible) to find the rational zero(s).

P(x) = 2x^3 + x^2 - 4x - 2
= (2x^3 + x^2) - (4x + 2)
= x^2 ( 2x + 1) - 2(2x + 1)
= (2x + 1)(x^2 - 2)
= (2x + 1)(x + sqrt2)(x - sqrt2)

.: the rational zero is -1/2 and the factorised P(x) is as shown.


:wave: