Polynomial Question Help Plz (1 Viewer)

csi

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Hi,

Let P(x) = qx^3 + rx^2 + rx + q where q and r are constants, q does not equal to 0. One of the zeros of P(x) is -1. Given that ɑ is a zero of P(x), ɑ does not equal to -1, which of the following is also a zero?

(A) -1/ɑ
(B) -q/ɑ
(C) 1/ɑ
(D) q/a

Thanks!!
 

shashysha

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Hi,

Let P(x) = qx^3 + rx^2 + rx + q where q and r are constants, q does not equal to 0. One of the zeros of P(x) is -1. Given that ɑ is a zero of P(x), ɑ does not equal to -1, which of the following is also a zero?

(A) -1/ɑ
(B) -q/ɑ
(C) 1/ɑ
(D) q/a

Thanks!!
So they give you one of the zeros as -1, another is α and let's say the last zero can be β (as it is a degree 3 so there are 3 zeros). Doing product of roots, which is -d/a, where d=q and a=q, gives us -1(α)(β) =-1. so by dividing by -1 you get αβ = 1, and rearranging you get the final zero to be β = 1/α, which gives you C.
 
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fan96

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We have



so if then .

Clearly cannot be zero, so is another root of .
Also since , and are indeed distinct roots.
 
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csi

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So they give you one of the zeros as -1, another is α and let's say the last zero can be β (as it is a degree 3 so there are 3 zeros). Doing product of roots, which is -d/a, where d=q and a=q, gives us -1(α)(β) =-1. so by dividing by -1 you get αβ = 1, and rearranging you get the final zero to be β = 1/α, which gives you C.
Thanks
 

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