Polynomial Question Help Plz (1 Viewer)

[csi]

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]

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.

 

[fan96]

We have



so if then .

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

 

[csi]

We have



so if then .

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

Thanks

 

[csi]

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