Question

A polynomial with integer coefficients is of the form \[x^4 + a_3 x^3 + a_2 x^2 + a_1 x + 18.\]You are told that the integer $r$ is a double root of this polynomial. (In other words, the polynomial is divisible by $(x - r)^2.$) Enter all the possible values of $r,$ separated by commas.

Answer 1

Answer:-2 and 2

Step-by-step explanation:

Since the polynomial given [x^4 + a_3 x^3 + a_2 x^2 + a_1 x + 18.\] is divisible by (x - r)^2, this means that (x - r)^2 is a factor of the polynomial just like 4 is divisible by 2, we say 2 is a factor.

Since (x - r)^2 is a factor, we equate the factor to zero in order to get the 'x' variable.

(x - r)^2 =0

Taking square root of both sides gives x-r =0 i.e x=r

Substituting x=r into the polynomial to get 'r' we have r^4 + a_3 r^3 + a_2 r^2 + a_1r + 18=0.

Let a be 1 since the coefficients are integers

We then have Answer:

Step-by-step explanation:

Since the polynomial given [x^4 + a_3 x^3 + a_2 x^2 + a_1 x + 18.\] is divisible by (x - r)^2, this means that (x - r)^2 is a factor of the polynomial just like 4 is divisible by 2, we say 2 is a factor.

Since (x - r)^2 is a factor, we equate the factor to zero in order to get the 'x' variable.

(x - r)^2 =0

Taking square root of both sides gives x-r =0 i.e x=r

Substituting x=r into the polynomial to get 'r' we have r^4 + r^3 + r^2 + r + 18=0.

Factorizing this, possible value of r will be -2 and 2