Question

If 3 and -3 are two zeroes of polynomial p(x)=x⁴+x³-11x²-9x+18 then find remaining two zeroes.​

Answer 1

Answer:

■(-2)&(1)

Step-by-step explanation:

alpha=3 & beta=-3

p(x)=x^2-(alpha+beta)x+(alpha×beta)

=x^2-[(3)+(-3)]x+[(3)(-3)]

=x^2-0x-9

=x^2-9

■ Divide the polynomial x^4+x^3-11x^2-9x+18 by x^2-9

x^2+x-2

x^2-9|x^4+x^3-11x^2-9x+18

|x^4 -9x^2

| x^3-2x^2-9x+18

| x^3 -9x

| -2x^2+18

| -2x^2+18

0

=x^2+x-2

=(x+2)(x-1)

☆(x+2)=0 | ☆(x-1)=0

x=-2 | x=1

Therefore,the other two zeroes are (-2) and (1).