Question

If 3 and -3 are the two zeros of polynomial p(x)=x⁴+x³-11x²-9x+18 Find the other two zeros.​

Answer 1

Step-by-step explanation:

(x+3)(x-3)=0

x²-9=0

x⁴+x³-11x²-9x+18/x²-9

when you divide you will get,

remainder=0

quotient= x²+x-2

x²+x-2=0

x²-x+2x-2=0

x(x-1)+2(x-1)=0

(x+2)(x-1)=0

∴The zeroes are -2, 1, 3 and -3

Answer 2

Answer:

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

Putting 3 in place of x, we get

=3^4+3^3-11×3^2-9×3+18

=81+27-99-27+18

=0

Let's put the 2nd zero that is -3 in place of x, we get

(-3)^4+(-3)^3-11×(-3)^2-9×(-3)+18

=81-27-99+27+18

=0

So the other two zeroes are 0 and 0

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