Question

Find the zeroes of the given polynomials.
(i) P(x) = 3x
(ii) p(x) = x²+ 5x + 6
(iii) p(x) = (x+2)(x+3)
(iv) p(x) = x⁴ - 16​

Answer 1

Answer:

(i) 0

(ii) -2 and -3

(iii) -2 and -3 again

(iv)+2 and -2

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

(i)p(x)=3x=>p(x)=03x=0x=0⟹the zero of the polynomial is 0.

(ii)p(x)=x2+5x+6=>p(x)=0x2+5x+6=0x2+2x+3x+6=0x(x+2)+3(x+2)=0(x+2)(x+3)=0x=−2,−3⟹−2 and −3 are the zeroes of the polynomial

(iii)p(x)=(x+2)(x+3)=>p(x)=0(x+2)(x+3)=0x=−2,−3⟹−2 and −3 are the zeroes of the polynomial

(iv)p(x)=x4−16=>p(x)=0x4−16=0x4=16x=416x=±2

\implies +2 \ and \ -2 \ are \ the \ zeroes \ of \ the \ polynomial⟹+2 and −2 are the zeroes of the polynomial