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 :

$(i) p(x)=3x \\ => p(x)=0 \\3x=0\\x=0 \\\\ \implies the \ zero \ of \ the \ polynomial \ is \ 0. \\\\ (ii)p(x)=x^2+5x+6 \\ =>p(x)=0\\x^2+5x+6=0\\x^2+2x+3x+6=0\\x(x+2)+3(x+2)=0 \\ (x+2)(x+3)=0\\x=-2,-3 \\\\ \implies -2 \ and \ -3 \ are \ the \ zeroes \ of \ the \ polynomial \\\\ (iii) p(x)=(x+2)(x+3) \\ =>p(x)=0\\ (x+2)(x+3)=0 \\ x=-2,-3 \\\\ \implies -2 \ and \ -3 \ are \ the \ zeroes \ of \ the \ polynomial \\\\ (iv)p(x) = x^4 - 16 \\ =>p(x)=0 \\ x^4 - 16=0\\x^4 =16 \\x=\sqrt[4]{16} \\x= \pm2$

$\implies +2 \ and \ -2 \ are \ the \ zeroes \ of \ the \ polynomial$