Question

find the products of zeroes of the polynomials p(x) = x²+3x+2​

Answer 1

$\bf{\underline{Given:-}}$

$\sf{p(x) = x^2 + 3x + 2}$

$\bf{\underline{To\:Find:-}}$

Product of zeroes.

$\bf{\underline{Solution:-}}$

$\sf{p(x) = x^2 + 3x + 2 = 0}$

$\sf{\underline{By\:splitting\:the\:middle\:term}}$

= $\sf{x^2 + 2x + x + 2 = 0}$

= $\sf{x(x+2)+1(x+2)}$

= $\sf{(x+2)(x+1)}$

Either,

$\sf{x+2=0}$

=> $\sf{x = -2}$

Or,

$\sf{x+1=0}$

=> $\sf{x=-1}$

$\sf{\therefore}$ The zeroes of $\sf{p(x)=x^2+3x+2}$ are -2 and -1

Hence,

Product of zeroes = $\sf{-2\times -1}$ = 2

$\bf{\underline{Verification:-}}$

$\sf{\underline{Sum\:of\:zeroes} = \dfrac{-Cofficient\:of\:x}{Coefficient\:of\:x^2}}$

$\sf{(-2) + (-1) = \dfrac{-3}{1}}$

= $\sf{-2-1 = -3}$

= $\sf{-3=-3}$ [Verified]

$\sf{\underline{Product\:of\:zeroes} = \dfrac{Constant\:term}{Coefficient\:of\:x^2}}$

$\sf{-2\times-1 = \dfrac{2}{1}}$

= $\sf{2 = 2}$ [Verified]

Answer 2

x^2 + 3x + 2

= x^2 + x + 2x + 2

= x(x+1)+2(x+1)

= (x+1)(x+2)

x = -1

x = -2

Product = -2×-1 = 2 (Answer)

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