Question

Verify the relationship between the zeroes

and coefficients of the following polynomials.

(a) x

2

+11x +18 (b) 2x2

+x-6​

Answer 1

Answer:

...........refer the attachment

Answer 2

$a) Let \: p(x) = x^{2} + 11x + 18$

We get zeroes of the polynomial p(x) we will take p(x) = 0

$i) x^{2} + 11x + 18 = 0$

/* Splitting the middle term,we get */

$\implies x^{2} + 2x + 9x + 18 = 0$

$\implies x( x + 2) + 9( x + 2 ) = 0$

$\implies (x+2)(x+9) = 0$

$\implies x+2 = 0 \: x+9= 0$

$\implies x = -2\: x = -9$

$\therefore The \: zeroes \:of \: polynomial \:-2,\:-9$

$i)Sum \:of \: zeroes$

$= -2 + (-9)$

$= -2 - 9$

$= -11$

$= \frac{-11}{1}$

$= \frac{- ( Coefficient \: of \:x )}{Coefficient\:of \:x^{2}}$

$ii) Product\:of \: zeroes$

$= (-2) \times (-9)$

$= 18$

$= \frac{18}{1}$

$= \frac{Constant \:term}{Coefficient\:of \:x^{2}}$

$i) Let \: p(x) = 2x^{2} + x - 6$

We get zeroes of the polynomial p(x) we will take p(x) = 0

$b) 2x^{2} + x -6= 0$

/* Splitting the middle term,we get */

$\implies 2x^{2} + 4x - 3x-6 = 0$

$\implies 2x( x + 2) -3( x + 2 ) = 0$

$\implies (x+2)(2x-3) = 0$

$\implies x+2 = 0 \: 2x-3= 0$

$\implies x = -2\: x = \frac{3}{2}$

$\therefore The \: zeroes \:of \: polynomial \:-2,\:\frac{3}{2}$

$i)Sum \:of \: zeroes$

$= -2 + \frac{3}{2}$

$= \frac{-4 +3}{2}$

$=\frac{ -1}{2}$

$= \frac{- ( Coefficient \: of \:x )}{Coefficient\:of \:x^{2}}$

$ii) Product \:of \: zeroes$

$= (-2 )\times \Big( \frac{3}{2}\Big)$

$= -3$

$=\frac{ -3}{1}$

$= \frac{- ( Coefficient \: of \:x )}{Coefficient\:of \:x^{2}}$

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