Question

find zeros of a polynomial x square - 8 x + 12 and verify the relationship between zeroes and coefficient of variables in the polynomials

Answer 1

${x}^{2} + 8x + 12$
${x}^{2} + 6x + 2x + 12$
$x(x + 6) + 2(x + 6)$
$(x + 6)(x + 2)$
Therefore zeros are -6 and -2

Putting x=-6 IN p(x)

${ - 6}^{2} + 8( - 6) + 12$
= 36 +(-48) + 12

=36-48+12
=48-48
=0

Similarly, you have to do with -2 as well...

Hope it helps

Answer 2

Given equation :- x² - 8x + 12

x² - 8x + 12

x² - 6x - 2x + 12

x ( x - 6 ) - 2 ( x - 6 )

( x - 2 ) ( x - 6 )


* ( x - 2 ) = 0
x = 2

* ( x - 6 ) = 0
x = 6


Verification :-

• Sum of zeros

$= \frac{ - coefficient \: of \: {x}}{coefficient \: of {x}^{2} }$

LHS :-

Sum of zeros :- 2 + 6 = 8

RHS :-

$\frac{ - b}{a} = \frac{8}{1} = 8$

LHS = RHS

• Product of zeros

$= \frac{constant \: term}{coefficient \: of \: {x}^{2} }$


LHS :-

Product of zeros :- 2 × 6 = 12


RHS :-

$= \frac{c}{a} = \frac{12}{1}$
= 12

LHS = RHS

hence verified !!