Question

the sum and product of the zeros of the polynomial x^2-6x+8 are respectively​

Answer 1

hope this helps u

x²-6x+8

x²-2x-4x+8

[x²-2x][-4x+8]

x[x-2] -4[x-2]

[x-2][x-4]

x-2=0                x-4=0

x=2                         x=4

x=2,4 are the zeros of the polynomial

sum of the zeros = 2+4=6

product of the zeros = 2*4=8

Answer 2

Given :

• Polynomial x²-6x+8.

To Find :

• Sum and Product of the Zeroes.

Solution :

$\longmapsto\tt\bold{{x}^{2}-6x+8}$

By Splitting Middle Term :

$\longmapsto\tt{{x}^{2}-(4x+2x)+8}$

$\longmapsto\tt{{x}^{2}-4x-2x+8}$

$\longmapsto\tt{x(x-4)-2(x-4)}$

$\longmapsto\tt{(x-2)(x-4)}$

• x = 2
• x = 4

So , 2 and 4 are the zeroes of polynomial x²-6x+8.

_______________________

Here :

• a = 1
• b = -6
• c = 8

Sum of Zeroes :

$\longmapsto\tt{\alpha+\beta=\dfrac{-b}{a}}$

$\longmapsto\tt{2+4=\dfrac{-(-6)}{1}}$

$\longmapsto\tt{6=6}$

$\longmapsto\tt\bold{L.H.S=R.H.S}$

Product of Zeroes :

$\longmapsto\tt{\alpha\beta=\dfrac{c}{a}}$

$\longmapsto\tt{2\times{4}=\dfrac{8}{1}}$

$\longmapsto\tt{8=8}$

$\longmapsto\tt\bold{L.H.S=R.H.S}$

HENCE VERIFIED