Question

7x^2-18x-8 find the zero of the polynomial​

Answer 1

it might came in fraction :

zero of polinomial=

$( - b + - \sqrt{(b {}^{2}) - 4ac }) \div 2a$

where a=7,b=-18 and c=-8

using this formula the root comes in fraction..actually roots are called zero of polinomial..

or the question might be this: 7x^2-18x+8

then,7x^2-(14+4)x+8

= 7x^2 -14x -4x +8

= 7x(x-2) -4(x-2)

=( 7x-4)(x-2)

then zeros are = 4/7 or 2

hope you understand and if not ask me again..