Question

find the zeroes of quadratic polynomial x2+13x+36 and variefy the relationship between zeroes and the coefficient of the polynomial

Answer 1

let x^2 + 3x + 36 be p(x)
to find other zeros put p(x) = 0
x^2 + 13x + 36 = 0
x^2 + [ 9 + 4 ]x + 36 = 0
x^2 + 9x + 4x + 36 = 0
x [1+9] + 4x [ 1 + 9 ]
( x + 4) ( 1+9)
x = -4. x = -9
relationship
sum of zeroes = - 4 + - 9 = -13 [ cofficent of x / cofficent of x^2]

product of zeroes = -4 x -9
= 36 [ constant term / cofficent of x^2 ]
hope this helps☺☺

Answer 2

f(x)=x2+13x+36
=x2+4x+9x+36
=x(x+4)+9(x+4)
=(x+4)(x+9)
f(x)=0
(x+4)(x+9)=0
x= -4 or x= -9
Sum of zeros = -4-9= -13
Product of zeros=(-4)(-9)=36