Question

if -4 is a zero of the polynomial $x^{3}$_$x^{2}$_14x+24, find other zeros

Answer 1

x³-x²-14x+24, zero = -4
x= -4
x+4=0
Dividing equation by x+4 

x³-x²-14x+24/x+4
After diving we will get 

x²-5x+6 = 0
Taking out its factors
x²-2x-3x+6 = 0
x(x-2)-3(x-2) = 0

(x-2) (x-3) = 0
 
x-2=0,        x-3=0
x=2,           x=3
 therefore, Zeros are -4,2,3 

I hope this will help you! 

Answer 2

if -4 is a zero of the polynomial P(x) then (x+4) is a factor of P(x). So

(x+4) (x² + a x + 6) = P(x) = x³ - x² - 14 x + 24

co-efficient of x² is 1 as coefficient of x³ = 1 in P(x),

constant term is 24 in P(x), so 24/4 = 6 is the constant term in the second factor.

So co-efficient of x² = 4 + a = -1          So a = -5

co-efficient of x  = 4a + 6 = -20 + 6 = -14. So it verifies.