Find the zeros of cubic polynomial 2x^3-5x ^2-14x+8 and verify the relationship between the zeroes and the coefficients
Answers
Step-by-step explanation:
2x^3-5x^2-14x+8=0
compare it with ax^3+bx^2+cx+d=0
a=2
b=-5
c-14
d=8
a×d=2×8=16
factors of 16=1,2,4,8,16,-1,-2,-4,-8,-16
use hit and trial method;
put x= 1, then ;
2 (1)^3-5 (1)^2-14 (1)+8
2-5-14+8
10-19 not equal to 0
then value of x=1 is not possible.
similarly x=-2,
2 (-2)^3-5 (-2)^2-14 (-2)+8
-16-20+28+8
-36+36
=0
hence the value of x=-2 is possible.
x=-2
x+2=0.
divide the given equation with x+2=0
(2x^3-5x^2-14x+8)/(x+2)
=2x^2-x-12
hence , 2x^3-5x^2-14x+8=(x+2)(2x^2-x-12)
=(x+2)(2x^2-x-12)
is it helpful??
2x^3-5x^2-14x+8=0
compare it with ax^3+bx^2+cx+d=0
a=2
b=-5
c-14
d=8
a×d=2×8=16
factors of 16=1,2,4,8,16,-1,-2,-4,-8,-16
use hit and trial method;
put x= 1, then ;
2 (1)^3-5 (1)^2-14 (1)+8
2-5-14+8
10-19 not equal to 0
then value of x=1 is not possible.
similarly x=-2,
2 (-2)^3-5 (-2)^2-14 (-2)+8
-16-20+28+8
-36+36
=0
hence the value of x=-2 is possible.
x=-2
x+2=0.
divide the given equation with x+2=0
(2x^3-5x^2-14x+8)/(x+2)
=2x^2-x-12
hence , 2x^3-5x^2-14x+8=(x+2)(2x^2-x-12)
=(x+2)(2x^2-x-12)