Find the zeroes of the polynomial x2 +-x-2, and
verify the relation between the coefficients and the
zeroes of the polynomial.
Answers
Explanation:
Let p(x) be the given polynomial
Let @ and be the zeros
Now,
p(x)=x² -x -2
=x² -2x+x-2
=x(x-2)+1(x-2)
=(x-2)(x+1)
We know that,
p(x)=0
=>(x-2)(x+1)=0
=>x=2 or -1
=>@=2 and ß= -1
Sum of zeros:
@ + ß
=2-1
=1
Product of zeros:
@ß
=(2)(-1)
=-2
x^2-x-2=0.
The first term is, x2 its coefficient is 1 .
The middle term is, -x its coefficient is -1 .
The last term, "the constant", is -2 .
Multiply the coefficient of the first term by the constant .
So,1×-2=-2.
Find the factors of -2 whose sum = -1 .
The product of (-2×1) is -2 And their product is -1.
Rewrite the polynomial splitting the middle term using the two factors found above.
x^2-2x+x-2=0.
=》x(x-2)1(x-2).
=》(x-2)(x+1)=0.
=》x-2= 0 (or) x+1=0.
=》x=2 (or) x=-1.
therefore 2 and -1 are the zeroes of the given equation.
Here α=2 β=-1. a=1,b=-1,c=-2.
Relationship between zeroes and the coefficients of the polynomial.
α+β = -b/a.
αβ =c/a.
Sum of zeroes and its relationship with coefficients.
α+β = -b/a.
=》2+(-1)=-(-1)/1
=》2-1=1/1.
=》1=1
Product of zeroes and its relationship with coefficients.
αβ =c/a.
(2)(-1)=-2/1
-2=2.