Someone Solve 15 Number Question....
Dear student,
Given,
a+b+c =0 .... (i)
We need to find a zero of the polynomial ax2 + bx + c =0
We can rewrite equation (i) as : a = -(b+c)
Now, we will substitute this value of 'a' into the polynomial,
-(b+c)x2 + bx + c =0
Multiplying with (-1) on both sides, we get
(b+c)x2 - bx - c = 0
bx2+ cx2 - bx -c = 0
bx(x-1) + c(x2 -1) =0
bx(x-1) + c (x+1)(x-1)=0
(x-1) (bx + cx + c) =0
We know (b+c) = -a, hence,
(x-1)[(b+c)x + c] =0
(x-1)(c - ax) =0
x=1, c/a
Putting both of the values, the polynomial ax2 + bx + c=0 gets satisfied. Hence, you have two zeroes --> x=1, c/a.
I hope it is clear now.
Given,
a+b+c =0 .... (i)
We need to find a zero of the polynomial ax2 + bx + c =0
We can rewrite equation (i) as : a = -(b+c)
Now, we will substitute this value of 'a' into the polynomial,
-(b+c)x2 + bx + c =0
Multiplying with (-1) on both sides, we get
(b+c)x2 - bx - c = 0
bx2+ cx2 - bx -c = 0
bx(x-1) + c(x2 -1) =0
bx(x-1) + c (x+1)(x-1)=0
(x-1) (bx + cx + c) =0
We know (b+c) = -a, hence,
(x-1)[(b+c)x + c] =0
(x-1)(c - ax) =0
x=1, c/a
Putting both of the values, the polynomial ax2 + bx + c=0 gets satisfied. Hence, you have two zeroes --> x=1, c/a.
I hope it is clear now.