Question

If a+c+e=0 and b+d=0 then find the zeros of the polynomial ax2+bx3+cx2+dx+e

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Answer 1

Answer:

Given, a+c+e=0 and b+d=0

⇒c=–(a+e) and d=–b

Now, ax

4

+bx

3

+cx

2

+dx+e

=ax

4

+bx

3

+[–(a+e)]x

2

+(–b)x+e

=ax

4

–ax

2

–ex

2

+e+bx

3

–bx

=ax

2

(x

2

–1)–e(x

2

–1)+bx(x

2

−1)

=(x

2

−1)(ax

2

–e+bx)

=(x+1)(x–1)(ax

2

–e+bx) ..........(1)

As (x+1) and (x–1), are the factors of (1)

so, it is divisible by both (x+1) and (x–1)

Hence the zeroes are −1,1 and x=

2a

−b±

b

2

+4ae