if a + C +e = 0 and b + d = 0 then find the zeros of the polynomial ax4+bx3+cx2+dx+e.
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Answered by
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Given, a + c + e = 0 and b + d = 0
⇒c = – (a + e) and d = – b
Now, ax4 + bx3 + cx2 + dx + e
= ax4 + bx3 + [– (a + e)] x2 + (– b)x + e
= ax4 – ax2 – ex2 + e + bx3 – bx
= ax2 [x2 – 1] – e [x2 – 1] + bx [x2 – 1]
= [x2 – 1] [ax2 – e + bx]
= (x + 1) (x – 1) [ax2 – 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).
⇒c = – (a + e) and d = – b
Now, ax4 + bx3 + cx2 + dx + e
= ax4 + bx3 + [– (a + e)] x2 + (– b)x + e
= ax4 – ax2 – ex2 + e + bx3 – bx
= ax2 [x2 – 1] – e [x2 – 1] + bx [x2 – 1]
= [x2 – 1] [ax2 – e + bx]
= (x + 1) (x – 1) [ax2 – 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).
Answered by
92
Answer:
Given, a + c + e = 0 and b + d = 0
⇒c = – (a + e) and d = – b
Now, ax4 + bx3 + cx2 + dx + e
= ax4 + bx3 + [– (a + e)] x2 + (– b)x + e
= ax4 – ax2 – ex2 + e + bx3 – bx
= ax2 [x2 – 1] – e [x2 – 1] + bx [x2 – 1]
= [x2 – 1] [ax2 – e + bx]
= (x + 1) (x – 1) [ax2 – 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).
I Hope It Will Help!
^_^
Given, a + c + e = 0 and b + d = 0
⇒c = – (a + e) and d = – b
Now, ax4 + bx3 + cx2 + dx + e
= ax4 + bx3 + [– (a + e)] x2 + (– b)x + e
= ax4 – ax2 – ex2 + e + bx3 – bx
= ax2 [x2 – 1] – e [x2 – 1] + bx [x2 – 1]
= [x2 – 1] [ax2 – e + bx]
= (x + 1) (x – 1) [ax2 – 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).
I Hope It Will Help!
^_^
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