If x² -1 is a factor of ax⁴+bx³+cx²+dx+e, show that a+c+e=b+d=0
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Answers
Step-by-step explanation :
Given,
- x² -1 is a factor of ax⁴+bx³+cx²+dx+e
To prove,
- a + c + e = b + d = 0
Solution,
Let p(x) = ax⁴+bx³+cx²+dx+e
=> x² - 1 is a factor
x² = 1
x = √1
x = ±1
- When we substitute x = +1, the result is zero.
Put x = +1,
p(x) = ax⁴+bx³+cx²+dx+e
p(1) = a(1)⁴ + b(1)³ + c(1)² + d(1) + e = 0
a(1) + b(1) + c + d + e = 0
a + b + c + d + e = 0 ----[ 1 ]
- When we substitute x = -1, the result is zero.
Put x = -1,
p(x) = ax⁴+bx³+cx²+dx+e
p(-1) = a(-1)⁴ + b(-1)³ + c(-1)² + d(-1) + e = 0
a(1) + b(-1) + c(1) - d + e = 0
a - b + c - d + e = 0
a + c + e = b + d ----[ 2 ]
Substitute [2] in [1],
a + b + c + d + e = 0
a + c + e + b + d = 0
b + d + b + d = 0
2b + 2d = 0
2( b + d ) = 0
b + d = 0/2
b + d = 0
=> a + c + e = b + d
= 0
∴ a + c + e = b + d = 0
Step-by-step explanation:
given x² -1 is a factor
x² -1 =0
x² =1
x=+or- 1 are factors ax⁴+bx³+cx²+dx+e
let f(x)=ax⁴+bx³+cx²+dx+e
f(-1)=0
a(-1)⁴+b(-1)³+c(-1)²+d(-1)+e=0
a-b+c-d+e=0 .... (1)
a+c+e=b+d
f(1)=0
a(1)⁴+b(1)³+c(1)²+d(1)+e=0
a+b+c+d+e=0
(a+c+e)+(b+d)=0
(b+d)+(b+d)=0 from (1)
2(b+d)=0
b+d=0 (2)
from (1) & (2)
a+c+e=0
therefore,a+c+e=b+d=0