Question

If (xsquare -1)is a factor of ax4+bxcube+cxsquar+dx+e, show that a+c+c=b+d=0.​

Answer 1

Step-by-step explanation:

x^2-1 is a factor means (x-1)(x+1) also Factors

so that means 1 and -1 are the zeros of the given polynomial

so put 1 and -1 in the place of x

so

a-b+c-d+e=0---eq1

a+c+e=b+d

if you put 1

then a+b+c+d+e=0----eq2

so from eq1 and eq2

you will get your answer

Answer 2

Step-by-step explanation:

Given :-

x²-1 is a factor of ax⁴+bx³+cx²+dx+e

To find :-

Show that a+c+e = b+d = 0

Solution :-

Given bi-quadratic polynomial is

P(x) = ax⁴+bx³+cx²+dx+e

Given factor = x²-1

=>(x+1) (x-1)

We know that

If (x-a) is a factor of P(x) then P(a) = 0

So If x²-1 is a factor of P(x) then P(1) = 0 and

P(-1) = 0

Now,

P(1) = 0

=> 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 -------(1)

and

P(-1) = 0

=> 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 = 0

=> a+c+e-(b+d) = 0

=> a+c+e = b+d ----------(2)

On Substituting (2) in (1) then

=> b+d+b+d = 0

=> 2b+2d = 0

=> 2(b+d) = 0

=> b+d = 0/2

=> b+d = 0 -----------(3)

From (3) we have

a+c+e = 0 ------------(4)

From above equations we get

a+c+e = b+d = 0

Hence, Proved.

Answer:-

If x²-1 is a factor of ax⁴+bx³+cx²+dx+e then a+c+e = b+d = 0

Used formulae:-

Factor Theorem:-

→ Let P(x) be a polynomial of the degree greater than or equal to 1 and x-a is another linear polynomial if x-a is a factor of P (x) then P(a) = 0 vice-versa.