Question

If x⁴- ax³+ bx -1 is entirely divisible by x²-1 then find the values of a and b. Whoever does it correct first will be brainliest...!!!!​

Answer 1

Answer:

Step-by-step explanation:

x²-1=0

x²=1

$x=\sqrt{1} \\x=+/- 1\\x=+1 x=-1$

As +1 and -1 are the zeroes of the polynomial $p(x)=x^{4} - ax^{3} +bx-1$

p(x)=0 where x= -1 or +1

$p(x)=1^{4} - a 1^{3} +b1-1=0$      ( here x= +1)

$p(x)=1 - a 1 +b1-1=0$

$p(x)= - a 1 +b1=0$

$p(x)= +b- a=0$     equation i

Solving with x= -1

$p(x)= -1^{4} -( a *-1^{3}) +(b*-1)-1=0$

$p(x)= 1 + a -b-1=0$

$p(x)= a -b=0$        (equation ii)

Adding i and ii

(a-b)+(a+b)=0+0

a-b+a+b=0

2a=0

a=0/2=0

a=0

substituting a in equation i

b-a=0

b-0=0

b=0

Therefore a=0 and b=0