If (x^2-1) is a factor of x^3-ax^2+bx+2 then findthevaluesof a and b.What i the remainingfactorofthepolynomial

Question

please say the answer with procedure.

Answer 1

x^3 -ax²+bx+2
x²-1 is a factor
x²-1 = (x-1)(x+1)

I will use synthetic division .
for x = 1 and x=-1

x=1
1 -a +b + 2
0 1 (-a+1) +(b+(-a+1) )
1-a+1 +(b+(-a+1) ) 0
2 + b + (-a+1) = 0
x=-1
1 -a +b +2
0 -1+(a+1) - (b+a+1)
1-(a+1)+(b+a+1) 0

2-(b+a+1)=0

You have two equations :-
2 + b + (-a+1) = 0
3+b = a==>(1)
2-(b+a+1)=0
1-b-a=0
Substitute (1)
1-b-(3+b) =0
1-b-3-b=0
-2-2b=0
2=-2b
b=-1

3+b = a
a=3+(-1)
a=2

a=2 and b=-1

Answer 2

The third factor is of degree 3-2=1. (That's the step that lets you avoid synthetic long division).

Call that factor cx+d.

The coefficient of x^3 in (x^2-1)(cx+d) is c. We know it must = 1.

The constant term is -d. We know it must equal 2.

So the third factor is (x-2), and the polynomial is (x^2-1)(x-2). Multiply that out and you get x^3 - 2x^2 - x + 2. So a = 2 and b = -1.

Answer 3

Blah!!!

All we know from what is given is that (x+1)(x-1) are factors and we are missing a third factor. The nice thing is that we are given the last number (notice the +2).

To get +2 from a -1, we need to multiply by -2. This means our missing factor is (x-2).

Answer 4

(x^2-1)=(x-1)(x+1)therefore the 2 factors are-1,1
use remainder theorem and you get
a+b=-1
a-b=3
a=1,b=-2