If the polynomial p(x) =x^4+ax^3+2x^2-3x+b is exactly divisible by x^2-1 find the value of a and b
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p(x) =x^4+ax^3+2x^2-3x+b
x^2-1=(x-1)(x+1)
x-1=0
x=1
p (1) =0
1^4+a×1^3+2×1^2-3×1+b=0
1+a+4-3+b=0
2+a+b=0
a+b=-2....... (1)
x+1=0
x=-1
p (-1)=0
-1^4+a×-1^3+2×-1^2-3×-1+b=0
1-a+2+3+b=0
4=a-b
a-b=4....... (2)
Adding (1) and (2)
a+b+a-b=-2+4
2a=2
a=1
a+b=-2
1+b=-2
b=-3
Hope it helps!
Thankyou ☆ ☆
p(x) =x^4+ax^3+2x^2-3x+b
x^2-1=(x-1)(x+1)
x-1=0
x=1
p (1) =0
1^4+a×1^3+2×1^2-3×1+b=0
1+a+4-3+b=0
2+a+b=0
a+b=-2....... (1)
x+1=0
x=-1
p (-1)=0
-1^4+a×-1^3+2×-1^2-3×-1+b=0
1-a+2+3+b=0
4=a-b
a-b=4....... (2)
Adding (1) and (2)
a+b+a-b=-2+4
2a=2
a=1
a+b=-2
1+b=-2
b=-3
Hope it helps!
Thankyou ☆ ☆
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