Find the value of A and B if (X + 1) and (x -1) are the factors of the polynomial x^4+ax^3-3x^2+2x+b by using factor theorem
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Given Equation is f(x) = x^4 + ax^3 - 3x^2 + 2x + b.
By remainder theorem, we get
x + 1 = 0
x = -1.
Now substitute x = -1 in f(x), we get
f(-1) = (-1)^4 + a(-1)^3 - 3(-1)^2 + 2(-1) + b = 0
= 1 - a - 3 - 2 + b = 0
= b - a - 4 = 0
b - a = 4. ------- (1)
By remainder theorem, we get
x - 1 = 0
x = 1.
Substitute x = 1 in f(x), we get
f(1) = x^4 + ax^3 - 3x^2 + 2x + b
= (1)^4 + a(1)^3 - 3(1)^2 + 2(1) + b = 0
= 1 + a - 3 + 2 + b = 0
= a + b = 0 -------------- (2)
On solving (1) & (2), we get
b - a = 4
b + a = 0
------------------
2b = 4
b = 2
Substiitute b = 2 in (1), we get
b - a = 4
2 - a = 4
-a = 2
a = -2.
Hope this helps!
By remainder theorem, we get
x + 1 = 0
x = -1.
Now substitute x = -1 in f(x), we get
f(-1) = (-1)^4 + a(-1)^3 - 3(-1)^2 + 2(-1) + b = 0
= 1 - a - 3 - 2 + b = 0
= b - a - 4 = 0
b - a = 4. ------- (1)
By remainder theorem, we get
x - 1 = 0
x = 1.
Substitute x = 1 in f(x), we get
f(1) = x^4 + ax^3 - 3x^2 + 2x + b
= (1)^4 + a(1)^3 - 3(1)^2 + 2(1) + b = 0
= 1 + a - 3 + 2 + b = 0
= a + b = 0 -------------- (2)
On solving (1) & (2), we get
b - a = 4
b + a = 0
------------------
2b = 4
b = 2
Substiitute b = 2 in (1), we get
b - a = 4
2 - a = 4
-a = 2
a = -2.
Hope this helps!
siddhartharao77:
:-)
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